matematika ii (mathematics ii) - Stavebná fakulta TU v Košiciach

Transcription

Moderné vzdelávanie pre vedomostnú spoločnosť
Projekt je spolufinancovaný zo zdrojov EÚ
MATEMATIKA II
(MATHEMATICS II)
Stavebná fakulta
RNDr. Pavol Purcz, PhD.
Doc.Ing. Roman Vodička, PhD.
NÁZOV:
AUTORI:
VYDAVATEĽ:
ROK:
NÁKLAD:
ROZSAH:
ISBN:
Matematika II (Mathematics II)
Purcz Pavol, Vodička Roman
Technická univerzita v Košiciach
2015
50 ks
171 strán
978-80-553-2046-5
Rukopis neprešiel jazykovou úpravou.
Za odbornú a obsahovú stránku zodpovedajú autori.
MATEMATIKA II
(MATHEMATICS II)
Stavebná fakulta
Introduction
Aim
To provide theoretical knowledge required for studying specialized subjects and to apply obtained knowledge to the solution of technically oriented
problems basically using integral calculus and differential equations.
Contents and objectives
The aim of the module is to familiarize students with the necessary mathematical theory and its applications. At the beginning of each chapter are
definitions of some terms and their properties required to solve the problems. As the publication is designed as a collection of problems, it neither contains
definitions of all notions and terms nor proofs of mathematical theorems. The publication also includes solved examples and problems for solving and
self-assessment according to the following content.
1. Function of several variables
2. Indefinite integral
3. Definite integral
4. Differential equations
5. Double integral
Prerequisities
Matematika I (Mathematics I. (P. Purcz, R. Vodička, TU Košice, 2012, ISBN 978-80-553-1279-8)
Introduction – 2
MATEMATIKA II
(MATHEMATICS II)
Stavebná fakulta
Táto publikácia vznikla za finančnej podpory z Európskeho sociálneho fondu v rámci Operačného programu VZDELÁVANIE.
Prioritná os 1 Reforma vzdelávania a odbornej prípravy.
Opatrenie 1.2 Vysoké školy a výskum a vývoj ako motory rozvoja vedomostnej spoločnosti.
Názov projektu: Balík inovatívnych prvkov pre reformu vzdelávania na TUKE
NÁZOV:
AUTORI:
VYDAVATEĽ:
ROK:
NÁKLAD:
ROZSAH:
ISBN:
Matematika II (Mathematics II)
Purcz Pavol, Vodička Roman
Technická univerzita v Košiciach
2015
50 ks
171 strán
978-80-553-2046-5
Rukopis neprešiel jazykovou úpravou.
Za odbornú a obsahovú stránku zodpovedajú autori.
MATEMATIKA II
(MATHEMATICS II)
Stavebná fakulta
Introduction
Aim
To provide theoretical knowledge required for studying specialized subjects and to apply obtained knowledge to the solution of technically oriented
problems basically using integral calculus and differential equations.
Contents and objectives
The aim of the module is to familiarize students with the necessary mathematical theory and its applications. At the beginning of each chapter are
definitions of some terms and their properties required to solve the problems. As the publication is designed as a collection of problems, it neither contains
definitions of all notions and terms nor proofs of mathematical theorems. The publication also includes solved examples and problems for solving and
self-assessment according to the following content.
1. Function of several variables
2. Indefinite integral
3. Definite integral
4. Differential equations
5. Double integral
Prerequisities
Matematika I (Mathematics I. (P. Purcz, R. Vodička, TU Košice, 2012, ISBN 978-80-553-1279-8)
Introduction – 2
Chapter 1. (Function of several variables)
Mission
Show to students the real function of two or several real variables, generally, its properties, so-called partial derivative as a way to calculate derivatives
in this case using calculus of one real functions of one real variable. Giving to students the basics of vector analysis, i.e. calculus of vector function,
both, its properties and derivatives.
Objectives
1. Learning to determine the domain of function, mainly of two (or three variables respectively) focused on its geometric interpretation in the plane (or
space respectively).
2. Learning to calculate partial derivatives of functions of several variables, as well as the composed functions and higher order partial derivatives.
3. Being able to apply the knowledge obtained from the previous point to calculate the equation of the tangent plane and its normal line to the graph
of the function of two variables as well as the local, and global extremes as well as extremes with respects to given sets.
4. Discover the concept of implicit function in the case of functions of one, or two variables, respectively, knowing calculate its derivative or partial
derivatives, respectively and apply them to determine the equations of tangents and normals or tangent plane and normal line to the graph of a
function, respectively.
5. Understand concepts such as vector function, directional derivatives, gradients, divergence and rotation and apply them to know the role of vector
analysis.
Prerequisite knowledge
differential calculus of the one function of the one real variables, analytical geometry, linear algebra
Chapter 1. (Function of several variables) – 1
Introduction
The subject of our further study will be a set of n- real numbers and real function, which can be defined on this set. Terms such as domain, boundedness,
operations between functions, continuity or limits are equivalent concepts for real functions of one real variable and their definitions are very similar.
Graph of the function of n variables is generally very difficult to see, because our imaging resources are quite limited. And therefore normally only chart
shows the graph of the function of one or two variables. It is still possible to easily view domain of the function of three variables simultaneously and it
is already everything that concerns our ability of the geometric visualization, since our existence in this world is given by the three-dimensional space.
Therefore, the practical problems encountered in this chapter include (with few exceptions), just a function of two variables. The term derivative of the
real function of several variables is limited to the term so called partial derivative, and its calculation proceeds the same way as in the case of functions
of one variable.
Function of several variables – Domain of the function of several variables.
In determining the domain proceed similarly as in the function of one variable. Since the result is a set of points in the 2- or 3-dimensional space,respectively,
it is suggested to reinforce the concepts of geometric sets and draw graphically formed situation.
A set of all n real numbers is called as n-dimensional Euclidean space En , if for each couple of the sets of n-real numbers A[a1 , a2 , ..., an ] and
B[b1 , b2 , ..., bn ] is given following value:
v
u n
uX
%(A, B) = t
(ai − bi )2 ,
i=1
which is called the difference (between the sets of n-real numbers).
Let’s M ⊂ En . Then each mapping of M to one-dimensional Euclidean space is called a real function y of n variables. We denote its as y =
f (x1 , x2 , ..., xn ), or y = f (X). The set M is a domain of this function.
Excercise 1. Let’s find domain of the function
1
f (x, y, z) = p
.
16 − x2 − y 2 − z 2
Solution.
The domain is possible to determine from inequality 16 − x2 − y 2 − z 2 > 0, or x2 + y 2 + z 2 < 16. the domain is created by the set of all points of X,
which satisfy following condition %(O, X) < 4, what represents the interior of sphere with midpoint O[0, 0, 0] and radius equals 4 geometrically. Chapter 1. (Function of several variables) – 2
Problems.
In problems 1 - 28 determine domain of the function several variables.
1. z = ln(−x − y).
3. z = √5xy .
2
2
2
−y −z )
√
.
5. u = ln(9−x
x2 +y 2 +z 2 −4
p
√
7. z = 1 − x2 + 1 − y 2 .
9. z = arcsin(x + y).
p
√
11. z = x − y.
q
x2 +y 2 −x
13. z = 2x−x
2 −y 2 .
1
.
y 2 −x2
p
4. z = (4 − x2 − y 2 )(x2 + y 2 − 1).
2. z =
6. z = arcsin x−1
.
y
8. z =
10. z =
12. z =
1
.
25−x
p 2 −y2
2
y−x +
√
1 − y.
ln(4−x2 −y 2 )
√
.
xy
14. z = x2 + 2xy − 3y 3 .
2
+ 1 .
√ y−1 6
17. z = 3x − √y .
x y
16. z = 2x+|y|
.
1
18. z = √
−
19. z = y + arccos x.
2
21. z = √ln x y .
|y−x|
√
4x−y 2
23. z = ln(1−x2 −y2 ) .
√
25. z = ln(x − y).
27. u = arcsin x + arcsin y + arcsin z.
1
20. z = ln(|x| + y) + √y−x
.
p
22. z = πy 2 x2 − y 2 + ln xy.
15. z =
1
x
x+|y|
1
.
x−|y|
24. z = ln(y 2 − 4x + 8).
1
1
26. z = √x+y
+ √x−y
.
28. z = arcsin[2y(1 + x2 ) − 1].
Function of several variables – Partial derivatives of the 1-st order of the function of several variables.
The term derivative of the real function of several variables is limited to a term so called partial derivatives, in computing which proceed in the same way
as in the case of functions of one variable. If we calculate derivatives with respect of one variable follow to fact that all other variables are considered
constant.
Let’s given function f (x1 , x2 , ..., xn ) defined in neighbourhood of the point A[a1 , a2 , ..., an ]. If there exist a finite limit
∂f
f (a1 , ..., ai−1 , xi , ai+1 , ..., an ) − f (a1 , ..., ai , ..., an )
= lim
x
→a
∂xi A
x i − ai
i
i
∂f
then its value is called the partial derivative of f (x1 , x2 , ..., xn ) with respect to xi at point A It is denoted by: ∂x
, ∂f∂x(A)
, fx0 i (A).
i
i
A
Excercise 2. Let’s find partial derivatives of the 1-st order of the function z = x3 + 7x2 y 5 + y 2 − 2.
Solution.
First, let’s calculate zx0 . the variable y is considered constant now. It means, we calculate derivatives of z only with respect one variable x. Thus
zx0 = 3x2 + 7.2xy 5 = 3x2 + 14xy 5 .
Similarly we calculate zy0 . The variable x we considered constant. We calculate again derivatives of z only with respect one variable y. Thus
zy0 = 7x2 5y 4 + 2y = 35x2 y 4 + 2y. y
Excercise 3. Let’s find partial derivatives of the 1-st order of the function z = e− x at point A[1, 0].
Solution.
The partial derivatives of the 1-st order are calculated as:
y
zx0 = e− x xy2 ,
y
y
zy0 = e− x − x1 = − x1 e− x .
After substitution of the coordinates of the point A we have zx0 (A) = 0, zy0 (A) = −1. Problems.
In problems 29 - 61 calculate partial derivatives of the
29. z = x4 + y 4 − 4x2 y 2 .
30.
32. z = √ x2 2 .
33.
x +y
36.
35. z = xy .
38. u = ( xy )z .
39.
4
2
42.
41. z = 3x y − 5xy + 2y.
44. u = x3 y 2 z − xy 2 z 3 .
45.
xy
47. z = y−x
.
48.
50. z = ln(x − yex ).
51.
53. u = z sin(xyz)
p − y cos(yz) + sin(xz).
54. z = ln(x + x2 + y 2 ).
55. z = xyesin(πxy) .
1-st order of the given functions.
3
z = x2 y + xy 4 .
2
z = cosyx .
z = ln(x + y 2 ).
y
u = xz .
2
√
z = x 3 y + √y x .
z = (5x − y)4 .
z = e−xy .
u = x2 ey sin z.
31. z =
x
.
y2
34. z = x sin(x + y).
37. z = arctg xy .
40. z = ln(x + ln y).
43.
46.
49.
52.
z
u = xy .
√
z = xyp2 − xy + 2 x.
z = x x2 + y 2 .
z = (sin x)cos y .
3
3
y
.
57. z = xx2 +y
56. z = xe x .
+y 2
x−y
59. z = ln x+y .
60. u = (y tg z)ln x .
62. Prove, that function z = f (x, y) satisfies the equation:
a) xyzx0 + x2 zy0 = 2yz, ak z = x2 sin(y 2 − x2 );
x
b) xzx0 + yzy0 = lnzy , ak z = e y ln y;
c) xzx0 + lnyx zy0 = 2yz, ak z = xy ;
d) yzx0 − xzy0 = 0, ak z = ln(x2 + y 2 );
x
e) 2xzx0 + yzy0 = 0, ak z = e y2 .
63. Prove, that function u = f (x,
p y, z) satisfies the equation:
a) u0x 2 + u0y 2 + u0z 2 = 1, ak u = x2 + y 2 + z 2 ;
b) u0x + u0y + u0z = 1, ak u = x + x−y
y−z
√
58. z = arctg xy .
11
61. u = (2xy 2 + z 3 ) .
Function of several variables – Total differential of function.
In the case of functions of two variables we can give partial derivatives at a given point some geometrical interpretation. There are tangent line at this
point in the direction of the coordinate axes. Both of these tangent lines then generate the tangent plane to the surface to the graph of the function of
two variables. Using analytical geometry we can easily find the equation of the normal line perpendicular to the plane at this point.
Let’s given the function of two variables z = f (X) = f (x, y), with its continual partial derivatives of the 1-st order at point A[a1 , a2 ] . Then total
differential of the function f (X) at point A is defined as
df (A, X) =
∂f (A)
∂f (A)
(x − a1 ) +
(y − a2 ),
∂x
∂y
or generally
dz =
∂f
∂f
dx +
dy.
∂x
∂y
To calculate the approximate true relationship we use
f (X) ≈ f (A) + df (A, X).
If the graph of the function z = f (x, y) is given and moreover f (x, y) is differentiable function, then the tangent plane to this graph at point M [x0 , y0 , z0 ]
is given by following equation
∂f (M )
∂f (M )
(x − x0 ) +
(y − y0 ) − (z − z0 ) = 0.
∂x
∂y
The system of parametric equations of the normal line to this tangent plane at point M is
(M )
x = x0 + ∂f∂x
t,
∂f (M )
y = y0 + ∂y t,
z = z0 − t.
Excercise 4. Let’s calculate total differential of function z = ln(x2 + y 2 ).
Solution.
First, let’s calculate the partial derivatives of the 1-st order. zx0 =
dz =
Excercise 5. Let’s calculate total differential of function z =
Solution.
First zx0 =
1
y2
x
y2
2x
x2 +y 2
a zy0 =
2x
dx
x2 +y 2
+
2y
.
x2 +y 2
2y
dy.
x2 +y 2
Then
at point A[1, 1].
and zy0 = − 2x
. After substitution zx0 (A) = 1 and zy0 (A) = −2. Then
y2
df (A, X) = (x − 1) − 2(y − 1). Excercise 6. Let’s find approximate value of the expression (using total differential)
p
1, 023 + 1, 973 .
Solution.
p
Let’s consider function z = x3 + y 3 and point A[1, 2]. Then it is possible calculate approximate value at point X[1, 02; 1, 97] using total differential
p
√
2
2
3y
3x
3
3
3
3
1, 02 + 1, 97 ≈ 1 + 2 + √ 3 3
(1, 02 − 1) + √ 3 3
(1, 97 − 2) =
2
x +y
A
= 3 + 0, 01 − 2.0, 03 = 2, 95. 2
x +y
A
Excercise 7. Let’s find the algebraic representation both, of the tangent plane and normal line to graph of the function z = x2 +y 2 at point M [1, −2, 5].
Solution.
First, let’s calculate the partial derivatives of the 1-st order at point M . We have zx0 (M ) = 2, zy0 (M ) = −4. Then the algebraic representation of the
tangent plan we can write as
2(x − 1) − 4(y + 2) − (z − 5) = 0,
or
2x − 4y − z − 5 = 0,
and the parametric representation of the normal line as
x = 1 + 2t,
y = −2 − 4t,
z = 5 − t. Problems.
In problems 64 - 69 find total differential of given functions. p
66. z = ln cotg xy .
64. z = x2 − 2xy + y 2 .
65. z = ln x2 + y 2 .
x−y
.
68. z = arctg 1+xy
69. u = sin(3x − 2y + 5z).
67. z = ex ln y.
In problems 70 - 78 find total differential of given functions at point A.
70. z = x2xy
, A[3, −1].
71. z = arcsin xy , A[−1, 3].
q−y2
72. u = z xy , A[1, 1, 1].
73. z = x arctg(2y), A[−1, 12 ].
74. u = x2 y + 2y 2 z + 3z 2 x, A[−1, 0, 1].
75. z = x3 + 2xy 2 − y 3 , A[2, −1], dx = −0, 1, dy = 0, 2.
2
76. z = ex y , A[1, 1], dx = 0, 15, dy = −0, 05.
77. z = arctg xy , A[1, 2], dx = 0, 05, dy = 0, 01.
78. u = 2x sin y arctg z, A[−2, π2 , 0], dx = 0, 03, dy = −0, 02, dz = 0, 04.
In problems
p 79 - 84 let’s find approximate value of the terms.2,01
81. sin 151◦ cotg 41◦ .
80. 1, 05 .
79. p3, 032 + 4, 012 .
84. sin 29◦ tg 46◦ .
83. 0, 971,05 .
82.
1, 023 + 1, 973 .
85. Find the approximate change both, in the diagonal and the surface area of a rectangle with sides x = 12 m, y = 9 m, if the first side is increased
by 2 cm and second is decreased by 4 cm.
86. There are given height of the cone h = 15 cm and its radius of the base r = 8 cm. Find the approximate change of the volume of this cone, if
the height is decreased by 0, 3 cm a radius of the base is increased by 0, 2 cm.
87. By deformation of the cylinder its radius is increased from 2 dm to 2, 05 dm, and its height is decreased from 10 dm to 9, 8 dm. Find the
approximate change of the volume of this cylinder.
88. Find the approximate change both, of the volume of the block and the area of its surface, if the lengths of its sides changes from 2 cm to 2, 04
cm, from 3 cm to 2, 97 cm and from 4 cm to 4, 02 cm.
89. Find the approximate change of the volume of the cone, of the area of its surface and of the area of its surface casing, if its radius of the base
changes from 30 cm to 30, 1 cm and its height changes from 60 cm to 59, 5 cm.
90. Let’s find the algebraic representation both, of the tangent plane and normal line to graph of the function at given point.
2
2
a) z = x
p + 4y , T [3, −1, ?];
b) z = x2 + y 2 − xy, T [3, 4, ?];
c) z = x4 − 2x2 y + xy + 2y, T [1, ?, 3];
d) x + y 2 + z 2 = 18, T [?, 3, 2];
e) z = exy , T [1, 2, ?];
f) z = ln(xy 2 ) + x2 y, T [ 14 , 2, ?];
g) z = sin xy , T [π, 1, ?];
1
h) z = xy
, T [1, 1, ?];
i) z = ln(x2 + y 2 ), T [1, 0, ?].
Function of several variables – Partial derivative of the composed function.
Derivative of a composite function was in case of a function of one variable, a very important practical proposition. Even in the case of functions of
more variables can it be decomposed into several components of the original function, which can be derivative separately. The structure functions can
be varied. We discuss now some of their variants.
A. If z = f (x, y) and x = ϕ(t) and y = ψ(t), then z = f (ϕ(t), ψ(t)) is a composed function of one variable t. If functions f , ϕ and ψ are differentiable,
then is differentiable the composed function too and we have
dz
∂z dx ∂z dy
=
+
.
dt
∂x dt
∂y dt
(1.1)
dz
∂z ∂z dy
=
+
.
dx
∂x ∂y dx
(1.2)
Especially, if t = x, we get
Excercise 8. Let’s calculate
dz
dt
of the function z = ex−2y , where x = sin t, y = t3 .
Solution.
Using (1.1) we have
dz
dt
dz
dx
3
= ex−2y cos t − 2ex−2y 3t2 = esin t−2t (cos t − 6t2 ) . of the function z = arctg(xy), where y = ex .
Solution.
Using (1.2) we get
dz
dx
=
1
y
1+x2 y 2
+
1
xex
1+x2 y 2
=
ex +xex
.
1+x2 e2x
B. If z = f (x, y) and x = ϕ(u, v) and y = ψ(u, v), then if functions f , ϕ and ψ are differentiable, then is differentiable the composed function too and
we have
∂z
∂z ∂x ∂z ∂y
=
+
,
(1.3)
∂u
∂x ∂u ∂y ∂u
∂z ∂x ∂z ∂y
∂z
=
+
.
∂v
∂x ∂v ∂y ∂v
(1.4)
Excercise 10. Let’s prove, that function z = arctg xy , where x = u + v, y = u − v satisfies the equation
u−v
∂z ∂z
+
= 2
.
∂u ∂v
v + u2
Solution.
Using (1.3) and (1.4) we have
∂z
∂u
=
∂z
∂v
=
1
2
1+ x2
y
1
2
1+ x2
1
2
1+ x2
y
1
2
1+ x2
−x
y2
=
−x+y
x2 +y 2
=
−v
,
u2 +v 2
x
y2
=
x+y
x2 +y 2
=
u
,
u2 +v 2
−v
v 2 +u2
+
u
v 2 +u2
1
y
+
1
y
+
y
y
thus
∂z
∂u
Given functions satisfies the equation.
+
∂z
∂v
=
=
u−v
.
v 2 +u2
Problems.
91. Let’s prove, that each differentiable function F (x, y) satisfies the equation:
2
a) y 2 Fx0 − xyFy0 = yx F , ak F (x, y) = xf (x2 + y 2 )
b) yFx0 + xFy0 = 0, ak F (x, y) = f (x2 − y 2 )
c) y 2 Fx0 + xyFy0 = xF , ak F (x, y) = f (x2y−y2 )
In problems 92 - 97 calculate dz
.
dt
2
2
92. z = x + xy + y , ak x = sin t, y = cos t.
√
93. z = xy, ak x = sin t, y = t2 .
94. z = exy ln(x + y), ak x = t3 , y = 1 − t3 .
y
3
2
95. z = xln
√ , ak x = t , y = t .
96. z = x y, ak x = ln t, y = 1 + et .
97. z = xy , ak x = et , y = 1 − e2t .
∂z
dz
In problems 98 - 103 calculate ∂x
and dx
.
y
2
98. z = arctg x , ak y = x .
99. z = ln(ex + ey ), ak y = x3 .
2
101. z = xey , ak y = ϕ(x).
100. z = arctg x+1
, ak y = e(x+1) .
y
103. z = xy , ak y = ϕ(x).
102. z = ln(x2 + y 2 ), ak y = ϕ(x).
∂z
∂z
In problems 104 - 109 calculate ∂x
and ∂y
.
2
2
104. z = u + v , u = x + sin y, v = ln(x + y).
105. z = u2 v − v 2 u, u = x cos y, v = x sin y.
x
106. z = uv , u = ln(x − y), v = e y .
107. z = u2 ln v, u = xy , v = 3x − 2y.
u
108. z = ue v , u = x2 + y 2 , v = xy.
2
109. z = uv , u = x − 2y, v = y + 2x.
In problems 110 - 112 calculate partial derivatives of the 1-st order of the following functions.
111. z = f (xy + xy ).
112. z = f (x + y, x − y).
110. z = f (x2 − y 2 , exy ).
Function of several variables – Higher order partial derivatives.
As with function of one variable, for its derivation we can look back an consider it as some other function, such as in the case of functions of several
variables. The only one difference is in the fact that we have partial derivatives, which are independent of each other and hence opportunities to develop
other partial derivatives is suddenly much more. The number of combinations increases exponentially. Under certain theoretical assumptions, on the
other hand, some combinations can be the same.
Let’s consider the function f (x1 , x2 , ..., xn ) with its domain M . Let’s suppose, that this function has (in domain M1 ⊂ M ) its partial derivative in
respect xi . fx0 i is again a function of n-variables. If the new function has (in domain M2 ⊂ M1 ) partial derivative in respect xj , then this new partial
2f
derivative is called as the 2-nd partial derivative of the origin function f (x1 , x2 , ..., xn ) in respect xi and xj . This fact we denote as: ∂x∂i ∂x
or fx00i xj . If
j
xi = xj , then
∂2f
.
∂x2i
There are exist account of the n2 partial derivatives of the 2-nd order, where n is a number of variables.
Theorem 1. If function f (x1 , x2 , ..., xn ) has at point A[a1 , a2 , ..., an ] differentiable partial derivatives fx0 i and fx0 j , then
fx00i xj = fx00j xi .
We call this interchangeability of the partial derivatives of the 2-nd order.
Similarly is possible to define the partial derivatives of the 3-th, 4-th and next higher orders.
Excercise 11. Let’ calculate partial derivatives of the 2-nd order of the function z = x3 − 3x4 y + y 5 + 9.
Solution.
First, partial derivatives of the 1-st order are fx0 = 3x2 − 12x3 y, fy0 = −3x4 + 5y 4 .
From last derivatives let’s calculate again next partial derivatives and we get:
00
= (3x2 − 12x3 y)0x = 6x − 36x2 y,
fxx
00
fxy = (3x2 − 12x3 y)0y = −12x3 ,
00
fyx
= (−3x4 + 5y 4 )0x = −12x3 ,
00
= (−3x4 + 5y 4 )0y = 20y 3 .
fyy
00
00
Actually is possible to see, that fxy
= fyx
. Problems.
In problems 113 - 127 calculate partial derivatives of the 2-nd order of the given
113. z = xy + xy .
114.
y2
115. z = 1+5x .
116.
117. z = ln(s3 + t).
118.
1
119. z = 3xy
.
120.
121. z = xy + cos(x − y).
122.
123. z = arctg x−y
.
124.
x+y
cos2 y
126.
125. z = x .
functions.
z=p
xy .
z = 3xy + x2 .
z = x3 − 3x4 y + y 5 .
z = e2y sin x.
z = x sin(x + y) + y cos(x + y).
z = exy .
√
y
z=x y+ √
3 x.
127. z = arctg xy .
128. Let’s prove, that given differentiable functions satisfies following equation:
00 2
00 00
) =0
zyy − (zxy
a) z = ln(ex + ey ), zxx
00
00
b) z = arctg(2x − y), zxx + 2zxy = 0
00
00
c) z = 2 cos2 (x − y2 ), 2zyy
=0
+ zxy
x
00
0
0
y
d) z = ye , zxy − zy + zx = 0
p
00
00
=0
+ zyy
e) z = ln x2 + y 2 , zxx
00
2
00
= 4y
f) z = 2xy + cos(x + y), zxy − zxx
Function of several variables – Extremes of the function of several variables.
Instead, the entire course of the investigation functions of several variables on the properties of the derivatives, and so on, we will now deal only gradually
finding local extreme, global extremes and local extreme with respect to a set, i.e. maximum and minimum of the functions. Procedure for determining
the local extremes is similar to the function of one variable and plays a key role here as the determination of stationary points. In determining the
extremes with respect to a set enters into the calculation condition that limits the search to a given set of extremes. The concept of a global extreme
is actually a maximum or minimum function on a closed set.
A. Let’s consider a function f (X) = f (x1 , x2 , ..., xn ), with its continued partial derivatives of the 2-nd order. Let’s denote
aij = aji = fx00i xj (A).
When searching for local extremes proceed as follows:
A. Calculate partial derivatives of the 1-st order fx0 1 , fx0 2 , ..., fx0 n .
B. Solve the system of equations fx0 1 = 0, fx0 2 = 0, ..., fx0 n = 0. Solutions
a11
a11 a12 > 0, a21
C. If at some stationary point A are values a11 > 0, a21 a22 a31
point A the function f has a local minimum.
a11
a11 a12 > 0, a21
If at some stationary point A are values a11 < 0, a21 a22 a31
f has a local maximum.
are stationary points.
a12 a13 a22 a23 > 0, ... (it means, all sub-determinants are positive), then at
a32 a33 a12 a13
a22 a23
a32 a33
< 0, ... (regularly sign changes), then at point A the function
a
a
By irregular sign changes (for example 11 12
a21 a22
< 0) at point A is no local extreme. In case det(A) equals 0 is impossible say any result.
Excercise 12. Find local extreme of the function z = x2 + y 2 + xy − 6x − 9y.
Solution.
Partial derivatives of the 1-st order must be zero.
zx0 = 2x + y − 6 = 0
zy0 = 2y + x − 9 = 0.
Solving previous system of equations we have x = 1, y = 4 These values are coordinates of stationary point A[1, 4] too.
00
00
00
00
= 2 and we have
= 1, zyy
= zyx
= 2, zxy
Next zxx
a11 a12 2 1 = 3 > 0.
a11 = 2 > 0, =
a21 a22 1 2 So it is clear, that at point A[1, 4] exist a local minimum of the given function and z(1, 4) = 21. B. Let’s consider function z = f (x, y) and let’s find its local extreme with respect to a set g(x, y) = 0. We proceed as follows:
A. If from equation g(x, y) = 0 is possible determine some variable, we calculate it and substitute to origin function. In next step we find local
extreme of the function of one variable only.
B. If this process from previous point is impossible to do, we arrange so called Lagrange function
L(x, y) = f (x, y) + λg(x, y)
and we find local extreme of this new function. Stationary points with connected parameter λ is possible to determine from following equations:
L0x = 0, L0y = 0, L0λ = g(x, y) = 0.
Next steps as in previous case A..
Excercise 13. Find local extreme of the function z = 6 − 4x − 3y with respect to a set x2 + y 2 = 1.
Solution.
Arrange Lagrange function and calculate its partial derivatives of the 1-st order
L(x, y, λ)
L0x
L0y
L0λ
=
=
=
=
6 − 4x − 3y + λ(x2 + y 2 − 1).
−4 + 2λx = 0,
−3 + 2λy = 0,
x2 + y 2 − 1 = 0.
Solving last system of equations we have two stationary points
pre λA = 25 , A[ 54 , 35 ],
pre λB = − 52 , B[− 45 , − 53 ].
Mme L00xx = 2λ, L00xy = 0, L00yy = 2λ. a11 a12 5 0 = 25 > 0 so at point A exist local minimum of the origin problem.
=
At stationary point A are a11 = 5 > 0, a 21 a22 0 5 a11 a12 −5 0 = 25 > 0 so at point B exist local maximum of the origin problem.
=
At stationary point B are a11 = −5 < 0, −5 a21 a22 0
Problems.
In problems 129 - 151 determine local extremes of the following functions.
129. z = 1 + 6y − y 2 − xy − x2 .
130. z = x2 + y 2 − xy − x − y + 2.
132. z = 4 − (x − 2)2 − (y + 3)2 .
131. z = x2 + y 2 + xy − 6x − 9y.
134. z = 2x2 − 6xy + 5y 2 − x + 3y + 2.
133. u = 5 + 6z − 4z 2 − 3t2 .
135. z = x2 − 2y 2 − 3x + 5y − 1.
136. z = x2 − y 2 + 2x − 2y.
137. z = x2 + (y − 1)2 .
138. z = x3 + y 3 − 18xy + 215.
2
2
2
3
139. z = 27x y + 14y − 69y − 54x.
140. z = e−x −y (2y 2 + x2 ).
141. z = 5xy + 25
+ y8 , x > 0, y > 0.
142. z = x2 − (y − 1)2 .
x
143. z = xy + x + y − x2 − y 2 + 2.
144. z = x2 + y 2 − xy − 2x + y.
3
2
145. z = x − 3xy + y + y − 7.
146. z = 6xy − x3 − y 3 .
147. z = e2x (x + y 2 + 2y).
148. z = e2x+3y (8x2 − 6xy + 3y 2 ).
20
50
149. z = xy + x + y .
150. z = 2x3 + xy 2 + 5x2 + y 2 .
2
3
151. z = 3x y + y − 18x − 30y.
In problems 152 - 159 determine local extremes of the following functions with respect to given sets.
153. z = xy, x + y = 1.
152. z = x2 + y 2 , 2x − y + 5 = 0.
1
1
1
1
1
154. z = x + y , x2 + y2 = 4 .
155. z = x + 12 y, x2 + y 2 = 1.
157. z = x + y, x12 + y12 = 12 .
156. z = x2 + 12xy + 2y 2 , 4x2 + y 2 = 25.
158. z = 9 − 8x − 6y, x2 + y 2 = 25.
159. z = 8 − 2x − 4y, x2 + 2y 2 = 12.
In problems 160 - 169 determine global extremes of the following functions.
160. z = x − 2y − 3, 0 ≤ x ≤ 1, 0 ≤ y ≤ 1, 0 ≤ x + y ≤ 1.
161. z = x2 − 2y 2 + 4xy − 6x − 1, x ≥ 0, y ≥ 0, y ≤ −x + 3.
162. z = x3 + y 3 − 3xy, y ≥ −1, y ≤ 2, x ≥ 0, x ≤ 2.
163. z = xy 2 (4 − x − y), x = 0, y = 0, x + y = 6.
164. z = x2 + y 2 − 12x + 16y, x2 + y 2 ≤ 25.
165. z = x2 − xy + y 2 , |x| + |y| ≤ 1.
166. z = x − 2y − 1, x ≤ 0, y ≥ 0, y − x ≤ 1.
167. z = x2 + 4xy + y 2 , |x| + |y| ≤ 4.
168. z = xy, x2 + y 2 ≤ 2.
169. z = x2 − y 2 , x2 + y 2 ≤ 4.
Next problems.
170. In plane σ : x + y − 2z = 0 find a point, the sum of the squares of the distances from the plane ρ1 : x + 3z = 6 and ρ2 : y + 3z = 2 is minimal.
171. Among all of the blocks surface P choose one block
√ with its maximal volume.
172. Among all blocks with a solid diagonal length 2 3 choose one with maximal volume.
173. Among all of the cylinder surface P = 6π choose one with maximal volume.
174. Among all of the cone casing S choose one with maximal volume.
175. Into hemisphere with a given radius R enter the block of the maximal volume.
176. Into cone of its both, base radius R and height h enter the block of the maximal volume.
177. Distribute a positive number a to three positive summands so, that their product will be maximal.
178. In the plane locate such a point that the sum of the squares of the distances from the lines x = 0, y = 0 and x + 2y − 16 = 0 is minimal.
179. Into ellipsoid with half-axis a, b, c enter a block of the maximum volume so that its edges are parallel to the axes of the ellipsoid.
180. Determine the dimensions of the concrete tank shape tetrahedral prism so that the concrete was minimal consumption for a given volume V
tank. Wall thickness are not considered.
181. In plane x + 2y − z + 3 = 0 determine a point, the sum of the squares of the distances from points [1, 1, 1] a [2, 2, 2] is minimal.
Function of several variables – Derivative of the function given by implicit form.
There are situations where it is not possible from the equation explicitly express one variable as a function of one or more variables, although we show
that the certain set can be such function defined. Then we talk about the function given by implicit form. With such expressed functions as a result of
any procedure we often meet, for example the analytical expression of the results of solving ordinary differential equations, which will be the subject of
study in one of the next chapter.
A. Let’s consider an equation F (x, y) = 0, with its point A[x0 , y0 ]. If function F (x, y) is in the neighbourhood of the point A continuous and moreover
exist continuous partial derivatives up to 2-nd order, where ∂F∂y(A) 6= 0, then using equation F (x, y) = 0 is determined an implicit function y = f (x) in
the neighbourhood of the point A. For derivatives of this function (first and second) holds:
Fx0
,
Fy0
(1.5)
00 0
00 0 2
Fxx + 2Fxy
y + Fyy
y
Fy0
(1.6)
y0 = −
y 00 = −
Excercise 14. Let’s determine both derivatives, y 0 and y 00 of the function y given implicitly by equation x2 − 2xy − y 2 − 16 = 0 with point A[4, 0].
Solution.
Calculate firstly both, partial derivatives of the 1-st and 2-nd order. We have:
Fx0 = 2x − 2y, Fy0 = −2x − 2y,
00
00
00
Fxx
= 2, Fxy
= −2, Fyy
= −2.
Partial derivatives are continuous across the all considered area and because Fy0 (A) = −8 6= 0, using both, (1.5) and (1.6) we get:
2x−2y
= x−y
,
y 0 = − −2x−2y
x+y
2
x−y
x−y
2
2
−2( x+y )
2−4
= 2y +4xy−2x
. y 00 = − x+y
−2x−2y
(x+y)3
B. Let’s consider an equation F (x1 , x2 , ..., xn , y) = 0, with its point A[a1 , a2 , ..., an , an+1 ]. If function F (x1 , x2 , ..., xn , y) is in the neighbourhood of the
point A continuous and moreover exist continuous partial derivatives of the 1-nd order, where ∂F∂y(A) 6= 0, then using equation F (x1 , x2 , ..., xn , y) = 0 is
determined an implicit function y = f (x1 , x2 , ..., xn ) in the neighbourhood of the point A. For its partial derivatives holds:
F0
∂f
= − x0i .
∂xi
Fy
(1.7)
Excercise 15. Let’s determine derivatives of the 1-st order of the function z given implicitly by equation 4x2 + 2y 2 − 3z 2 + xy − yz + x − 4 with point
A[1, 1, 1].
Solution.
Calculate firstly partial derivatives. We have:
Fx0 = 8x + y + 1, Fy0 = 4y + x − z, Fz0 = −6z − y.
Using (1.7) we get:
zx0 = − 8x+y+1
= 8x+y+1
,
−6z−y
6z+y
4y+x−z
4y+x−z
zy0 = − −6z−y = 6z+y . C. Let’s consider an equation of the surface in implicit form F (x, y, z) = 0, then the equation, represent the tangent plane to this surface at point
M [x0 , y0 , z0 ] is possible write in form:
Fx0 (M )(x − x0 ) + Fy0 (M )(y − y0 ) + Fz0 (M )(z − z0 ) = 0.
(1.8)
Parametric representation of normal line to considered tangent line is possible write in form:
x = x0 + Fx0 (M )t,
y = y0 + Fy0 (M )t,
z = z0 + Fz0 (M )t.
(1.9)
Excercise 16. Let’s determine an representation of the tangent plane and its normal line to graph of the function given implicitly x2 − y 2 + z 2 − 6 = 0
at point A[1, 2, −3].
Solution.
We have Fx0 (A) = (2x)A = 2, Fy0 (A) = (−2y)A = −4, Fz0 (A) = (2z)A = −6.
Using both, (1.8) and (1.9) we get following representation of the tangent plane
x − 2y − 3z − 6 = 0,
and Parametric representation of normal line
x = 1 + 2t
y = 2 − 4t
z = −3 − 6t. Chapter 1. (Function of several variables) – 17
Problems.
In problems 182 - 185 calculate derivative y 0 at point A of the function given implicitly by equation F (x,
√ y) = 0.
2
2
y
2
182. x − xy − y − 1 = 0, A[2, 1].
− 2x + y = 0, A[ 3, 2].
183. 2 p
184. x sin y − y cos x = 0, A[ 14 π, − 14 π].
185. ln x2 + y 2 − arctg xy = 0, A[1, 0].
In problems 186 - 193 calculate derivative y 0 of the function given implicitly by equation F (x, y) = 0.
187. ex cos y + ey cos x − 1 = 0.
186. xy − ln(exy + e−xy ) + ln 2 = 0.
188. xex − y 2 − xy = 0.
189. x2 − 2xy − y 2 − 16 = 0.
191. x − y + 4 sin y = 0.
190. x − ln y − y 2 = 0.
3
2
193. xy − y x = 0.
192. y − 4xy + x = 0.
In problems 194 - 197 calculate partial derivatives zx0 and zy0 at point A of the function given implicitly by equation F (x, y, z) = 0.
194. 3x2 − 4y 2 + 2z 2 − xy + xz − y = 0, A[1, 1, 1].
195. cos2 x + cos2 y + cos2 z − 2 = 0, A[ 31 π, 21 π, 16 π].
196. z 4 − 4xyz − 1 = 0, A[0, 2, 1].
197. xez − yz − xy 2 = 0, A[2, 1, 0].
In problems 198 - 205 calculate partial derivatives zx0 and zy0 of the function given implicitly by equation F (x, y, z) = 0.
199. x cos y + y cos z + z cos x − 1 = 0.
198. 4x2 + 2y 2 − 3z 2 + xy − yz + x − 4 = 0.
z
2
200. e + x y + z + 5 = 0.
201. x3 + 2y 3 + z 3 − 3xyz − 2y + 27 = 0.
202. sin xy + sin yz + sin zx − 1 = 0.
203. z − xy sin xz = 0.
204. z + ez − xy − 1 − 0.
205. arctg x + arctg y + arctg z − 5 = 0.
206. Determine the equation of the tangent line and normal line at point T to the graph of the function y = f (x) given implicitly by equation
F (x, y) = 0.
a) x5 + y 5 − 2xy = 0, T [1, 1];
b) x2 − y 2 + 2x − 4y − 6 = 0, T [1, −1];
c) 2x3 − x2 y + 3x2 + 4xy − 5x − 3y + 6 = 0, T [0, 2];
d) x3 + y 3 − 2xy = 0, T [1, 1];
207. Determine the equation of the tangent plane and normal line to the graph of the function z = f (x, y) at point T given implicitly by equation
F (x, y, z) = 0.
a) x2 −p
y 2 + z 2 − 6 = 0, T [1, 2, −3];
b) 4 + x2 + y 2 + z 2 − x − y − z = 0, T [2, 3, 6];
c) x2 + y 2 + z 2 − 49 = 0, T [2, −6, ?];
d) (z 2 − x2 )xyz − y 5 − 5 = 0, T [1, 1, 2];
e) z − y − ln xz = 0, T [1, 1, 1];
f) x3 + y 3 + z 3 + xyz − 6 = 0, T [−1, ?, 1];
g) ez − z + xy − 3 = 0, T [2, 1, ?];
y
x
h) 8 − 2 z − 2 z = 0, T [2, ?, 1].
Function of several variables – Basics of vector analysis.
The motivation for this part of the chapter is a number of physical applications, where we can find practical use of the following descriptions to describe
various physical phenomenons.
A. Scalar vector function. Let’s M is a set of real numbers. Then a function f , which assigned for each number t from M a vector f (t) is called the
vector function of one real variable or scalar vector function. The set of M is called a domain of the function f and f (t) the value of the scalar vector
function f at value equals t.
Let’s define in 3-D space rectangular coordinate system, then for vector function f holds:
f (t) = x(t)i + y(t)j + z(t)k,
or
f (t) : x = x(t),
y = y(t),
z = z(t),
where i, j, k are unit vectors. The function x(t) is called as 1-st component, the function y(t) as 2-nd component and the function z(t) as 3-th
component of the vector function f .
At considered rectangular coordinate system f (t) = x(t)i + y(t)j + z(t)k has a function its derivative f at value equals t0 if and only if exist derivatives
of the functions x(t), y(t) and z(t) at value equals t0 and holds:
f 0 (t0 ) = x0 (t0 )i + y 0 (t0 )j + z 0 (t0 )k.
(1.10)
Excercise 17. Let’s determine the equation of the tangent plane and normal line to the curve f (t) : x = t4 , y = t − 3t2 , z = t − 1 at point T [1, −2, 0].
Solution.
First, let’s determine a value t0 corresponding to point T . Since the point T lies on the curve f (t), next holds:
f (t) : 1
= t4 ,
−2 = t − 3t2 ,
0
= t − 1.
Solving last equations we have t0 = 1. Because (1.10) holds, it possible to calculate derivative of the scalar function at t0 = 1.
f 0 (1) = x0 (1)i + y 0 (1)j + z 0 (1)k = (4t3 )1 i + (1 − 6t)1 j + 1k = 4i − 5j + 1k.
This vector (4, −5, 1) is a directional vector of the tangent line and normal vector of the normal plane together. Parametric representation of tangent
line is system:
x = 1 + 4t
y = −2 − 5t
z = t.
And finally a normal plane to given curve is represents by equation:
4(x − 1) − 5(y + 2) + 1(z − 0) = 0,
4x − 5y + z − 14 = 0. B. Directional derivative. Let’s consider a function of several variables f (X) = f (x1 , x2 , ..., xn ), where X[x1 , x2 , ..., xn ] with point A[a1 , a2 , ..., an ]
and unit vector l = (l1 , l2 , ..., ln ). Then derivative of the function f (X) at point A in direction of vector l is called following limit
lim+
t→0
f (A + tl) − f (A)
t
(A)
and we denote this (definite) limit (if exist) as dfdl
.
If f is a differentiable function of two variables f (x, y) at point A and vector l creates (in rectangular coordinate system) an angle α with coordinate
ox (in 2-D space), l = (cos α, sin α), then
df (A)
= fx0 (A) cos α + fy0 (A) sin α.
(1.11)
dl
If f is a differentiable function of three variables f (x, y, z) at point A and for vector l (in rectangular coordinate system) in 3-D space holds l =
(cos α, cos β, cos γ), then
df (A)
= fx0 (A) cos α + fy0 (A) cos β + fz0 (A) cos γ.
(1.12)
dl
Excercise 18. Let’s calculate derivative of the function f (x, y, z) = xy 2 + z 3 − xyz at point A[1, 1, 2] in direction of vector l, which creates with
coordinates in rectangular system angles α = π3 , β = π4 , γ = π3 .
Solution.
First, calculate partial derivatives of the function f at point A. We have
fx0 (A) = (y 2 − yz)A = −1, fy0 (A) = (2xy − xz)A = 0, fz0 (A) = (3z 2 − xy)A = 11.
Using term (1.12) we have
df (A)
dl
= −1 cos π3 + 0 cos π4 + 11 cos π3 = 5. C. Gradient. Let’s consider define a real function f (X) with its domain M , this situation is denoted as scalar field too. Let’s consider define a real
function f (X) with its domain M , this situation is denoted as vector field.
Let’s f (x) is a differentiable function at point A (in 3-D space). Then a gradient of the function f (X) at point A is called a vector grad f (A), which
satisfies following conditions:
grad f (A) = fx0 (A)i + fy0 (A)j + fz0 (A)k.
(1.13)
Let’s f is a differentiable function with its domain M . Then a gradient of the function f (X) (or gradient of the scalar field) f (X) is called a vector
function (with the same domain M ), which satisfies following conditions:
grad f (X) = fx0 (X)i + fy0 (X)j + fz0 (X)k,
(1.14)
for each point X ∈ M . Let’s f (X) is a differentiable function at point A In 3-D (or 2-D) space, respectively. Then for derivative of the function f (X)
at point A in direction determined by unit vector l holds:
df (A)
= grad f (A).l,
(1.15)
dl
df (A)
dl
is a projection of the vector grad f (A) into vector l.
Excercise 19. Let’s determine a gradient of the function f (x, y, z) = x2 + y 2 + z 2 at point A[2, −2, 1].
Solution.
First, calculate partial derivatives f at point A. We have
fx0 (A) = (2x)A = 4, fy0 (A) = (2y)A = −4, fz0 (A) = (2z)A = 2.
Using (1.13) we get
grad f (A) = 4i − 4j + 2k. Chapter 1. (Function of several variables) – 21
Excercise 20. Let’s determine the derivative of the function f (x, y, z) = 3x3 − 4y 3 + 2z 4 at point A[2, 2, 1], in direction of vector l, if vector l = AB,
where B[5, 4, 6].
Solution.
From partial derivatives of the functions f at point A
fx0 (A) = (9x2 )A = 36, fy0 (A) = (−12y 2 )A = −48, fz0 (A) = (8z 3 )A = 8.
is possible determine the gradient
grad f (A) = 36i − 48j + 8k.
Next calculate coordinates of the vector l and its length
l = AB = B − A = (3, 2, 5), |l| =
√
9 + 4 + 25 =
√
38.
Using (1.15) we get
df (A)
dl
√
= (36i − 48j + 8k) (3i+2j+5k)
= (36.3 − 48.2 + 8.5) √138 =
38
√52 .
38
D. Divergence a rotation. Let’s consider rectangular coordinate system of three vectors ordered in positive orientation and real function of three
variables:
f (X) = f1 (X)i + f2 (X)j + f3 (X)k,
where functions f1 (X), f2 (X), f3 (X), X[x, y, z] have their partial derivatives for set M in Euclidean space E3 .
A divergence of the vector function f (X) is called a vector function
div f =
∂f1 ∂f2 ∂f3
+
+
.
∂x
∂y
∂z
(1.16)
A rotation of the vector function f (X) is called a vector function
∂f1 ∂f3
∂f2 ∂f1
∂f3 ∂f2
−
i+
−
j+
−
k,
rot f =
∂y
∂z
∂z
∂x
∂x
∂y
(1.17)
This situation is possible write in form of one determinant symbolically so:
i
j
k
∂ ∂ ∂ rot f = ∂x ∂y ∂z .
f1 f2 f3 Chapter 1. (Function of several variables) – 22
Excercise 21. Let’s determine both, divergence and rotation of the vector function f (X) = 5x2 yi + (x2 − z 2 )j +(2x − z 2 )k.
Solution.
div f = (5x2 y)0x + (x2 − z 2 )0y + (2x − z 2 )0z = 10x + 0 − 2z.
rot f = (2x − z 2 )0y − (x2 − z 2 )0z i + ((5x2 y)0z − (2x − z 2 )0x ) j + (x2 − z 2 )0x − (5x2 y)0y k =
= 2zi − 2j + (2x − 5x2 )k. Problems.
In problems 208 - 219 determine the equation of the tangent and normal plane to the curve.
208. x = at, y = 21 at2 , z = 31 at3 , T [6a, 18a, 72a].
209. x = t, y = t2 , z = t3 , t0 = −1. √
210. x = sin 2t, y = 1 − cos 2t, z = 2 cos t, t0 = π4 .
√
211. x = t − sin t, y = 1 − cos t, z = 4 sin 2t , T [ π2 − 1, 1, 2 2].
√
√
212. x = 2 cos t, y = 2 sin t, z = 4t, t0 = π4 .
213. x = t2 , y = 1 −√t, z = t3 , T [1,
√0, 1].
214. x = 3 tg t, y = 2 cos t, z = 2 sin t, t0 = π4 .
215. x = t3 − t2 − 5, y = 3t2 + 1, z = 2t3 − 16, t0 = 2.
216. x = z, y = z 2 , T [−1, 1, −1].
217. x2 + y 2 + z 2 = 50, x2 + y 2 = z 2 , T [3, −4, −5].
218. y 2 + z 2 = 25, x2 + y 2 = 10, T [1, 3, 4].
219. x2 + y 2 + z 2 = 3, x2 + y 2 = 2, T [1, 1, 1].
220. Find a point on the curve x = t3 , y = t2 − 2t, z = 2t + t2 at which the tangent is parallel to the plane 2x − 3y − 3z − 5 = 0.
221. Find a point on the curve x = 2t2 , y = 3t − t3 , z = 3t + t3 at which the tangent is parallel to the plane 3x + y − z + 7 = 0.
222. Find a tangent plane to graph of the ellipsoid x2 + 2y 2 + z 2 = 1, parallel to one another plane x − y + 2z = 0.
223. Find a tangent plane to graph of the function x2 − y 2 − 3z = 0 passing through point A[0, 0, −1] parallel to line x2 = y1 = z2 .
224. Find a tangent plane to graph of the ellipsoid 3x2 + 2y 2 + z 2 − 21 = 0 parallel to one another plane 6x + 4y + z = 0.
225. Find a tangent plane to graph of the function z = xy perpendicular to line x+2
= y+2
= z−1
.
2
1
−1
In problems 226 - 231 calculate derivative in the direction of vector I given function at given point, if
226. f (x, y) = 3x4 − x2 y 3 + y 2 , A[1, −1] and vector I create to rectangular coordinates so the angles α = π6 , β = π3
227. f (x, y) = 3x2 − 6xy 4 + 11y 5 , A[1, 1] and vector I is determined by vector B − A, where B[4, 5].
228. f (x, y) = 5x2 − 6xy + 10y 3 − 7, A[0, 1] and vector I = −i.
229. f (x, y, z) = xy 2 + z 3 − xyz, A[1, 1, 2] and vector I creates with the coordinate axis angles α = π3 , β = π4 , γ = π3 .
230. f (x, y, z) = 5x2 − 2y 2 , A[2, 5, 0], B[−2, 2, 0] and vector I determined by vector B − A.
231. f (x, y, z) = 3x3 − 4y 3 + 2z 4 , A[2, 2, 1] and vector I is determined by vector B − A, where B[5, 4, 6].
In problems 232 - 237 find the gradient of the given function.
2
2
232. z = x
233. z = x2 − 2xy + y 2 + 4x − 8y − 7.
p − 6xy + y − 10x − 2y + 9.
235. u = xyzex+y+z .
234. z = x2 + y 2 .
2
2
3
236. u = arctg (x + 3y + z ).
237. u = x2 + 2y 2 + 3z 2 − xz + yz − xy.
In problems 238 - 245 find the gradient of the given function at a given point.
2
238. z = x2 +3y
2 , A[2, −1].
x
239. z = arctg x+y
, A[1, 0].
240. u = p
x sin z − y cos z, A[2, 1, 0].
241. z = 4 + x2 + y 2 , A[1, 2].
242. u = x3 + y 3 + z 3 − 3xyz, A[2, 1, −1].
243. u = x2 +yx2 +z2 , A[1, 2, 2].
244. z = x2 − 2xy + 3y 2 + 4x − 2y + 5, A = [7, 4].
245. u = x2 + 3y 2 + 4z 2 + xy − 2yz + 5xz, A[1, 0, 1].
p
246. Locate an angle between gradients of the function z = ln x2 + y 2 at points M [1, 1] and N [1, −1].
247. Locate an angle between gradients of the functions u = x3 + y 3 + z 3 a v = x2 − y 2 + z 2 at point M [1, −1, 1].
248. Locate points in the plane in which the gradient of the function z = ln (x − y1 ) equals i + j.
√
x
249. Locate an angle between gradients of the functions u = x2 + y 2 − z 2 and v = arcsin x+y
at point A[1, 1, 7].
250. Locate points in the plane in which the gradient of the function u = ln 1+xy
equals − 16
i + j.
x
9
In problems 251 - 254 find the divergence of the vector function.
252. u = xy 2 z 3 (2i + 3j − k).
251. u = x2 i − xyj + xyzk.
253. u = (yi + zj + xk) √ 2 1 2 2 .
254. u = xyzi + (2x + 3y + z)j + (x2 + z 2 )k.
x +y +z
In problems 255 - 258 find the divergence of the vector function at a given point.
y
255. u = xyi + (x2 − z 2 )j + (x+z)
k, A[1, 0, 2].
y
x
z
256. u = y i + z j + x k, A[1, 1, 1].
257. u = exyz i + sin x sin y sin zj + ln (x + y + z)k, A[0, 2, −1].
258. u = (2x2 y − 3xz 3 + 5x3 yz)i + (4y 3 x + xyz + 8z 2 )j + (6z 3 xy 2 − 7z 2 x + 9zy)k, A[1, 1, 1].
In problems 259 - 262 find the rotation of the vector function.
259. u = x2 i + y 2 j + z 3 k.
260. u = xy 2 z 3 (2i + 3j − k).
261. u = (yi + zj + xk) √ 2 1 2 2 .
x +y +z
262. u = (2x − 3y + 5z)i + (x2 + 4y 2 − 8z 2 )j + (3x3 − y 3 + 2z 3 )k.
In problems 263 - 266 find the rotation of the vector function at a given point.
y
263. u = xyi + (x2 − z 2 )j + (x+z)
k, A[1, 0, 2].
y
z
x
264. u = y i + z j + x k, A[1, 1, 1].
265. u = exyz i + sin x sin y sin zj + ln (x + y + z)k, A[0, 2, −1].
266. u = y 2 z 2 i + x2 z 2 j + x2 y 2 k, A[1, 2, 3].
Self-assessment questions
1. Which one equivalent exist at the theory of function of one variable equivalent to the term closed or open set, respectively, which is defined in the
theory of function of several variables?
2. It is possible at the theory of function of several variables to define terms equivalent to term one-side limit (from the right and left) used in the theory
of function of one variable?
3. Which conditions much be satisfied for equality to mixed partial derivatives of higher order?
4. Under what conditions is sufficient for the calculation of global extremes functions of several variables simultaneously consider cases only local extremes
(or local extremes with respect to given sets)?
Conclusion
Since we live in three-dimensional space, all the events going on around us are physical point of view generally described functions of several variables.
Therefore, it is important to study not only the real functions of one real variable, but several variables. Knowledge acquired in this chapter are necessary
for studying other chapters of mathematical analysis, working with functions of several variables that have an irreplaceable role in describing and solving
physics problems in engineering practice.
Bibliography
[1] Siagova, J.: Advanced Mathematics, Bratislava, STU, 2011.
[2] Bubenik, F.: Problems to Mathematics for Engineers, Praha, CVUT, 1999.
[3] Kreyszig, E.: Advanced Engineering Mathematics, New York 1993.
[4] Neustupa, J.: Mathematics I, Praha, CVUT, 1996.
[5] Neustupa, J.: Mathematics II, Praha, CVUT, 1998.
Solutions
Answers to self-assessment questions (if you are not able to formulate this answers), as well as many other answers can be found in the recommended
literature.
1.
3.
5.
7.
9.
11.
13.
15.
17.
19.
21.
23.
24.
26.
28.
half-plane x + y < 0
the interior both, of the first and third quadrant
4 < x2 + y 2 + z 2 < 9
< −1, 1 > × < −1, 1 >
−1 − x ≤ y ≤ 1 − x
x ≥ 0, 0 ≤ y ≤ x2
2
(x − 21 )2 + y 2 ≥ 14 (x − 1) + y 2 < 1
all plane except for lines x = 0 and y = 1
x ≥ 0, y > 0
|x| ≤ 1, y ∈ R
x < 0, y > 0; x > 0, y > x
part of plane inside parabola y 2 = 4x between parabola and circle x2 + y 2 = 1
outside of the parabola y 2 = 4(x − 2)
y < x, y > −x
1
y ≥ 0, y ≤ 1+x
2
30. zx0 = 2xy −
2
32. zx0 = √ y
4y 3
x5 ,
(x2 +y 2 )
3y 2
x4
xy
(x2 +y 2 )3
+ y), zy0
zy0 = x2 +
, zy0 = − √
3
zx0 = sin(x + y) + x cos(x
= x cos(x + y)
1
2x
0
0
zx = x+y2 , zy = x+y2
z
z
z
u0x = xz ( xy ) , u0y = − yz ( xy ) , u0z = ( xy ) ln xy
1
1
0
zx0 = x+ln
y , zy = y(x+ln y)
2
√
2y
x
√
42. zx0 = 3 y − 2xy√x , zy0 = 3 √
3 y2 +
x
34.
36.
38.
40.
2. all plane except for lines y = x, y = −x
4. 1 ≤ x2 + y 2 ≤ 4
6. |x − 1| ≤ |y|, y 6= 0
8. [x, y] : x2 + y 2 6= 25
10. y ≥ x2 , y ≤ 1
12. x2 + y 2 < 4, xy > 0
14. R3
16. all plane except for lines y = 2x, y = −2x, x < 0
18. x > 0, |y| < x
20. y > x
22. x < 0, x ≤ y < 0; x > 0, 0 < y ≤ x
except the top of parabola and circle points
25. y < x2 , y ≤ 0
27. < −1, 1 >3
29. zx0 = 4x3 − 8xy 2 , zx0 = 4y 3 − 8x2 y
31. zx0 = y12 , zy0 = − 2x
y3
2
2
x
, zy0 = − cosx
33. zx0 = − 2x sin
y
y2
35. zx0 = yxy−1 , zy0 = xy ln x
y
x
0
37. zx0 = − x2 +y
2 , zy = x2 +y 2
yu
u ln x
0
0
39. u0x = yu
xz , uy =
z , uz = − z 2 ln x
41. zx0 = 12x3 y − 5y 2 , zy0 = 3x4 − 10xy + 2
43. u0x =
yz
x u,
u0y = zy (z−1) u ln x, u0z = y z u ln x ln y
44. u0x = 3x2 y 2 z − y 2 z 3 , u0y = 2x3 yz − 2xyz 3 , u0z = x3 y 2 − 3xy 2 z 2
46. zx0 = y 2 =
y
x2
+
√1 ,
x
zy0 = 2xy −
1
x
47. zx0 =
49. zx0 =
48. zx0 = −ye−xy , zy0 = −xe−xy
50. zx0 =
x
1−ye
x−yex ,
zy0 =
3
−ex
x−yex
cos y−1
3
45. zx0 = 20(5x − y) , zy0 = −4(5x − y)
51. u0x =
2
2
y
x
, zy0 = − (y−x)
2
(y−x)2
2x2 +y 2
xy
0
√
√
, zy =
x2 +y 2
x2 +y 2
y
2xe sin z, u0y = x2 ey sin z,
u0z = x2 ey cos z
cos y
52. zx0 = cos x cos y(sin x)
, zy0 = − sin y ln sin x(sin x)
53. u0x = yz 2 cos(xyz) + z cos(xz), u0y = xz 2 cos(xyz) − cos(yz) + yz sin(yz), u0z = sin(xyz) + xyz cos(xyz) + y 2 sin(yz) + x cos(xz)
54. zx0 = √ 21 2 , zy0 = 2 2 y√ 2 2
55. zx0 = ( x1 + πy cos(πxy))z, zy0 = ( y1 + πx cos(πxy))z
x +y
x +y +x x +y
y
y
56. zx0 = e x (1 − xy ), zy0 = e x
58. zx0 =
u0x
√
√
y xy
ln x xy
0
2x(1+xy ) , zy = 2(1+xy )
u ln(y tg z)
x
, u0y = u ln
x
y ,
59.
u0z
=
60.
=
64. df = 2(x − y)dx + 2(y − x)dy
66. df = −2(ydx − xdy)/y 2 sin 2x
y
2
u ln x
cos z sin z
61.
65.
67. df = e (y ln ydx + dy)/y
2
)dx−(1+x )dy
68. df = (1+y
1+x2 +y 2 +x2 y 2
5
(dx + 3dy)
70. 32
π
73. 4 dx − dy
76. 41 e
79. 9, 506
82. 2, 95
85. −0, 8cm, −0, 3m2
88. 0, 36cm3 ; 0, 4cm2
89. −94, 25cm3 ; −16, 86cm2 ; 1, 99cm2
y+1
z−13
90. a) 6x − 8y − z − 13 = 0, x−3
6 = −8 = −1 ;
y−4
x−3
z+7
b) 17x + 11y + 5z − 60 = 0, 17 = 11 = 5 ;
y−2
z−3
c) 2x − y + z − 3 = 0, x−1
2 = −1 = 1 ;
y−3
x−5
d) x + 6y + 4z − 31 = 0, 1 = 6 = z−2
4 ;
2
y−2
x−1
2
2
2
e) 2e x + e y − z − 3e = 0, 2e2 = e2 = z−e
−1 ;
x− 1
f) 80x + 17y − 16z − 52 = 0, 804 = y−2
17 =
y−1
z
x−π
g) x − πy + z = 0, −1 = π = −1 ;
y−1
z−1
h) x + y + z − 3 = 0, x−1
−1 = −1 = −1 ;
z
i) 2x − z − 2 = 0, x−1
2 = −1 , y = 0.
92. cos 2t
94. 0 t
(2+t ln t)
96. 2+e2t√
1+et
4
2 2
x4 +3x2 y 2 −2xy 3
y −2x3 y
, zy0 = y +3x
(x2 +y 2 )2
(x2 +y 2 )2
−2x
0
zx0 = x22y
−y 2 , zy = x2 −y 2
10
u0x = 22y 2 (2xy 2 + z 3 ) , u0y = 44xy(2xy 2
y
x
df = x2 +y
2 dx + x2 +y 2 dy
x
57. zx0 =
z− 18
−16
71.
74.
77.
80.
83.
86.
1
12
√
10
+ z 3 ) , u0z = 33z 2 (2xy 2 + z 3 )
10
69. df = (3dx − 2dy + 5dz) cos(3x − 2y + 5z)
72.
75.
78.
81.
84.
87.
2(3dx + dy)
3dx + dy − 6dz
−0, 018
1, 10
0, 97
70, 37cm3
dx − dy
−3, 6
0, 01
0, 555
0, 502
3, 77dm3
;
√
t
93. 2t√cos
+ sin t
sin t
95. 12t6 ln t−1 ln t
97. −et − e−t
98.
100.
102.
104.
105.
106.
107.
1
1+x2 ,
115.
116.
117.
118.
119.
120.
121.
122.
123.
124.
125.
126.
127.
129.
132.
135.
138.
139.
140.
99.
1−2(x+1)2
y
y, y2 +(x+1)
2
y 2 +(x+1)2
0
2x+2yϕ (x)
2x
, x2 +y2
x2 +y 2
2
2
0
zx = 2x + x+y
ln(x + y), zy0 = cos y + x+y
ln(x + y)
0
2
0
zx = 3x sin y cos y(cos y − sin y), zy = x3 (sin y + cos y)(1 − 3 sin y cos y)
x
x
1
1
y
+ uv ln u y1 e y , zy0 = vuv−1 y−x
+ uv ln u −x
zx0 = vuv−1 x−y
y2 e
2
2
−2x
3x
2x2
0
zx0 = 2x
y 2 ln(3x − 2y) + y 2 (3x − 2y), zy = y 3 ln(3x − 2y) − y 2 (3x−2y)
x2 +y 2
xy
2
2
x
ex +3ex x2
, exe+ey
ex +ex3
y
y 0
y
101. e + xe ϕ (x), e , kde y = ϕ(x)
103. yxy−1 + xy ln xϕ0 (x), yxy−1
2
x +y
4
4
3
x4 −y 4 +2x3 y
, zx0 = e xy y −xxy+2xy
2
x2 y
zx0 = 2(x−2y)(x+3y)
, zy0 = (2x−y)(9x+2y)
(y+2x)2
(y+2x)2
zx0 = 2xfu0 + yexy fv0 , zy0 = −2yfu0 + xexy fv0 , u = x2 − y 2 , v = exy
zx0 = y(1 − x12 )fu0 , zy0 = (x + x1 )fu0 , u = xy + xy
zx0 = zu0 + zv0 , zy0 = zu0 − zv0 , u = x + y, v = x − y
00
00
00
= 1 − y12 , zyy
= 0, zxy
zxx
= 2x
y3
00
y−2
00
00
= xy ln2 x
, zxy = xy−1 (1 + y ln x), zyy
zxx = y(y − 1)x
2
50y
−10y
2
00
00
00
= (1+5x)
zxx
= 1+5x
, zyy
3 , zxy =
(1+5x)2
−9y 2
9xy
−9x2
00
00
00
zxx
=
3 , zxy =
3 , zyy =
3
4(3xy+x2 ) 2
4(3xy+x2 ) 2
4(3xy+x2 ) 2
3
2
)
−1
00
00
00
zss
= 3s(2t−s
, zst
= (s−3s
3 +t)2 , ztt = (s3 +t)2
(s3 +t)2
00
00
00
zxx
= 6x − 36x2 y, zxy
= −12x3 , zyy
= 20y 3
2
1
2
00
00
00
zxx = 3x2 y , zxy = 3x2 y2 , zyy = 3xy3
00
00
00
zxx
= −e2y sin x, zxy
= 2e2y cos x, zyy
= 4e2y sin x
00
00
00
zxx = −cos(x − y), zxy = 1 + cos(x − y), zyy
= − cos(x − y)
00
00
zxx = (2 − y) cos(x + y) − x sin(x + y), zxy = (1 − y) cos(x + y) −
x2 −y 2
2xy
00
00
00
2
2
zxx
= −2xy
v 2 , zxy =
v 2 , zyy = v 2 , v = x + y
00
00
00
zxx
= y 2 exy , zxy
= (1 + xy)exy , zyy
= x2 exy
2 cos2 y
−2 cos 2y
00
00
00
zxx = x3 , zyy =
, zxy = sinx22y
x
−x
1
1
00
00
00
= √
, zxy
= 2√
zxx
= 4y7 , zyy
y − 3x 34
4 y3
9x 3
108. zx0 = e
109.
110.
111.
112.
113.
114.
3
y
− x2 +y
2
00
(1 + x) sin(x + y), zyy
= −(2 + x) sin(x + y) − y cos(x + y)
2
−x
−2xy
00
00
00
zxx
, zxy
= (x22xy
= (xy2 +y
2 )2 , zyy = (x2 +y 2 )2
+y 2 )2
max = z(−2, 4) = 13
130. min = z(1, 1) = 1
max = z(2, −3) = 4
133. max = u( 43 , 0) = 29
4
∅
136. ∅
min = z(6, 6) = −1
min = z(1, 1) = −82, max = z(−1, −1) = 82
min = z(0, 0) = 0, max = z(0, 1) = z(0, −1) = 2e
131. min = z(1, 4) = −21
134. max = z(−2, − 32 ) = 34
137. max = z(0, 1) = 0
141.
144.
147.
150.
151.
152.
154.
155.
156.
157.
158.
159.
160.
161.
162.
163.
164.
165.
166.
167.
168.
169.
min = z( 25 , 45 ) = 30
142. ∅
max = z(1, 0) = −1
145. min = z(1, 1) = −7
min = z( 21 , −1) = − 2e
148. min = z(0, 0) = 0
min = z(0, 0) = 0, max = z(− 53 , 0) = 125
27
min = z(1, 3) = −72, max = z(−1, −3) = 72
min = z(−2, 1) = 5
153. max = z( 12 , 12 ) = 14
√
√
√ √
√
√
max = z(2 2, 2 2) = 22 , min = z(−2 2, −2 2) = − 22
√
√
√
√
√
√
max = z( 52 5, 15 5) = 21 5, min = z(− 25 5, − 15 5) = − 12 5
max = z( 32 , 4) = z(− 32 , −4) 425
4 , min = z(2, −3) = z(−2, 3) = −50
min = z(2, 2) = 4, max = z(−2, −2) = −4
max = z(−4, −3) = 59, min = z(4, 3) = −41
min = z(2, 2) = −4, max = z(−2, −2) = 20
min z = −5, max z = −2
min = z(0, 3) = −19, max = z(0, 0) = −1
max = z(2, −1) = 13, min = z(0, −1) = z(1, 1) = −1
max = z(1, 2) = 4, min = z(2, 4) = −64
min z = −75, max z = 125
min z = 0, max z = 1
max = z(0, 0) = −1, min = z(0, 1) = −3
max = z(2, 2) = z(−2, −2) = 24, min = z(2, −2) = z(−2, 2) = −8
max = z(1, 1) = z(−1, −1) = 1, min = z(1, −1) = z(−1, 1) = −1
max = z(2, 0) = z(−2, 0) = 4, min = z(0, −2) = z(0, 2)
q= −4
170. (3, −1, 1)
173. 1, 2
√
√
176. 23 2R, 23 2R,
2a
2b
2c
179. √
, √
, √
3
3
3
182.
171.
174.
1
3h
3
4
185. 1
188. y 0 =
0
191. y =
183.
186.
ex +xex −y
2y+x
1
1−4 cos y
− 56 ,
194.
2
197. 0, 4
sin x−cos y
0
199. zx0 = zcos
x−y sin z , zy =
201. zx0 =
203. zx0 =
177.
180.
x sin y−cos z
cos x−y sin z
2
2
x −yz
6y −3xz−2
0
xy−z 2 , zy = 3(xy−z 2 )
y sin xz+xyz cos xz
x sin xz
, zy0 = 1−x
2 y cos xz
1−x2 y cos xz
189.
192.
143. max = z(1, 1) = 3
146. max = z(2, 2) = 8
149. max = z(5, 2) = 30
P
q6
√S ,
√2S
3π
3π
1
1
1
a,
a,
a
3
3 √3
√
√
3
2V , 3 2V , 21 3 2V
√
4 3
1+4 ln 2
0
y = − xy
y 0 = x−y
x+y
4y−2x
y 0 = 3y
2 −4x
195. −1, 0
172. 2, 2, 2
√
175. 23 3R,
q
2
3
√
3R,
1
3
√
3R
8 16
5, 5
( 21 , − 21 , 25 )
π+4
π−4
y
x−ex cos y
y 0 = eey sin
cos x−ex sin y
y
y 0 = 1+2y
2
2
y 1−ln x
0
y = x2 1−ln y
178.
181.
184.
187.
190.
193.
196. 2, 0
x+4y−z
0
=
198. zx0 = 8x+y+1
,
z
y
6z+y
6z+y
2
x
0
200. zx0 = − e2xy
z +1 , zy = − ez +1
202. zx0 =
204. zx0 =
y cos xy+z cos xz
0
x cos xz+y cos yz , zy
y
x
0
ez +1 , zy = ez +1
xy+z cos yz
= − xx cos
cos xz+y cos yz
2
2
1+z
1+z
0
205. zx0 = − 1+x
2 , z y = − 1+y 2
206. a) x + y − 2 = 0, x − y = 0;
b) 2x − y − 3 = 0, x + 2y + 1 = 0;
c) x + y − 2 = 0, x − y = 0;
d) x − y = 0, x + y − 2 = 0.
y−2
z+3
207. a) x − 2y − 3z − 6 = 0, x−1
1 = −2 = −3 ;
y−3
x−2
z−6
b) 5x + 4y + z − 28 = 0, 5 = 4 = 1 ;
y+6
z−3
c) 2x − 6y + 3z − 49 = 0, x−2
4 = −12 = 6 , 2x − 6y − 3z − 49 = 0,
y−1
z−2
d) 2x + y + 11z − 25 = 0, x−1
2 = 1 = 11 ;
y−1
x−1
z−1
e) x + y − 2z = 0, 1 = 1 = −2 ;
y−2
z−1
f) 5x + 11y + z − 18 = 0, x+1
5 = 11 = 1 ;
y−1
x−2
g) x + 2y − 4 = 0, 1 = 2 , z = 0;
y−2
z−1
h) x + y − 4z = 0, x−2
1 = 1 = −4 .
x−6a
208. x + 6y + 36z − 2706a = 0, 1 = y−18a
= z−72a
6
36
y−1
z+1
209. x − 2y + 3z + 6 = 0, x+1
1 = −2 = 3
z−1
210. 2y − z − 1 = 0, y−1
2 = −1 , x − 1 = 0
√
√
x− π
z−2
2 +1
√ 2
211. x + y + 2z − π2 − 4 = 0,
= y−1
1
1 =
2
212.
214.
216.
218.
220.
y−1
z−π
x − y − 4z + 4π = 0, x−1
−1 = 1 = 4
y−1
x−3
z−1
6x − y + z − 18 = 0, 6 = −1 = 1
y−1
z+1
x − 2y + z + 4 = 0, x+1
1 = −2 = 1
y−3
x−1
12x − 4y + 3z − 12 = 0, 12 = −4 = z−4
3
(0, 0, 0), (8, 0,q
8)
x−2
4
=
y+6
−12
=
z+3
−6 ;
213.
215.
217.
219.
221.
y
z−1
2x − y + 3z − 5 = 0, x−1
2 = −1 = 3
y−13
x+1
2x + 3y + 6z − 37 = 0, 2 = 3 =
y+4
4x + 3y = 0, x−3
4 = 3 , z+5=0
y−1
x−1
x − y = 0, 1 = −1 , z − 1 = 0
(0, 0, 0), (8, −2, 14)
4x − 2y − 3z − 3 = 0
2x + y − z − 2 = 0
24, 8
5
222.
224.
226.
228.
230.
232.
234.
x − y + 2z ± 11
2 =0
6x
√+ 4y + z ± 21 = 0
7 3 − 2, 5
6
−4
(2x − 6y − 10)i + (2y − 6x − 2)j
√ 21 2 (xi + yj)
223.
225.
227.
229.
231.
233.
235.
236.
1
(2xi
1+(x2 +3y 2 +z 3 )2
4
49 (−2, 3)
237.
239.
241.
243.
245.
247.
249.
238.
240.
242.
244.
246.
248.
x +y
(0, −1, 2)
15i + 9j − 3k
10i + 8j
ϕ = π2
(0, 1), (2, 1))
+ 6yj + 3z 2 k)
z
6
√52
38
(2x − 2y + 4)i + (2y − 2x − 8)j
ex+y+z [(1 + yz)i + (1 + xz)j + (1 + xy)k]
(2x − y − z)i + (4y − x + z)j + (6z − x + y)k
(0, − 21 )
1
2
3i + 3j
7i−4j−4k
81
7i − j + 13k
ϕ=0
ϕ = π2
250. ( 34 , − 13 ), (− 34 , 37 )
252.
254.
256.
258.
260.
261.
2 3
3
2 2
2y z + 6xyz − 3xy z
z(y + 2) + 3
3
42
(−2xyz 3 − 9xy 2 z 2 )i + (6xy 2 z 2 − y 2 z 3 )j + (3y 2 z 3 − 4xyz 3 )k
− √ 2 1 2 2 3 [(x2 + y 2 + xy)i + (y 2 + z 2 + zy)j + (x2 + z 2 + xz)k]
251. x(1 + y)
253. − √ 2 1 2
(x +y +z 2 )3
(xy + yz + xz)
255. 0
257. −1
259. 0
(x +y +z )
262. (−3y 2 + 16z)i + (5 − 9x2 )j + (2x − 3)k
263. − 11
3 i−k
265. i + j − (sin 2 sin 1)k
264. i + 2j + k
266. −2i + 16j − 18k
Chapter 1. (Indefinite integral)
Mission
Explain students concepts as antiderivative and indefinite integral, the fundamental formulas for the integration of real function of one real variable and
teach students various methods and techniques of integration.
Objectives
1. Understand the concepts of antiderivative and indefinite integral, learn the basic formula for integrating and be able to apply them to solve simple
problems.
2. Learning to use both, the substitution method and method by parts as two basic methods used for finding any antiderivative.
3. Being able to integrate all types of so called partial fractions and then manage the integration of rational functions of fraction decomposition method
into partial fractions.
4. Learn the methods of integrating certain types functions containing square root (or higher) and finally, trigonometric functions.
Prerequisite knowledge
differential calculus of the function of one real variable, linear algebra, algebraic equations
Chapter 1. (Indefinite integral) – 1
Introduction
In this chapter we first introduce the notion of antiderivative and indefinite integral and basic formulas and rules for calculating the antiderivative or
indefinite integral, respectively. Then we will discuss next methods for finding the antiderivative for some of the more complex functions. This process is
also called as integration process. It is the reverse task to finding the derivative of the function. Integration is much more difficult to perform than the
process of differentiation and there are functions that can not be integrated naturally. Mastery learning content of this chapter is the key to all other
chapters that follow behind her. The problem of finding antiderivatives to the function is a leader for mostly of physician problems. E.g. if we fix some
coordinates of a moving point, the speed we know, among other things, we must establish a primitive function to function, which indicates the speed
at each time point. With many other problems where it is necessary to determine the antiderivative we will meet in next three chapters.
Indefinite integral – Antiderivative, indefinite integral, basic rules
Basic rules for the process of integration is possible to get simply using the base rules for process of differentiation.
A. We say that a functionF (x) on the interval (a, b) is an antiderivative to the function f (x), if for each x ∈ (a, b) holds that
F 0 (x) = f (x).
Each function F (x) + C, where C is any real number is an antiderivative to f (x) too.
R
B. The set of all antiderivatives to f (x) is called as indefinite integral of the function f (x) which is denoted as f (x) dx. Thus
Z
f (x) dx = F (x) + C,
where C is any real number, which is denoted as constant of integration.
C. Basic rules
R
α+1
a) xα dx = xα+1 + C (α real, α 6= −1),
R
b) x1 dx = ln |x| + C,
R
x
c) ax dx = lna a + C,
R
d) ex dx = ex + C,
R
e) sin x dx = − cos x + C,
f)
R
cos x dx = sin x + C,
g)
R
1
cos2 x
dx = tg x + C,
h)
R
1
sin2 x
dx = − cotg x + C,
i)
R
√ 1
a2 −x2
j)
R
1
a2 +x2
k)
R
√ 1
x2 +a
l)
R
f 0 (x)
f (x)
dx = arcsin xa + C,
arctg xa + C,
√
dx = ln |x + x2 + a| + C,
dx =
1
a
dx = ln |f (x)| + C.
Basic properties of the indefinite integral
Z
Z
Z
(f (x) ± g(x)) dx = f (x) dx ± g(x) dx,
Z
Z
kf (x) dx = k f (x) dx, (k is a constant).
Excercise 1. Let’s evaluate following indefinite integrals
√
Z
Z
3x x − x6 + x3 cos x
cos 2x
dx, b)
dx,
a)
x3
cos2 x sin2 x
Z
c)
√
1
dx.
4 − 4x2
Solution.
Z
Z
Z
cos 2x
cos2 x − sin2 x
cos2 x
sin2 x
a)
dx
=
dx
=
dx
−
dx =
cos2 x sin2 x
cos2 x sin2 x
cos2 x sin2 x
cos2 x sin2 x
Z
Z
1
1
dx = − cotg x − tg x + C,
=
dx −
2
cos2 x
sin x
√
√
Z
Z
Z 6
Z 3
3x x − x6 + x3 cos x
3x x
x
x cos x
b)
dx =
dx −
dx +
dx =
3
3
3
x
x
x
x3
Z
Z
Z
6
x4
3
− 32
= 3 x dx − x dx + cos x dx = − √ −
+ sin x + C,
4
x
Z
Z
c)
1
√
dx =
4 − 4x2
Z
1
1
p
dx =
2
4(1 − x2 )
Z
√
1
1
dx = arcsin x + C. 2
1 − x2
Problems
Using basic
determine following
indefinite integrals.
R rules
1
2
1.
3x + 2x + 1 + 3x dx.
R (1+√x)2
√
3.
dx.
x
R √x4 +4x2 +4
5.
dx.
3
R x x e−x 7.
e 1 − x2 dx.
R x x
9.
2 5 dx.
R 3.2x −2.8x
11.
dx.
6x
R (4x −5x )2
13.
dx.
R 220x
15. R tg x dx.
2x
17. R coscos
2 x sin2 x dx.
19.
sin2 x dx.
R cos 22x
21.
dx.
R cosx2x−sin x
23.
2 dx.
R 1+x
4−x2
25.
2 dx.
R 1+x
x5 −16x+2
27.
dx.
x2 +4
Using rule
determine next indefinite integrals.
R No.12
x
29.
dx.
R x2 −32x−5
31.
2 −15x+22 dx.
R 3xsin
2x
33.
2 x+3 dx.
R 2 cos
1
35.
dx.
x ln x
11x9 +12
dx. x4
1
1
√ − √
dx.
4 3
x
x
2x(x2 − 5 + x32 ) dx.
−x
ax 1 − ax4 dx.
x
x 2
2.
R
4.
R
6.
R
8.
R
R
10.
(2 + 3 ) dx.
R (3x −4x )2
12.
dx.
x
√
R 12
14.
5 cos x − 3x5 +
R
16. R cotg2 x dx.
18. R cos2 x1sin2 x dx.
20.
cos2 x dx.
R 1+cos22 x
dx.
22.
2x
R 1+cos
x2 −1
24.
dx.
R x2x+1
4
26.
2 dx.
R 1+x
(1−x)2
dx.
28.
x2
3
1+x2
dx.
R x2
30.
dx.
R x3 +1
32.
(tg x + cotg x) dx.
R e2x
34.
dx.
R 1−3e2x1
√
dx.
36.
1−x2 arcsin x
Indefinite integral – Integration by substitution
Rarely is satisfactory to use the basic formulas for integration only. Even small deviations that change sign, and the constant into the formulas of the
functions leads to the necessity of introducing substitution. This method is very important and very often used in the integration process. Accurate
application and use of substitution in indefinite integral are described by following two basic rules.
A. The 1st rule of substitution ϕ(x) = z.
An indefinite integral
Z
f (ϕ(x))ϕ0 (x) dx
(1)
is evaluated formally using a new variable z by substitution ϕ(x) = z and ϕ0 (x) dx = dz and replacing into origin problem. We get
Z
Z
0
f (ϕ(x))ϕ (x) dx = f (z) dz.
Finally we evaluated the indefinite integral on the right hand side and replace again z = ϕ(x).
Excercise 2. Let’s evaluate following indefinite integrals
Z √
arctg x
dx,
a)
1 + x2
Z
b)
cos x
dx.
3 + sin x
Solution.
a) The problem is according to 1st rule, where ϕ(x) = arctg x. So we use the substitution arctg x = z;
Z √
arctg x
dx =
1 + x2
Z
√
3
z dz =
z2
3
2
+C =
1
1+x2
dx = dz and we get
2p
arctg3 x + C.
3
b) The nominator is a derivatives of the denominator. Using last basic rule (No.12) we have
Z
cos x
dx = ln |3 + sin x| + C. 3 + sin x
B. The 2nd rule of substitution x = ϕ(z).
An indefinite integral
Z
f (x) dx
is evaluated formally using substitution x = ϕ(z) and dx = ϕ0 (z)dz. After replacing into origin problem we have
Z
Z
f (x) dx = f (ϕ(z))ϕ0 (z) dz.
We evaluate the indefinite integral on the right hand side and finally is necessary express z using x, z = g(x) and put retrospectively.
Excercise 3. Let’s evaluate indefinite integral
Z
x2
Solution. Using substitution x =
Z
1
t
√
1
dx.
1 + x2
and dx = − t12 dt we get
dx
√
=
2
x 1 + x2
Z
− t12 dt
q
=−
1
1
1
+
t2
t2
Z
√
=−
dt
q
Z
=−
t2 +1
t2
√
√
t dt
= − t2 + 1 + C =
t2 + 1
1 + x2
+ C. x
Problems
Using
37.
39.
41.
43.
45.
47.
49.
51.
methods
of substitution determine following indefinite integrals.
R
9
R (x1 + 2) dx.
dx.
R 5−x2
dx.
R (x+3)2 2
11
R x(x x− 4) dx.
) dx.
3
R cos(
1
2 +1 dx.
R 9x−x
R e√ dx.
ex dx.
38.
40.
42.
44.
46.
48.
50.
52.
R
R
R
R
R
R
R
R
√
x + 3 dx.
5
dx.
6x−1
2
dx.
(2x+3)4
sin(2x) dx.
tg(4x − 1) dx.
√ 1
dx.
1−4x2
2x
e dx.
3
x2 ex dx.
53.
55.
57.
59.
61.
R
R
R
R
R
R
63.
R
65.
x2
dx.
x6 +4
2
2x
dx.
cos2 (x3 +1)
sin
x
√
4 cos x dx.
√1+x dx.
1−x2
arcsin
√ x−x dx.
1−x2
1/x
e
dx.
x2
x
2
√
dx.
1−4x
54.
R
x2
(1−x)3
2
dx.
R
56.
sin x cos x dx.
R ln x−2
58.
dx.
x)2
R x(ln
1−2 sin x
dx.
60.
2x
R cos
tg x
dx.
62.
3
(cos x)
√
R ln
x
e√
64.
dx.
R x 1
√
dx.
66.
cos2 x tg x−1
Indefinite integral – Integration by parts
The method of integration by parts we obtain easily from the formula for differentiation of the product of two functions. Using this method is possible
find an antiderivatve of the some functions in a special format. Principle of the method is contained in the following theorem.
If functions u and v of the one real variable x have continuous derivatives u0 and v 0 on interval (a, b) then following formula holds on this interval:
Z
Z
0
uv dx = uv − u0 v dx.
(2)
The rule (2) is possible to use for finding the indefinite integral in some cases of product of two functions. One of them we
R choose as u and the other
as v 0 . The choice needs to be done so that we know to calculate v by integration of v 0 and moreover the new problem u0 v dx will be simpler than
the origin one. It’s dependent on own experience, which function is better
choose as u and v 0 , respectively. However, in some cases, we can see some
R
principles. If P (x) is a polynomial, then, for example by computing P (x) arcsin x dx we put u = arcsin x, v 0 = P (x). We can make the choice of
the
x by any another cyclometric function or logarithmic one. On the other side, for example, by computing
R replacing similarly, if we replace arcsin
0
P (x) sin x dx we choose u = P (x), v = sin x. We can make the choice of the replacing similarly, if we replace sin x by any another goniometric
function or exponential one.
Sometimes is necessary the formula (2) to reuse several times. It may be that we get back to the original integral. In this case, we have the integral
equation, from which is possible express the finding result.
Excercise 4. Let’s calculate following indefinite integrals
Z
Z
x
a) (2x − 1)e dx, b) (x2 + 3x − 3) cos x dx,
Z
c)
Z
ln(2x + 3) dx,
d)
ex cos x dx.
Solution.
R
a) We replace u = 2x − 1, v 0 = ex . Then u0 = 2, v = ex dx = ex . We get
Z
Z
x
x
(x − 1)e dx = (x − 1)e − 2ex dx = (x − 1)ex − 2ex + C = (x − 3)ex + C.
b) We replace u = x2 + 3x − 3, v 0 = cos x. Then u0 = 2x + 3, v = sin x and
Z
Z
2
2
(x + 3x − 3) cos x dx = (x + 3x − 3) sin x − (2x + 3) sin x dx.
This method we use again the second times. Now, we replace u = 2x + 3, v 0 = sin x. Then u0 = 2, v = − cos x and
Z
Z
2
2
(x + 3x − 3) cos x dx = (x + 3x − 3) sin x − (2x + 3)(− cos x) − −2 cos x dx =
= (x2 + 3x − 3) sin x + (2x + 3) cos x − 2 sin x + C.
2
c) We replace u = ln(2x + 3), v 0 = 1. Then u0 = 2x+3
, v = x and
Z
Z
ln(2x + 3) dx = x ln(2x + 3) −
Z
2x
dx =
2x + 3
Z
2x
dx.
2x + 3
Z Z
2x + 3 − 3
3
1
dx =
1−
dx = x − 3
dx =
2x + 3
2x + 3
2x + 3
Z
3
2
3
=x−
dx = x − ln |2x + 3| + C.
2
2x + 3
2
After replacing we have
Z
ln(2x + 3) dx = x ln(2x + 3) − x +
3
ln |2x + 3| + C.
2
d) We replace u = ex , v 0 = cos x. Then u0 = ex , v = sin x and
Z
Z
x
x
e cos dx = e sin x − ex sin dx.
This method we use again the second times. Now, we replace u = ex , v 0 = sin x. Then u0 = ex , v = − cos x and
Z
Z
x
x
x
x
e cos x dx = e sin x − e (− cos x) − e (− cos x) dx =
x
x
Z
= e sin x + e cos x −
ex cos x dx.
On the right side of the last equation we have the origin problem again. From this is possible to express finding indefinite integral:
Z
Z
1
x
x
x
2 e cos x dx = e sin x + e cos x,
ex cos x dx = ex (sin x + cos x) + C. 2
Problems
Using
67.
69.
71.
73.
75.
77.
79.
81.
83.
85.
87.
89.
91.
93.
95.
97.
99.
101.
103.
method
by parts determine following indefinite integrals.
R
x
ln
x
dx.
R
−x
R x e2x dx.
R x e dx.
R x arctg x dx.
x dx.
R x sin
x
R cos2 x dx.x
1) e dx.
R (x2 +−x
R x e 2dx.
R 4x2 cos x dx.
x
R (x3 + 2x) e dx.
R x sin x dx.
R ln x dx.
x dx.
R arcsin
2
ln
x
dx.
R
ln(x2 + 3) dx.
R xarcsin
x
dx.
R e2x
e
sin
x
dx.
R 3x
R e xsin 2x dx.
√
arcsin x dx.
1−x2
68.
70.
72.
74.
76.
78.
80.
82.
84.
86.
88.
90.
92.
94.
96.
98.
100.
102.
104.
R
x
R x e x dx.
R x 3 2 dx.
R x ln2 x dx.
R (9x + 4x) ln x dx.
R x cos2 2x dx.
R x tg x dx.
R (x2 − 3) sin x dx.
R x 2sin 2x dx.
R (x2 − 2) cos x dx.
R (x3 + 6x + 3) cos 2x dx.
R x arctg x dx.
R arctg x dx.
x)2 dx.
R (arcsin
2
R ln(x + 1) dx.
x) dx.
R cos(ln
x
ex dx.
R ex arctg
2
R e2xsin x dx.
e cos 5x dx.
R x2
arctg x dx.
1+x2
105.
R
x
x cotg
dx.
sin2 x
106.
R
ln3 x
x2
dx.
Indefinite integral – Integration of the partial fractions
Integration of the so called partial fractions is the preparatory phase of the solving of a much more complex problem, such as the integration of rational
functions.
As a partial fraction we call the rational function of the form
Mx + N
A
or
,
(x − α)n
(x2 + px + q)n
where n is a natural number, A, α, M, N, p, q are any real number and the discriminant of the quadratic trinomial is negative (D < 0).
A
A. Integration of the partial fraction (x−α)
n.
a) If n = 1, then
Z
Excercise 5. Let’s solve
R
5
x−7
Z
R
A
(x−α)n
Z
1
dx = A ln |x − α| + C.
x−α
dx.
Solution.
b) If n > 1, then
A
dx = A
x−α
5
dx = 5
x−7
Z
1
dx = 5 ln |x − 7| + C. x−7
dx we calculate by replacing x − α = z.
Excercise 6. Let’s solve
R
4
(x+5)7
dx.
Solution. We introduce the substitution x + 5 = z, dx = dz and we get
Z
Z
Z
4
4
z −6
−2
−2
−7
dx
=
dz
=
4
z
dz
=
4
+
C
=
+
C
=
+ C. (x + 5)7
z7
−6
3z 6
3(x + 5)6
B. Integration of the partial fraction
Mx + N
.
+ px + q)n
(3)
(x2
a) For n = 1 we have
R
M x+N
x2 +px+q
dx. This integral (3) is possible calculate by following way:
1) The trinomial x2 + px + q modify to the form of full square, x2 + px + q = (x + p2 )2 − ( p2 )2 + q.
2) Next we introduce the substitution x +
p
2
=z
3) After simplifying the integral is possible to divide into two integrals, which each of them can be calculated using
knowing formulas or methods.
Excercise 7. Let’s calculate I =
R
4x−9
x2 −6x+13
dx.
Solution. D = 36 − 52 = −16 < 0.
Z
I=
4x − 9
dx =
2
x − 6x + 13
Z
4x − 9
dx.
(x − 3)2 − 9 + 13
After substitution x − 3 = z, x = z + 3, dx = dz we have
Z
Z
Z
Z
4(z + 3) − 9
4z + 3
2z
1
I=
dz =
dz = 2
dz + 3
dz.
2
2
2
2
z +4
z +4
z +4
z +4
1
z
3
x−3
I = 2 ln(z 2 + 4) − 3 √ arctg √ + C = 2 ln(x2 − 6x + 13) − arctg
+ C. 2
2
4
4
b) If n > 1, is possible to apply a similar procedure as in case a) but in addition we need next recurrent formula
Z
1
1
x
2n − 3
In =
dx = 2
+
In−1 .
(x2 + a2 )n
a 2(n − 1)(x2 + a2 )n−1 2n − 2
Excercise 8. Let’s calculate I3 =
R
1
(x2 +4)3
(4)
dx.
Solution. Using last recurrent formula (4) we have
1
x
6−3
x
3
I3 =
+
I
=
+
I2 .
2
4 2(3 − 1)(x2 + 4)3−1 6 − 2
16(x2 + 4)2 16
I2 we calculate again using (4).
x
x
1
1
4−3
I1 =
+ I1 .
I2 =
+
2
2−1
2
4 2(2 − 1)(x + 4)
4−2
8(x + 4) 8
Z
1
1
x
I1 =
dx = arctg .
2
x +4
2
2
After replacing to I1 and I2 we get
x
3
x
11
x
I3 =
+
+
arctg
+C =
16(x2 + 4)2 16 8(x2 + 4) 8 2
2
=
Solution.
R
2x−5
(x2 +2x+2)2
3x
3
x
x
+
+
arctg
+ C. 16(x2 + 4)2 128(x2 + 4) 256
2
dx.
Z
2x − 5
2x − 5
I=
dx =
dx.
2
2
[(x + 1) − 1 + 2]
[(x + 1)2 + 1]2
We introduce the substitution x + 1 = z, x = z − 1, dx = dz. Next:
Z
Z
Z
Z
2z − 7
2z dz
dz
2(z − 1) − 5
dz =
dz =
dz − 7
.
I=
2
2
2
2
2
2
2
(z + 1)
(z + 1)
(z + 1)
(z + 1)2
Z
The first integral we calculate by substitution z 2 + 1 = t, 2zdz = dt.
Z
Z
2z
dt
1
1
1
dz =
=− =− 2
=− 2
.
2
2
2
(z + 1)
t
t
z +1
x + 2x + 2
The second integral we calculate using the recurrent formula (4) and we get
Z
x+1
1
1
dz
=
+
arctg(x + 1).
(z 2 + 1)2
2(x2 + 2x + 2) 2
After re-substituting we get
x+1
1
1
I=− 2
−7
+ arctg(x + 1) + C =
x + 2x + 2
2(x2 + 2x + 2) 2
7x + 9
7
=− 2
− arctg(x + 1) + C. x +x+1 2
Problems
Determine
R 2following indefinite integrals of the partial fractions.
107.
dx.
R x+7
1
109.
dx.
R 2x−1
4
111.
2 dx.
R (4x−3)
1
113.
dx.
R x2 −2x+5
1
115.
dx.
R 2x2 +2x+5
2x
dx.
117.
R x2 +6x+10
12x
119.
2 +x+6 dx.
R xx+2
121.
dx.
+9
R x210x+2
123.
dx.
R x2 −4x+5
8
125.
dx.
(x2 +1)2
108.
110.
112.
114.
116.
118.
120.
122.
124.
126.
R
R
R
R
R
R
R
R
R
R
5
dx.
3−2x
−8
dx.
3x+2
1
dx.
4(2x+1)3
1
dx.
x2 +10x+34
1
dx.
5x2 +2x+10
−x
dx.
2x2 −2x+1
3x
dx.
x2 +2x+10
2x−1
dx.
x2 +4
2−x
dx.
x2 −2x+5
2x+6
dx.
(x2 +9)2
Indefinite integral – Integration of rational functions
Let us now solve fully the general problem of integrating of the rational functions. In this step is sufficient when we restrict the general problem to the
problem of integrating so called simple rational functions, where the degree of the numerator is less than the degree of the denominator. Now follows
a fundamental point, which is the decomposition of simple rational functions of the sum of partial fractions. The whole procedure is described in the
following steps.
As a rational function f (x) is called the ratio of two polynomials Pm (x) and Pn (x)
f (x) =
Pm (x)
,
Pn (x)
(5)
where m and n are their degrees.
If m < n, then the the rational function f (x) is called as a simple rational function. Now we specify the method as of a rational function, which
is not simple rational is possible get a simple rational and next how is possible the simple rational function decompose to the sum of partial fractions.
Let’s consider a rational function f (x) that m ≥ n. In this case is necessary to divide the numerator by denominator and we get
Pm (x)
R(x)
= Q(x) +
,
Pn (x)
Pn (x)
(6)
where Q(x) is a polynomial and R(x) is a rest polynomial after dividing. Then polynomial R(x) has its degree less n, it means deg(R(x)) < n. Thus
R(x)
is a simple rational function. In the next step by decomposition of the simple rational function PR(x)
to the sum of partial fractions we proceed as
Pn (x)
n (x)
follows:
a) We find all real roots of the denominator Pn (x) and decompose it on the product of the numbers an (an of the coefficient at the
highest power of the polynomial Pn (x)) and next root factors of the real roots of a quadratic expressions whose discriminant is
negative (D < 0), (e.g. these quadratic expressions do not have real roots).
Pn (x) = an . . . (x − α)k . . . (x2 + px + q)l . . .
Note that these different factors are the denominators of partial fractions.
b) An expression (x − α)k corresponds to k of the partial fractions in form:
A2
Ak
A1
,
,
.
.
.
,
.
x − α (x − α)2
(x − α)k
An expression (x2 + px + q)l corresponds to l of the partial fractions in form:
M2 x + N2
Ml x + Nl
M1 x + N1
, 2
,..., 2
.
2
2
x + px + q (x + px + q)
(x + px + q)l
c) Then the rational function
R(x)
Pn (x)
is necessary to decompose to the sum of all partial fractions mentioned in previous step.
d) Obtained equality is multiplied by polynomial Pn (x) and we get the equality of two polynomials.
e) Then calculate coefficients Ai , Nj , Mj using one of the following methods:
a) method of comparing coefficients at the same powers,
b) substitute of (real) roots of the denominator,
c) combined method.
f) In the last step is possible to integrate the decomposed rational function at the one of the two form (5) or (6).
Z 2
x2 − 2x + 3
x − 8x − 2
a) I =
dx, b) I =
dx,
2
x +x−2
x3 − 3x + 2
Z 5
x + x4 + 3x3 + x2 − 2
dx.
c) I =
x4 − 1
Z
Solution.
a) (x2 − 2x + 3) : (x2 + x − 2) = 1 +
−(x2 +x−2)
−3x+5
−3x+5
x2 +x−2
Z
I=
Function
−3x+5
x2 +x−2
Z
1 dx +
−3x + 5
dx.
x2 + x − 2
is a simple rational.
1. D = 9 > 0, so the denominator has two real roots. We get x2 + x − 2 = (x − 1)(x + 2).
2. An expression x − 1 corresponds to one partial fraction of the form
B
.
to another partial fraction of the form x+2
A
.
x−1
Similarly, an expression x + 2 corresponds
3. we decompose the rational function to sum of two partial fractions:
A
B
−3x + 5
=
+
.
x2 + x − 2
x−1 x+2
4. The equality is multiplied by x2 + x − 2 and we get −3x + 5 = A(x + 2) + B(x − 1).
5. Both coefficients, A and B we can calculate for example substitute of real roots of the denominator.
x=1:
−3 + 5 = A(1 + 2) + B(1 − 1),
A = 23 ,
.
x = −2 : −3(−2) + 5 = A(−2 + 2) + B(−2 − 1), B = − 11
3
6.
Z I =x+
2
3
− 11
3
+
x−1 x+2
dx = x +
2
11
ln |x − 1| −
ln |x + 2| + C.
3
3
b) This is an integration of the simple rational function.
1. At first, we decompose the denominator. It is easy to find one root x1 = 1. Next we divide this polynomial by
term (x − 1). We have
(x3 − 3x + 2) : (x − 1) = x2 + x − 2,
x2 + x − 2 = (x − 1)(x + 2).
Then
x3 − 3x + 2 = (x − 1)(x − 1)(x + 2) = (x − 1)2 (x + 2).
2. An expression (x−1)2 correspond two partial fractions of the forms
B
one fraction of the form x+2
.
A1
, A2
x−1 (x−1)2
and an expression x+2 correspond
3. Fully decomposition is possible write in this form:
A1
A2
B
x2 − 8x − 2
=
+
+
.
3
2
x − 3x + 2
x − 1 (x − 1)
x+2
4. Now, the last equality is multiplied by x3 − 3x + 2 and we get
x2 − 8x − 2 = A1 (x − 1)(x + 2) + A2 (x + 2) + B(x − 1)2 .
5. To calculate all coefficients A1 , A2 and B we use the combined method. First we use the method of substitute
x=1:
12 − 8 − 2 = A2 (1 + 2),
A2 = −3,
of real roots of the denominator. We have
2
2
x = −2 : (−2) − 8(−2) − 2 = B(−2, −1) , B = 2.
To calculate all coefficients we use the method of comparing coefficients at the same powers. The right side we
modify and next compare coefficients at these powers.
x2 : 1 = A1 + B,
1 = A1 + 2, A1 = −1.
Z
Z
Z
−1
−3
2
I=
dx +
dx +
dx.
2
x−1
(x − 1)
x+2
To the middle integral we introduce the substitution x − 1 = z, dx = dz and we get
Z
Z
−3
1
1
3
3
dx
=
−3
dz
=
−3
−
=
=
.
(x − 1)2
z2
z
z
x−1
6. Finally,
I = − ln |x − 1| +
3
(x + 2)2
3
+ 2 ln |x + 2| + C = ln
+
+ C.
x−1
|x − 1|
x−1
c) The degree of nominator is greater then denominator, so is necessary to divide.
3x3 + x2 + x − 1
.
(x + x + 3x + x − 2) : (x − 1) = x + 1 +
x4 − 1
Z
Z
Z
3x3 + x2 + x − 1
I = x dx + dx +
dx.
x4 − 1
5
4
3
2
4
1. x4 − 1 = (x2 − 1)(x2 + 1) = (x − 1)(x + 1)(x2 + 1).
2. Both expressions, x − 1 and x + 1 correspond partial fractions
.
the partial fraction Mx2x+N
+1
A
x−1
and
B
.
x+1
Quadratic term x2 + 1 correspond
3. Final form of decomposition is:
3x3 + x2 + x − 1
A
B
Mx + N
=
+
+ 2
.
4
x −1
x−1 x+1
x +1
4. Last equality is multiplied by x4 − 1.
3x3 + x2 + x − 1 = A(x + 1)(x2 + 1) + B(x − 1)(x2 + 1) + (M x + N )(x2 − 1).
5. Coefficients A, B, M and N is possible to compute using combined method. First, we replace the roots of
denominator.
x=1:
3 · 1 + 1 + 1 − 1 = A(1 + 1)(1 + 1),
A = 1,
x = −1 : 3 · (−1)3 + (−1)2 − 1 − 1 = B(−1 − 1)((−1)2 + 1), B = 1.
Compare the coefficients M a N in the powers, where are found, for exaple by x3 and x0 .
x3 :
3 = A + B + M,
3 = 1 + 1 + M, M = 1,
0
x : −1 = A − B − N, −1 = 1 − 1 − N, N = −1.
6. Calculated coefficients substitute and complete calculation
Z
Z
Z
x2
1
1
x+1
+x+
dx +
dx +
dx.
I=
2
x−1
x+1
x2 + 1
Z
x+1
dx =
x2 + 1
I=
Z
x
dx +
x2 + 1
Z
1
1
dx
=
ln(x2 + 1) + arctg x.
x2 + 1
2
x2
1
+ x + ln |x − 1| + ln |x + 1| + ln(x2 + 1) + arctg x + C =
2
2
2
√
x
+ x + ln |x2 − 1| x2 + 1 + arctg x + C. =
2
Problems
Make a decomposition to sum of partial fractions.
2x
x−5
127. (x−1)(x−2)
.
128. x(x−2)(x+2)
.
2
2x+11
129. (x−1)2 .
130. x24−x
.
(x−3)
7x−3
12
131. x(x2 +1) .
132. (x−5)(x2 +25) .
−6
133. x2 (x2 +4) .
134. (x−1)(x12 +2x+5) .
135. (x2 +4)(xx2 −8x+17) .
136. x(x23+1)2 .
In following
problems calculate integrals of rational functions. The denominator contains
R 3x−4
R 2 only different real roots.
137.
dx.
138.
dx.
−2x
−1
R x2x+13
R x23x−5
139.
dx.
140.
dx.
R x2 +x−6
R x2 −3x+2
2
5x
141.
142.
dx.
2 +2x dx.
R 2x2 −3x−2
R x3 +3x
84
x
144.
dx.
143.
dx.
2
R (x−1)(x−4)(x+3)
R 2x +3x+1
x+3
6x2
dx.
146.
145.
3
2 +4 dx.
R xx42 −5x
R xx2 −x
−5x+9
+2x−25
147.
dx.
148.
dx.
2 +5x
R xx23 −5x+6
R x3x−7x−10
−x2 −4x+12
149.
dx.
150.
2 −2x−3 dx.
R 2x3 x2 −4
R x−3x
4
151.
152.
2 −1 dx.
2 +x−2 dx.
x
x
R 5x3 −15x2 +15x−3
R x3 −1
153.
dx.
154.
3
2
3 −x dx.
+17x−10
R xx3−8x
R 4x
−1
x5 +x4 −8
155.
dx.
156.
dx.
9x3 −x
x3 −4x
The denominator
has
only
real
roots,
some
are
multiple.
R 2x+3
R 3x2 +2
157.
dx.
158.
2 dx.
x2 +2x+1
R x2 +4
R xx32−2x
−3x+2
159.
160.
dx.
2 +4x dx.
R x3 +4xx−1
R x3 +2xx22+x
161.
dx.
162.
dx.
x3 +5x2 +8x+4
x3 +5x2 +8x+4
163.
R
4x2 +1
dx.
4x3 +4x2 +x
2
x −8x+6
dx.
x4 −x3
x3 +4
dx.
x4 −4x3 +4x2
3
x +1
dx.
x3 −x2
x4 +8
dx.
x4 −4x2
4x4 −27x2 +27x+9
4x3 −12x2 +9x
4x3 −4x+4
dx.
x3 −x2 −x+1
164.
R
R
R
166.
165.
R
R
167.
168.
R
R
170.
169.
R
R
172.
171.
R
R
173.
dx.
174.
R
R
175.
176.
The denominator
has varios roots, some are complex numbers.
R 1
R
dx.
177.
178.
R
R x31+x
180.
179.
dx.
3
+1
R x 8x
R
181.
dx.
182.
4
R x −16
R
4x−12
183.
dx.
184.
3
2
R
R x +xx2 +3x−5
−13
186.
185.
3 −3x2 +7x−5 dx.
x
R
R
4x2
188.
187.
4 +10x2 +9 dx.
x
R x2 +3x+2
R
189.
dx.
190.
2
R
R xx3 +2x+2
−4x2 +14x−13
dx.
192.
191.
3 −4x2 +13x
x
R x3 −x2 +10
R
193.
194.
3 −x2 −7x+15 dx.
x
R
R
x4 +1
195.
dx.
196.
3
2
x −x +x−1
The denominator
has various roots, some are multiple complex numbers.
R
R
12
197.
dx.
198.
2
2
R
R x(x +1) 6
200.
199.
2 −4x+5)2 dx.
(x−2)(x
R
R
x
201.
202.
2 +3x+3)2 dx.
(x
R
R
4x
203.
dx.
204.
2
2
)
R (x+1)(1+x
R
1
205.
dx.
206.
(x−1)(x2 −2x+3)2
x2 −2x+3
dx.
(x−1)(x3 −4x2 +3x)
2
3x +1
dx.
(x2 −1)2
2x3 +x+2
dx.
x5 +2x4
3
x +x+2
dx.
x3 −2x2 +x
x3 −28
dx.
x3 +3x2 −4
3x3 −x2 +1
dx.
3x3 −x2
3
2
x −10x +38x−50
dx.
x3 −10x2 +25x
−x2
dx.
x4 −1
x
dx.
x3 −1
18
dx.
x3 +9x
20
dx.
x3 −x2 +4x−4
3x+6
dx.
x4 +5x2 +4
x3 −16
dx.
x4 +16x2
4
x
dx.
x2 +3
4
x −2x3 +11x2 +50
dx.
x4 −2x3 +10x2
x4 −3x3 +20x2 −x−18
dx.
x4 +5x2 −36
5
3
x +2x +4x+4
dx.
x4 +2x3 +2x2
x4 −x3 +5x2 −3x
dx.
(x+2)(x2 +1)2
6x+1
dx.
(x2 +1)2
x+1
dx.
x4 +4x2 +4
4
3
x +3x
dx.
(x−1)(x2 +1)2
1
dx.
x4 (x3 +1)2
Indefinite integral – Integration of irrational functions
In this section we describe the procedures for integrating certain functions containing roots of linear, respectively. quadratic expressions.
√
R
A. Integral of the type
R(x, n ax + b) dx
This type of integral is possible modify using substitution ax + b = z n to integral of rational function.
Z
I=
√
2x − 3
p
dx.
√
4
3 2x − 3 + 4 (2x − 3)3
2x +
Solution. In this case n = 4. So we put substitution 2x − 3 = z 4 , x =
z 4 +3
,
2
dx = 12 4z 3 dz = 2z 3 dz.
(z 4 + 3 + z 2 )2z 3
dz = 2
3z + z 3
Z
I=
Z
z 6 + z 4 + 3z 2
dz.
3 + z2
Now we have known integral of rational function. It is necessary divide polynomials in this fraction. We get
z 6 + z 4 + 3z 2
27
= z 4 − 2z 2 + 9 −
.
2
3+z
3 + z2
z5
z3
1
z
− 2 + 9z − 27 · √ arctg √ + C =
5
3
3
3
p
√
4
4
√
(2x − 3)5 2 p
27
2x − 3
4
4
3
√
−
=
(2x − 3) + 9 2x − 3 −
arctg √
+ C. 5
3
3
3
q
R
) dx
R(x, n ax+b
cx+d
I=
B. Integral of the type
Similarly, is possible modify this type using substitution
ax+b
cx+d
= z n again to integral of rational function.
Z
I=
Solution. In this case n = 2. So we put substitution
Z
I=
x−2
x
1
x
r
x−2
dx.
x
= z 2 , x − 2 = xz 2 , x(1 − z 2 ) = 2, x =
2
,
1−z 2
dx =
4z
(1−z 2 )2
dz. Then
Z
Z 2
1 − z2
4z
z2
z −1+1
z
dz = 2
dz = 2
dz =
2
2
2
2
(1 − z )
1−z
1 − z2
Z
Z 1
1
=2
−1 +
dz
=
−2z
+
2
dz.
1 − z2
1 − z2
Due to 1 − z 2 = (1 − z)(1 + z), the function
1
1−z 2
is necessary decompose to partial fractions.
1
B
A
+
,
=
2
1−z
1−z 1+z
1 = A(1 + z) + B(1 − z).
x=1:
1 = A(1 + 1), A = 21 ,
x = −1 : 1 = B(1 + 1), B = 21 .
Z
Z
Z
1
1
1
1
1
1+z
1
2
2
dz
=
dz
+
dz = − ln |1 − z| + ln |1 + z| = ln |
|.
2
1−z
1−z
1+z
2
2
2
1−z
1+z
I = −2z + ln |
| + C.
1−z
q
Finally, we come back to result origin variable, it means, replace z = x−2
. x
R
Dx+E
C. Integral of the type √Ax2 +Bx+C dx
This type of integral is possible calculate following way:
a) Ax2 + Bx + C modify by full square method.
b) The term in brackets replace by z.
c) Function modify and divide the integral into two integrals, which have some form of type (α) − (δ).
Next calculations:
Z
Z
Z
1
x
1
√
√
√
dx,
(α)
dx, (β)
dx, (γ)
2
2
2
2
2
a −x
a −x
x +a
The integral (α) calculate by substitution x = az. After editing we get
Z
1
x
√
dx = arcsin .
2
2
a
a −x
Z
(δ)
√
x
dx.
+a
x2
Both integrals, (β) and (δ) calculate by substitution a2 − x2 = z 2 and x2 + a = z 2 . Integral (γ) we calculate using basic rule (No. 11).
Z
a) I =
x−5
√
dx,
x2 + 10x
Z
b) I =
√
x+5
dx.
6 − 2x − x2
Solution.
a) term x2 + 10x we modify to full square term and next use substitution
x2 + 10x = (x + 5)2 − 25,
x + 5 = z,
x = z − 5,
dx = dz.
We get
Z
I=
x−5
Z
z − 10
z−5−5
p
√
√
dz =
dz =
dx =
z 2 − 25
z 2 − 25
(x + 5)2 − 25
Z
Z
z
1
√
=
dz − 10 √
dz.
2
2
z − 25
z − 25
Z
The first integral is as (δ) and the second as (γ). Into first integral we put substitution z 2 − 25 = t2 , 2zdz = 2tdt and we have
Z
Z
√
z
tdt
√
dz =
= t = z 2 − 25.
t
z 2 − 25
Then
I=
√
z 2 − 25 − 10 ln |z +
√
z 2 − 25| + C =
√
x2 + 10x − 10 ln |x + 5 +
√
x2 + 10x| + C. b) term 6 − 2x − x2 we modify to full square term
Z
Z
Z
x+5
z−1+5
z+4
p
√
√
I=
dx =
dz =
dz =
7 − z2
7 − z2
7 − (x + 1)2
Z
Z
z
1
√
=
dz + 4 √
dz.
2
7−z
7 − z2
The first integral is as (β) and the second as (α). Into first integral we put substitution 7 − z 2 = t2 , −2zdz = 2tdt, zdz = −tdt
and we get
Z
Z
√
z
tdt
√
dz = −
= −t = − 7 − z 2 .
t
7 − z2
Finally
√
√
z
x+1
I = − 7 − z 2 + 4 arcsin √ + C = − 6 − 2x − x2 + 4 arcsin √ + C. 7
7
D. Method of undetermined coefficients
R
This method is possible calculate integral of type
Z
√
Pn (x)
ax2 +bx+c
dx, where Pn (x) is a polynomial of degree n ≥ 1. Then holds following equality:
Z
√
Pn (x)
1
2
√
dx = Qn−1 (x) ax + bx + c + k √
dx,
2
2
ax + bx + c
ax + bx + c
(7)
where Qn−1 (x) is a polynomial of degree n − 1 (with undetermined coefficients A, B, C, · · · and k as constant. These undetermined coefficients of the
polynomial
Qn−1 (x) and constant k is possible calculate by following process. The equality (7) derivative and obtained new equality multiply by term
√
2
ax + bx + c. We get equality of two polynomials. Compare coefficients by same powers.
The integral on the right-hand side of equality (7) calculate using process described at point C.
R √
Excercise 14. Let’s calculate I = x x2 + 2x + 2 dx.
√
Solution. This integral is not among the types listed above, but it can be adapted to such, if integrated function multiply and divide by x2 + 2x + 2.
We get
Z 3
Z
x + 2x2 + 2x
x(x2 + 2x + 2)
√
√
dx =
dx.
I=
x2 + 2x + 2
x2 + 2x + 2
Using (7) we have
Z 3
Z
√
x + 2x2 + 2x
1
2
2
√
dx = (Ax + Bx + C) x + 2x + 2 + k √
dx.
2
2
x + 2x + 2
x + 2x + 2
After derivative we get
√
x3 + 2x2 + 2x
2x + 2
√
= (2Ax + B) x2 + 2x + 2 + (Ax2 + Bx + C) √
+
x2 + 2x + 2
2 x2 + 2x + 2
1
+k √
.
2
x + 2x + 2
√
Multiplying by x2 + 2x + 2 we have
x3 + 2x2 + 2x = (2Ax + B)(x2 + 2x + 2) + (Ax2 + Bx + C)(x + 1) + k.
Compare
x3 :
x2 :
x1 :
x0 :
coefficients:
1 = 2A + A,
2 = B + 4A + B + A,
2 = 2B + 4A + C + B,
0 = 2B + C + k,
A = 31
B = 16
C = 16
k = − 12 .
Thus
1
1
1 √
1
I = ( x2 + x + ) x2 + 2x + 2 −
3
6
6
2
Z
1
dx.
+ 2x + 2
The integral on the right-hand include the type A. The trinomial x2 + 2x + 2 modify the full square, e.g. x2 + 2x + 2 = (x + 1)2 − 1 + 2 = (x + 1)2 + 1.
After substitution x + 1 = z, dx = dz we get
Z
Z
√
√
1
1
√
√
dx =
dz = ln |z + z 2 + 1| = ln |x + 1 + x2 + 2x + 2|.
x2 + 2x + 2
z2 + 1
Finally,
1
I=
3
Problems
√
R
A. R(x, n√ax + b) dx
R 1− x
√ dx.
207.
R 1+√xx
√
209.
3 x+1 dx.
√
R √x+
4 x+ √
3x
√
dx.
211.
6 7
x )
R 2(x+
√1
213.
dx.
R x x−4x
√
√
dx.
215.
x+1+ 3 x+1
q
R
B. R x, n ax+b
dx
cx+d
R q 1−x 1
217.
· dx.
1+x x
R q 1−x
219.
dx.
1+x
R
Dx+E
C. √Ax2 +Bx+C dx
R
1
√
221.
2 dx.
R 1−2x−x
x+3
√
223.
dx.
R x2 +2x
1
√
225.
dx.
R x2 +x+1
1
√
227.
dx.
5x2 +8x−3
R 2x−10
√
dx.
229.
1+x−x2
√
x2
√
1
1
1 √ 2
2
x + 2x + 2 − ln |x + 1 + x2 + 2x + 2| + C. x + x+
2
2
2
√
x−4
√ dx.
x+4−4
x
√
3x
√
dx.
6
x+√ x5
6 x+1
√
dx.
6 7 √
4
x + x5
x+1
√
dx.
3
3x+1√
3
x−1
√
√
2 x−1− 3 (x−1)2
208.
R
210.
R
212.
R
214.
R
216.
R
218.
R q 1+x
220.
R 1p
222.
224.
226.
228.
230.
R
R
R
R
R
dx.
1
1−x (1−x)(1+x)2
x
x
x−1
dx.
dx.
√ 3x+1
dx.
x2 +5x−10
1
√
dx.
12x−9x2 −2
1
√
dx.
9x2 −6x+2
1
√
dx.
3x2 −7x+5
8x−11
√
dx.
5+2x−x2
D.
R
231.
233.
√
Pn (x)
dx
ax
R 2 +bx+c
2
√2x −3x dx.
R x2 −2x+5
3
√ x
2 dx.
1−2x−x
R
2
√ x
dx.
R √1−2x−x2
x2 + 4x + 13
232.
R
234.
R
x2
dx.
x2 +2x+2
x3√
+5x2 +8x+3
dx.
2
√ x +4x+3
x2 + 4x + 3 dx.
√
R
236.
R√
2
238. R 1 − 2x −
√ 3x dx.
240.
(4x − 10) 2 + 3x − x2 dx.
235.
237. R √
dx.
239.
3 − 2x − x2 dx.
Indefinite integral – Integration of trigonometric functions
In the process of integration of trigonometric functions are often used substitution method, which usually prevent some modifications of the integral
functions using several well-known formulas valid for trigonometric functions and relations between them. In solving certain specific types of problems
can also be used some recurrent formulas.
R
R
A. sinn x dx, cosn x dx, n is integer
These types of integrals is possible to calculate using following recurrent formulas
Z
1
n−1
In = sinn x dx = − cos x sinn−1 x +
In−2 ,
(8)
n
n
Z
n−1
1
In−2 .
(9)
In = cosn x dx = sin x cosn−1 x +
n
n
R
R
R
Re-using these formulas is possible to reach the integral I0 or I1 (in case (8) I0 = sin0 x dx = dx = x, I1 = sin x dx = − cos x; in case (9)
analogically).
Z
a)
5
sin x dx,
Z
b)
cos4 x dx.
Solution.
a) We use the recurrent formula (8) for n = 5 and next again for n = 3
Z
1
1
4
1
4
2
5
4
4
2
I5 = sin x dx = − cos x sin x + I3 = − cos x sin x +
− cos x sin x + I1 =
5
5
5
5
3
3
Z
4
8
1
4
2
cos x sin x +
= − cos x sin x −
sin x dx =
5
15
15
4
8
1
cos x sin2 x −
cos x + C.
= − cos x sin4 x −
5
15
15
b) We use the recurrent formula (9) for n = 4 and next again for n = 2
1
3
1
3 1
1
3
3
I4 = sin x cos x + I2 = sin x cos x +
sin x cos x + I0 =
4
4
4
4 2
2
Z
3
3
1
3
3
1
3
cos0 x dx = sin x cos3 x + sin x cos x + x + C. = sin x cos x + sin x cos x +
4
8
8
4
8
8
Note that integrals of this type can be also calculate without the use of recurrent formulas.
a) if n is odd, the function modify and introduce appropriate substitution.
b) If n is even, we reduce the power of successively applying double angle formulas for both, sine and cosine functions
sin2 α =
1 − cos 2α
,
2
cos2 α =
1 + cos 2α
.
2
Let’s calculate again Excercise 15 using last methods.
a)
Z
5
sin x dx =
Z
Z
4
sin x sin x dx =
2
2
Z
(sin x) sin x dx =
(1 − cos2 x)2 sin x dx.
We use substitution cos x = z, − sin x dx = dz, sin x dx = −dz. Then
Z
Z
Z
z3 z5
5
2 2
sin x dx = − (1 − z ) dz = − (1 − 2z 2 + z 4 )dz = −(z − 2 + ) + C =
3
5
= − cos x +
2
1
cos3 x − cos5 x + C.
3
5
b)
Z
4
Z
cos x dx =
2
Z 2
(cos x) dx =
2
1
dx =
4
Z
(1 + 2 cos 2x + cos2 2x) dx =
Z
1
1
1
1 + cos 4x
cos 2x dx) = x + sin 2x +
dx =
4
4
4
2
1
1
1
1
3
1
1
= x + sin 2x +
x + sin 4x + C = x + sin 2x +
sin 4x + C. 4
4
8
4
8
4
32
1
= (x + sin 2x +
4
B. Integrals of types
There are two cases:
R
Z
1 + cos 2x
2
2
sinn x cosm x dx, n, m are integer
a) both integer numbers, n and m are even. Then is possible modify the function this way, that we get only powers of functions sin x
or cos x. These integrals are calculated using recurrent formulas.
R
Excercise 16. Let’s calculate I = sin4 x cos6 x dx.
Solution. It is preferable to modify the power of the smaller exponent, it means sin4 x.
Z
Z
Z
2
2
6
2
2
6
I = (sin x) cos x dx = (1 − cos x) cos x dx = (1 − 2 cos2 x + cos4 x) cos6 x dx =
Z
=
6
cos x dx − 2
Z
8
Z
cos x dx +
cos10 x dx.
Each of these integrals can be calculated using the recursive formula (9). b) At least one of the numbers n, m is odd. If n (m) is odd, after editing is possible use substitution cos x = z (sin x = z).
R
Excercise 17. Let’s calculate I = sin2 x cos5 x dx.
Solution. Now, it is possible to use similarly modification as in Problem 15a) (with the difference that now modify cos5 x).
Z
Z
2
4
I = sin x cos x cos x dx = sin2 x(1 − sin2 x)2 cos x dx.
After substituting sin x = z, cos x dx = dz we have
Z
Z
z3
z5 z7
2
2 2
I = z (1 − z ) dz = (z 2 − 2z 4 + z 6 )dz =
−2 +
+C =
3
5
7
=
1 3
2
1
sin x − sin5 x + sin7 x + C. 3
5
7
R
R
C. Integrals of types R(sin x) cos x dx, R(cos x) sin x dx
These indefinite integrals is preferable to calculate using substitution (after first rule about substitution)
sin x = z,
resp.
cos x = z.
Z
a) I =
1
dx,
2
sin x cos3 x
Z
b) I =
sin3 x
dx.
cos4 x
Solution.
a) First, the origin function modify by following way:
Z
Z
1
1
I=
cos x dx =
cos x dx.
2
2
4
sin x cos x
sin x(1 − sin2 x)2
Last integral include to first type and we use substitution sin x = z, cos x dx = dz. We get
Z
1
I=
dx.
2
z (1 − z 2 )2
We got known integral of rational function, which we know decompose to partial fractions and finalize calculation (make in-house).
b)
Z
1 − cos2 x
sin2 x
sin
x
dx
=
sin x dx.
I=
cos4 x
cos4 x
Last integral include to second type and we use substitution cos x = z, − sin x dx = dz and we have
Z
Z 1 − z2
1
1
1
1
1
1
I=−
dz = −
− 2 dz = 3 − + C =
−
+ C. 4
4
3
z
z
z
3z
z
3 cos x cos x
Z
R
D. Indefinite integrals of type R(sin x, cos x) dx
Indefinite integrals of this type is possible modify using substitution tg x2 = z to indefinite integral of rational function. Functions sin x, cos x a dx we
modify using new variable z. Plat
1 − z2
2
2z
,
cos
x
=
, dx =
dz.
sin x =
2
2
1+z
1+z
1 + z2
R
dx
Excercise 19. Let’s calculate I = 4 cos x+3
.
sin x
Solution. We use substitution tg x2 = z and we have
Z
I=
2
dz
1+z 2
2
2z
+ 3 1+z
4 1−z
2
1+z 2
Z
=−
1
dz = −
2
2z − 3z − 2
Z
1
dz,
(z − 2)(2z + 1)
1
A
B
=
+
.
(z − 2)(2z + 1)
z − 2 2z + 1
Using elementary method of finding roots we get z = 2 and z = − 12 . Then A = 15 , B = − 25 and we have
2 tg x2 + 1
1
21
1
ln |z − 2| −
ln |2z + 1| + C = ln |
I=−
| + C. 5
52
5
tg x2 − 2
If in some integral we can found only even powers of functions sin x and cos x, then we use the substitution tg x = z. Thus
sin2 x =
R
3+sin2 x
2 cos2 x−cos4 x
z2
,
1 + z2
cos2 x =
1
,
1 + z2
dx =
1
dz.
1 + z2
dx.
Solution. We use substitution tg x = z and we have
Z
I=
z2
1+z 2
− (1+z1 2 )2
3+
1
2 1+z
2
1
dz =
1 + z2
Z
4z 2 + 3
dz.
2z 2 + 1
After dividing the numerator of the denominator we have
√
Z
Z √
1
1
1
2
I=
dz = 2z +
dz
=
2z
+
arctg(
2z) + C =
2+ 2
2z + 1
2
2
z 2 + 12
√
√
2
arctg( 2 tg x) + C. 2
R
Integrals A-C are also of type R(sin x, cos x) dx. Although we do not solve them by substituting tg x2 = z, because its use leads to complicated
calculations.
E. Use trigonometric substitution to calculate integrals of irrational functions
Indefinite integrals of types:
Z
Z
Z
√
√
√
2
2
2
2
R(x, a + x ) dx,
R(x, x2 − a2 ) dx
R(x, a − x ) dx,
= 2 tg x +
is possible also modify using so called goniometric substitutions
x = a sin z,
to known type
R
x = a tg z,
x=
a
sin z
R(sin x, cos x) dx.
Z
a) I =
dx
√
,
( x2 + 4)3
b) I =
Z √
−x2 + 10x − 16 dx.
Solution.
a) This integral include to first type and, a = 2. Therefore, we use the substitution x = 2 tg z, dx = cos22 z dz,
2
sin2 z + cos2 z
4
sin z
2
2
2
+1 =4
=
.
x + 4 = 4 tg z + 4 = 4(tg z + 1) = 4
2
2
cos z
cos z
cos2 z
Then
Z
I=
2
dz
cos2 z
3
2
cos z
1
=
4
Z
cos zdz =
1
sin z + C.
4
We come back to origin variable x. We don’t express z, but sin z.
tg z =
x
,
2
x
tg z
sin z = p
=q 2
1 + tg2 z
1+
=√
x2
4
x
.
x2 + 4
Then
1
x
√
+ C. 4 x2 + 4
I=
b) The expression under the square root we will add to the full square.
−x2 + 10x − 16 = −(x2 − 10x + 16) = −[(x − 5)2 − 25 + 16] = 9 − (x − 5)2 .
Then
I=
Z p
9 − (x − 5)2 dx.
Now we use the substitution x − 5 = z, dx = dz and we have
Z √
I=
9 − z 2 dz.
Last integral include to first type and a = 3. Therefore, we use the substitution z = 3 sin t, dz = 3 cos tdt, 9 − z 2 = 9 − 9 sin2 t =
9(1 − sin2 t) = 9 cos2 t. Thus
Z
Z
1 + cos 2t
9
1
2
I = 9 cos tdt = 9
dt =
t + sin 2t + C.
2
2
2
Because sin t =
z
3
then
√
z2
z
9 − z2
1−
=
, t = arcsin ,
9
3
3
√
z 9 − z2
2 √
sin 2t = 2 sin t cos t = 2
= z 9 − z2.
3
3
9
p
cos t = 1 − sin2 t =
Finally we get
r
z 1 √
arcsin + z 9 − z 2 + C =
3 9
√
9
x−5 1
=
arcsin
+ (x − 5) −x2 + 10x − 16 + C. 2
3
9
9
I=
2
Problems
R
R
A. sinRn x dx, cosn x dx
241. R sin2 x dx.
243. R sin3 x dx.
245. R sin4 x dx.
247.
sin7 x dx.
R
B. sinRn x cosm x dx.
249. R sin x cos2 x dx.
251. R sin3 x cos2 x dx.
257.
sin4 x cos2 x dx.
R
R
C. R(sin
x)
cos
x
dx,
R(cos x) sin x dx
R sin x
dx.
259.
2x
R cos
sin3 x
261.
3 x dx.
R cos
cos x
263.
5 x dx.
R sin
sin2 x
265.
dx.
cos
R cos34 xx
267.
dx.
sin x
R sin
2x
269. R cos2 x dx.
cos x
dx.
271.
R sin2 x+9
sin x
273.
2 x+2 cos x dx.
R cos
sin3 x
275.
dx.
x
R 2+cos
1
277. R cos3 x dx.
1
279.
dx.
sin2 x cos x
R
D. R(sin
x,
cos
x)
dx
R
1
281.
dx.
cos x
R 5−3
2−sin x
283.
dx.
x
R 2+cos
1
285.
dx.
cos x
R 9+4sin
x
287.
dx.
R sin x+cos1 x
289.
dx.
sin x+7 cos x
R 8−4
1−tg x
291.
dx.
1+tg x
R
242. R cos2 x dx.
244. R cos3 x dx.
246. R cos4 x dx.
248.
cos5 x dx.
250.
252.
254.
256.
258.
R
R
R
R
R
sin2 x cos3 x dx.
sin7 x cos x dx.
sin4 x cos3 x dx.
sin2 x cos2 x dx.
sin4 x · cos4 x dx.
260.
262.
264.
266.
268.
270.
272.
274.
276.
278.
280.
R
R
R
R
R
R
R
R
R
R
R
sin3 x
dx.
cos2 x
sin3 x
dx.
cos4 x
cos5 x
dx.
sin2 x
cos2 x
dx.
sin x
sin2 x
dx.
cos x
sin4 x
dx.
cos2 x
cos x
dx.
sin2 x+6 sin x+5
sin x
dx.
cos2 x−2 cos x+1
cos x
dx.
(1−sin x)3
1
dx.
cos x
1
dx.
sin x cos x
282.
284.
286.
288.
290.
292.
R
R
R
R
R
R
1
dx.
5−4 sin x+3 cos x
x+sin x
dx.
1+cos x
cos x
dx.
1+cos x
1+sin x
dx.
1−sin x
1
dx.
cos x+2 sin x+3
1
dx.
1+8 cos2 x
R
1
293.
2 x dx.
R 4 sin2 x+9 cos
1
dx.
295.
x−5 cos2 x
R 1−3 sin x·cos
1
dx.
297.
R sin12 x−5 sin x·cos x
dx.
299.
cos
√
√4 x
√
R
R
R
2 − x2 ) dx,
2 + x2 ) dx,
E. R(x,
a
R(x,
a
R(x,
x2 − a2 ) dx
R
1
√
301.
2 2+x2 dx.
R x√x2 −9
303.
dx.
x
R √4−x
2
dx.
305.
R x2 1
√
307.
dx.
2 3
(9+x )
R
294.
R
296.
R
298.
R
300.
1
4−3 cos2 x+5 sin2 x
cos x
dx.
sin3 x−cos3 x
1
dx.
sin2 x−tg2 x
1
dx.
sin4 x
dx.
√ 1
dx.
R x xa22 +x2
√
304.
dx.
R x2 −4
1
√
306.
dx.
(1−x2 )3
R
√1
308.
dx.
x2 x2 −9
302.
R
Indefinite integral – Next different problems
Earlier in this chapter, we had the opportunity to meet with various ways and methods of calculation of indefinite integrals. That’s when you use such a
method, we learn the only way to calculate the number of different tasks. We must learn to ably use a variety of methods and appropriate combinations
of them. Also important is the modification of functions to integrate, if so required by the procedure. It often happens that in one example we must
use more methods.
Calculate
R ex −1following indefinite integrals.
309.
dx.
+1
R e2ex3x
+3ex
311.
dx.
R e12x +1
313.
dx.
R ax +1
3
315. R (ln x + ln x) dx.
ln x
317.
dx.
2
R x(1−ln 1x)
√
dx.
319.
1−x2 arccos2 x
R q 1−ex
321.
dx.
1+ex
R arctg
x
323. R x4 dx.
1
325.
4 x dx.
R sin4 x+cos
1√
327.
x 1−x2 dx.
R earcsin
ln arctg x
329.
dx.
(1+x2 ) arctg x
310.
312.
314.
316.
318.
R
R
R
R
R
ex −2
dx.
e2x +4
e4x
dx.
e8x +4
e3x +ex
dx.
e4x −e2x +1
arcsin x
dx.
x2
x·arctg
x
√
dx.
p1+x2
R
√
320.
1 − x dx.
R
sin x
√
322.
dx.
3
1+2 cos x
R
1
324.
dx.
sin x
R sin 2x−2
1
√
326.
x
2x dx.
√ +e x
R ex1+e
arctg e
328.
dx.
2x
R p1+e
arccos x
330.
dx.
1−x2
Self-assessment questions
1. What is the difference between the concepts of antiderivative and indefinite integral?
2. Why is the process of integration in general, much harder than the process of derivation, which is actually the opposite procedure?
3. List three basic uses methods by parts.
4. Try to find at least three elementary functions whose indefinite integral can not be expressed by elementary functions.
5. Determine the types of elementary functions, for which we use to integrating process some recurrent formula.
Bibliography
[1] Siagova, J.: Advanced Mathematics, Bratislava, STU, 2011.
[2] Bubenik, F.: Problems to Mathematics for Engineers, Praha, CVUT, 1999.
[4] Neustupa, J.: Mathematics I, Praha, CVUT, 1996.
[5] Neustupa, J.: Mathematics II, Praha, CVUT, 1998.
Solutions
Answers to self-assessment questions (if you are not able to formulate this answers), as well as many other answers can be found in the recommended
literature.
1.
3.
5.
7.
9.
3
x√
+ x2 + x + 13√ln |x| + C
2 x + 2x + 23 x x + C
ln |x| − x12 + C
ex + x1 + C
10x
ln 10 + C
( 1 )x
( 4 )x
11. 3 ln3 1 − 2 ln3 4 + C
3
( 45 )x
ln 45
3
13.
− 2x +
15. tg x − x + C
( 54 )x
ln 54
+C
4
6
2. 11
3 + C
6√x − x√
4. 2 x − 4 4 x + C
6. 12 x4 − 5x2 + 6 ln |x| + C
x
8. lna a + 3x1 3 + C
x
x
x
10. ln4 4 + 2 ln6 6 + ln9 9 + C
12.
( 34 )x
ln 34
− 2x +
( 34 )x
ln 43
+C
√ √
2 3
x7
7
14. 5 sin x −
+ 3 arctg x + C
16. − cotg x − x + C
17.
19.
21.
23.
25.
27.
29.
31.
33.
35.
− cotg x − tg x + C
1
1
2 x − 2 sin x + C
sin x − cos x + C
x − arctg x + C
−x + 5 arctg x + C
1 4
x
2
4 x − 2x + arctg 2 + C
1
2
2 ln |x − 3| + C
1
2
3 ln |3x − 15x + 22| + C
1
− 2 ln |2 cos2 x + 3| + C
ln | ln x| + C
37.
1
10 (x
39.
41.
43.
45.
47.
49.
51.
53.
55.
57.
59.
61.
63.
65.
67.
69.
71.
73.
75.
77.
79.
81.
83.
85.
87.
89.
91.
− ln |5 − x| + C
2
+C
− x+3
1
2
12
+C
24 (x − 4)
x
3 sin 3 + C
1
3 arctg 3x + C
−e−x + C
√
2 ex + C
1
x3
6 arctg 2 + C
2
3
+ 1) + C
3 tg(x
√
4
4
− 3 cos3 x + C
√
− x2 + C
arcsin x − 1√
2
1
arcsin
x
+
1 − x2 + C
2
1
x
−e + C
1
x
ln 2 arcsin 2 + C
1 2
1 2
2 x ln |x| − 4 x + C
−xe−x − e−x + C
1
2x
− 41 e2x + C
2 xe
1
1
2
2 x + 1 arctg x − 2 x + C
−x cos x + sin x + C
x tg x + ln | cos x| + C
xex + C
−e−x x2 + 2x + 2 + C
2x2 cos2 x − (x2 − 21 ) cos 2x + x sin 2x + C
x2 ex + C
(−x3 + 6x) cos x + (3x2 − 6) sin x + C
x ln x − x + √
C
x arcsin x + 1 − x2 + C
+ 2)
10
+C
18.
20.
22.
24.
26.
28.
30.
32.
34.
36.
38.
40.
42.
44.
46.
48.
50.
52.
54.
56.
58.
60.
62.
64.
66.
68.
70.
72.
74.
76.
78.
80.
82.
84.
86.
88.
90.
92.
tg x − cotg x + C
1
1
2 x + 2 sin x + C
1
1
2 tg x + 2 x + C
x − 2 arctg x + C
1 3
3 x − x + arctg x + C
x − x1 − 2 ln |x| + C
1
3
3 ln |x + 1| + C
ln | sin x| − ln | cos x| + C
− 61 ln |1 − 3e2x | + C
lnq
| arcsin x| + C
2
3
5
6
3
(x + 3) + C
ln |6x − 1| + C
1
− 3(2x+3)
3 + C
1
− 2 cos 2x + C
− 41 ln | cos(4x − 1)| + C
1
2 arcsin 2x + C
1 2x
+C
2e
1 x3
e
+
C
3
2
1
2(1−x)2 − 1−x − ln |1 − x| + C
3
1
3 sin x + C
ln | ln x| + ln2x + C
tg x − cos2 x + C
1
+C
2
2 ln
√ cos x
x
2e√ + C
2 tg x − 1 + C
xex − ex + C
1
1
x
x
ln 3 x.3 − ln2 3 3 + C
1 2 2
1 2
1 2
2 x ln x − 2 x ln |x| + 4 x + C
3
2
3
(3x + 2x ) ln x − x − x2 + C
1
1
2 x sin 2x + 4 cos 2x + C
x tg x + ln | cos x| − 21 x2 + C
(3 − x) cos x+ sin x + C
− 41 2x2 − 1 cos 2x + 12 x sin 2x + C
(x2 − 2) sin x +2x cos x − 2 sin x + C
1
5
1
2
2 x + 6x + 2 sin 2x + 2 (x + 3) cos 2x + C
1
1 3
1
4
4 (x − 1) arctg x − 12 x + 4 x + C
1
2
x arctg x − 2 ln(1 + x ) + C
√
x arcsin2 x + 2 1 − x2 arcsin x − 2x + C
93.
95.
97.
99.
101.
103.
105.
107.
109.
111.
113.
115.
117.
119.
121.
123.
125.
127.
129.
x ln2 x − 2x ln x + 2x
+C
1
2
3) ln x2 + 3 − 12 (x2 + 3) + C
2 (x +√
1
1 − x2 earcsin x + C
2 x+
1 2x
5 e (2 sin x − cos x) + C
1 3x
13 e √(3 sin 2x − 2 cos 2x) + C
x − 1 − x2 arcsin x + C
− 21 x cotg2 x − 12 cotg x − 12 x + C
2 ln |x + 7| + C
1
2 ln |2x − 1| + C
1
+C
− 4x−3
1
x−1
2 arctg 2 + C
1
2x+1
+C
3 arctg 3
2
ln(x + 6x + 10) − 6 arctg(x + 3) + C
√
6 ln(x2 + x + 6) − √1223 arctg 2x+1
+C
23
1
2
x
2
ln(x
+
9)
+
arctg
+
C
2
3
3
5 ln(x2 − 4x + 5) + 22 arctg(x − 2) + C
4x
x2 +1 + 4 arctg x + C
A
B
x−1 + x−2
B
A
x−1 + (x−1)2
131.
133.
135.
137.
139.
141.
143.
145.
147.
149.
A
Bx+C
x + x2 +1
A
B
Cx+D
x + x2 + x2 +4
Cx+D
Ax+B
x2 +4 + x2 −8x+17
2 ln |x| + ln |x − 2| + C
3 ln |x − 2| − 2 ln |x + 3| + C
ln |x| − 2 ln |x + 1| + ln |x + 2| + C
ln |x + 1| − 12 ln |2x + 1| + C
2 ln |x − 2| − ln |x − 1| + ln |x + 1| − 2 ln |x + 2| + C
x + 3 ln |x − 3| − 3 ln |x − 2| + C
1 2
2 x − x + 2 ln |x − 2| − 2 ln |x + 2| + C
155.
157.
159.
161.
x ln(x2 + 1) − 2x + 2 arctg x + C
1
1
2 x cos(ln |x|) + 2 x sin(ln |x|) + C
ex arctg ex − 12 ln(1 + e2x ) + C
ex sin2 x − 15 sin 2x + 52 cos 2x + C
1 2x
(5 sin 5x + 2 cos 5x) + C
29 e
x arctg x − 21 arctg2 x − 12 ln(1 + x2 ) + C
− x1 ln3 x + 3 ln2 x + 6 ln x + 6 + C
− 25 ln |3 − 2x| + C
− 38 ln |3x + 2| + C
1
− 16(2x+1)
2 + C
1
x+5
3 arctg 3 + C
1
5x+1
+C
7 arctg 7
1
− 4 ln(x2 − x + 12 ) − 21 arctg(2x − 1) + C
3
x+1
2
2 ln(x + 2x + 10) − arctg 3 + C
1
x
ln(x2 + 4) − 2 arctg 2 + C
− 21 ln(x2 − 2x + 5) + 12 arctg x−1
2 +C
x−3
1
x
3(x2 +9) + 9 arctg 3 + C
A
B
C
x + x−2 + x+2
A
B
C
x + x2 + x−3
A
Bx+C
x−5 + x2 +25
A
Bx+C
x−1 + x2 +2x+5
A
Bx+C
Dx+E
x + x2 +1 + (x2 +1)2
ln |x − 1| − ln |x + 1| + C
2 ln |x − 1| + ln |x − 2| + C
2 ln |x − 2| + 12 ln |2x + 1| + C
−7 ln |x − 1| + 4 ln |x − 4| + 3 ln |x + 3| + C
−3 ln |x| + 2 ln |x − 1| + ln |x + 1| + C
x − 5 ln |x| + 2 ln |x + 5| + C
1 2
2 x + 2x − ln |x − 3| + ln |x + 1| + C
16
152. −x3 + 23 x2 − 9x + ln (x+2)
|x−1| + C
151. x2 + ln |x − 1| + ln |x + 1| + C
1
161 2
6
+C
153. 5x + ln (x−1) (x−5)
7
(x−2) 3
x
1
13
14
9 + ln |x| − 27 ln |(3x − 1) (3x + 1) |
1
2 ln |x + 1| − x+1
+C
4
2 ln |x| + ln |x + 2| + x+2
+C
3
2 ln |x + 2| − 2 ln |x + 1| − x+2
+C
94.
96.
98.
100.
102.
104.
106.
108.
110.
112.
114.
116.
118.
120.
122.
124.
126.
128.
130.
132.
134.
136.
138.
140.
142.
144.
146.
148.
150.
+C
154.
x
4
156.
x3
x2
3 + 2
− 21 ln |x|
1
16
ln |(2x − 1)7 (2x + 1)9 | + C
2 (x−2)5 +C
+ 4x + ln x(x+2)
3
+ ln |x| −
+ 72 ln |x − 2| + x1 + C
158.
6
160. 2 ln |x| − ln |x + 1| + x+1
+C
4
162. ln |x + 1| + x+2 + C
163. ln |x| +
165.
167.
169.
171.
173.
175.
2
2x+1
√
(x−1)(x−3)
1
ln
+ x−1
+C
|x|
1
1
1
1
ln
|x
−
1|
−
ln
|x
+
1| − x−1
− x+1
+C
2
2
1
ln |x| − ln |x + 2| − 3x3 + C
4
2 ln |x| + x − x−1
+C
12
−3 ln |x − 1| + x − x+2
+C
3 ln |3x − 1| − 3 ln |x| + x + x1 + C
3
2 ln |x − 5| − 2 ln |x| + x − x−5
+C
1
x+1
1
178. 4 ln x−1 − 2 arctg x + C
+C
164.
166.
168.
170.
172.
174.
176.
ln |x| − ln |x − 1| + x32 − x2 + C
3
ln |x| − x−2
− x1 + C
2 ln |x − 1| − ln |x| + x + x1 + C
3
x−2
2
2 ln | x+2 | + x + x + C
x
1 2
3
ln | 2x−3 | + 2 x + 3x − 2x−3
+C
2
+C
3 ln |x − 1| + ln |x + 1| + 4x − x−1
177. ln √x|x|
+C
2 +1
179.
181.
183.
185.
187.
189.
191.
193.
195.
197.
199.
201.
203.
205.
(x+1)2
1
2x−1
√1
√
+C
6 ln x2 −x+1 + 3 arctg
3
2
1
x −4
+C
2
2 ln
√ x +4
x2 +2x+5
ln |x−1| + 3 arctg x+1
2 +C
(x2 −2x+5)2
x−1
ln |(x−1)3 | + arctg 2 + C
− 12 arctg x + 32 arctg x3 + C
x + 21 ln(x2 + 2x + 2) − arctg(x + 1) + C
2
x + 12 ln x −4x+13
− 13 arctg x−2
x2
3 +C
1
x − ln |x + 3| + 2 ln(x2 − 4x + 5) + 2 arctg(x
|x−1|
1 2
√
2 x + x + ln x2 +1 − arctg x + C
6
2
x2 +1 − 6 ln(x + 1) + 12 ln x + C
(x−2)2
3
x2 −4x+5 + 3 ln (x2 −4x+5) + C
x+2
√
− √23 arctg 2x+3
− x2 +3x+3
+C
3
√
x2 +1
x−1
ln |x+1| + x2 +1 + C
|x−1|
1
1
√
4 ln x2 −2x+3 − 4(x2 −2x+3) + C
√
180.
√ +C
221. arcsin x+1
2
arctg
2x+1
√
3
+C
184. 4 ln |x − 1| − 2 ln(x2 + 4) − 2 arctg
186.
188.
190.
− 2) + C
√
6
x
2
+C
2
1
x +1
x
2 ln x2 +4 + 2 arctg x − arctg 2 + C
1
1
2
+ 16) + 14 arctg x4 + C
x + 2 ln(x √
x3
x
√
3 − 3x + 3 3 arctg 3 + C
x2
x − x5 + 12 ln x2 −2x+10
− arctg x−1
3
2
192.
+C
194. x + ln |x − 2| − 2 ln |x + 2| − ln(x + 9) + 3 arctg
196.
198.
200.
202.
204.
206.
208.
2
2
x
2
2 − 2x − x + 2 ln(x + 2x + 2) − 2 arctg(x
x
1
2
x2 +1 + 2 ln(x + 1) − 2 ln(x + 2) + C
x−6
1
2(x2 +1) + 2 arctg x + C
1
− 2(x21+2) + 4(x2x+2) + 4√
arctg √x2 + C
2
2x−1
ln |x − 1| + 2 arctg x + 2(x
2 +1) + C
3
2
x
+1
1
1
3 − 3 −
3 ln
3(x3 +1) + C
√ x 8 3x
2 x + √x−2 + C
210. 3
√
6
x
3
+C
+ 1) + C
√
√
x2 − 2 6 x + 2 ln 6 x − 1 + C
√
12
x
1
1
212. 2 ln 1+ 12√
+ 12√
− 2√
6 x + C
x
x
p
p
1 3
1 3
5
214. 15 (3x + 1) + 3 (3x + 1)2 + C
√
+
√1
3
182. ln x2x+9 + C
207. −x + 4 x − 4 ln | x + 1| + C
√
√
6 7
6 5
√
√
√
209. 6 7x − 6 5x + 2 x − 6 6 x + 6 arctg 6 x + C
√
√
√
211. 32 3 x + 6 12 x − 6 arctg 12 x + C
x+1
6
ln √x|x−1|
+
2 +x+1
2
√
x−4
213. arctg
√ 2 + C√
√
3
6
(x+1)3
(x+1)4
(x+1)7
215. 6
−
+
−
9
8
7
q
√1−x−√1+x 217. ln √1−x+√1+x + 2 arctg 1−x
1+x + C
q
√
1 − x2 − arctg 1−x
219.
1+x + C
1
3
(x+1)5
5
√
6
−
(x+1)4
4
+C
5
4
3
2
216. − 18
5 t − 9t − 24t − 72t − 288t − 576 ln |t − 2| + C, kde t =
218.
√ x
1−x2
√
6
x−1
+C
p
220. ln(2x − 1 + 2 x(x + 1)) + C
√
√
5
222. 3 x2 + 5x − 10 − 13
x2 + 5x − 10 + C
2 ln x + 2 +
223.
225.
227.
229.
231.
233.
235.
237.
239.
241.
243.
245.
247.
249.
251.
253.
255.
257.
259.
261.
263.
265.
267.
269.
271.
273.
275.
277.
279.
281.
283.
285.
√
x2 + 2x + 2 ln x + 1 + x2 + 2x + C
√
ln x + 12 + x2 + x + 1 + C
q
2 + 8x − 3 + C
√1 x + 4 +
x
5
5
5
5√
√
−2 1 + x − x2 − 9 arcsin 2x−1
+C
5 √
√
2 − 2x + 5 − 5 ln x − 1 +
2 − 2x + 5 + C
x
x
x
√
2
√ +C
− x3 + 56 x − 19
1 − 2x − x2 − 4 arcsin x+1
6
2
√
1
x+1
2
√ +C
2 (3 − x) 1 − 2x − x + 2 arcsin
2
√
√
x+2
2 + 4x + 13 + 9 ln x + 2 +
x
x2 + 4x + 13 + C
2
2
√
x+1
3 − 2x − x2 + 2 arcsin x+1
2
2 +C
1
− 2 sin x cos x + x2 + C
− 31 sin2 x cos x − 23 cos x + C
1
1
3
8 x − 4 sin 2x + 32 sin 4x + C
3
− cos x + cos x − 35 cos5 x + 17 cos7 x + C
− 13 cos3 x + C
1
3
2
15 cos x(3 cos x − 5) + C
2t5
t7
t3
− 3 + 5 − 7 + C, kde t = cos x
t6
t8
8 − 6 + C, kde t = cos x
1
1
7
1
3
5
16 x + 32 sin 2x − 18 sin x cos x + 6 sin x cos x + C
1
cos x + C
1
ln | cos x| + 2 cos
2x + C
1
− 4 sin4 x + C sin x
1
sin x−1 C
2 cos2 x + 4 ln sin x+1 +
cos x−1 1
1
3
3 cos x + cos x + 2 ln cos x+1 + C
tg x − x + C
1
sin x
3 arctg
3 + C
1
cos x+2 +C
2 ln
cos x
1
2
cos
x
−
2
cos
x + 3 ln | cos x + 2| + C
2
sin x+1 sin x
1
2 cos2 x + 4 ln sin x−1 + C
1
1
sin
x+1
− sin x + 2 ln sin x−1 + C
x
1
2 arctg 2 tg 2 + C
tg2 x +3 ln tg2 x2 +1 + √43 arctg √13 tg x2 + C
2
q
5
x
√2 arctg
tg
13
2 +C
65
√
224.
226.
228.
230.
232.
234.
236.
238.
240.
242.
244.
246.
248.
250.
252.
254.
256.
258.
260.
262.
264.
266.
268.
270.
272.
274.
276.
278.
280.
282.
1
3
√
arcsin 3x−2
+C
2
q
√
1
2 − 6x + 2| + C = 1 ln |x − x−1 +
9x
x2 − 23 x + 29 | + C
ln
|3x
−
1
+
3
3
3
q
√1 ln x − 7 +
x2 − 72 x + 35 + C
6
3√
√ +C
−8 5 + 2x − x2 − 3 arcsin x−1
6
√
√
x−3
1
2 + 2x + 2 + ln x + 1 +
2 + 2x + 2 + C
x
x
2
2
√
2
√
5
x
x2 + 4x + 3 − 32 ln x + 2 + x2 + 4x + 3 + C
3 + 6x + 1
√
√
x+2
x2 + 4x + 3 − 12 ln x + 2 + x2 + 4x + 3 + C
2
√
2
√
arcsin 3x+1
+ 12 x + 61
1 − 2x − 3x2 + C
2
3 3
√
1
3−2x
4 2
√
2 + 3x − x2 + 17
+C
3 x − 6x + 3
2 arcsin
17
1
x
2 sin x cos x + 2 + C
1
2
2
3 sin x cos x + 3 sin x + C
3
1
sin 4x
8 x + 4 sin 2x + 32 + C
5
3
2
sin x − 3 sin x + sin5 x + C
3
1
2
15 sin x(3 cos x + 2) + C
8
1
8 sin x + C
z5
z7
5 − 7 + C, kde z = sin x
1
sin 2x(− 81 cos2 x + 16
) + x8 + C
1
1
1
1
sin3 2x + C
− 4 − 4 sin 2x − 16 sin 4x + 24
cos x + cos1 x + C
1
1
3 cos3 x − cos x + C
3
1
1
3 sin x − 2 sin x − sin x + C
1
cos x−1 cos x + 2 ln cos x+1 + C
sin x+1 − sin x + 21 ln sin
x−1 + C
1
tg x + 2 sin x cos x − 32 x + C
1
sin x+1 4 ln sin x+5 + C
1
cos x−1 + C
1
2(1−sin
x)2 + C
1
sin x+1 2 ln sin x−1 + C
ln | tg x| + C
1
2−tg x + C
2
284. x tg
x
2
286. x − tg
+C
x
2
+C
287. 14 ln (tg2 x + 1) − 21 ln (tg x + 1) +
289. ln (tg x2 − 5) − ln (tg x2 − 3) + C
291. ln | sin x + cos x| + C
x
293. 61 arctg ( 2 tg
3 )+C
295.
297.
299.
301.
303.
305.
307.
309.
1
5
1
5
ln | tg x − 4| −
ln (tg x − 5) −
3
tg x + tg3 x
−1
2 sin arctg √x2
√
1
5
1
5
x
2
+C
ln | tg x + 1| + C
ln (tg x) + C
+C
+C
x2 − 9 + 3 arctg √x32 −9 +
√
2
− arcsin x2 + C
− 4−x
x
x
√
+C
9 9+x2
(t+1)2
ln |t| + C, kde t = ex
x
x
C
1
319. arccos
x +C √
321. − ln e−x + e−2x − 1 − arcsin ex + C
2
1
6
329.
(ln | arctg x|)2
2
ln 1+x
x2 −
arctg x
3x3
− 6x1 2 + C
√
325. C − 22 arctg( 2 cotg 2x)
1
327. − earcsin
x + C
√
+C
tg x tg x+1
298.
300.
302.
311. 2e + arctg e + C
x
313. ln1a ln axa+1 + C
315. ln3 x − 3 ln2 x + 7 ln x − 7 x + C
1
317. ln √1−t
+ C, kde t = ln x
2
323.
4
288. 1−tg
x − x + C
2
290. arctg (tg x2 + 1) + C
292. 31 arctg tg3x + C
294. 13 arctg(3 tg x) + C
√
√
3
tg x−1
3
296. ln √
−
6
3 arctg
2
304.
306.
308.
310.
312.
314.
316.
318.
320.
322.
324.
2 tg√x+1
3
1
3 tg3 x + C
−1
1
3 tg3 x − tg x + C
x 1
arctg a + C
a ln
2
√
√
1
2
x2 − 4
2 x x − 4 + 2 ln x +
√ x
+C
2
√1−x
2
x −9
+C
9x
x
1
2x
+ 4 − 12 x + 21 arctg e2
4 ln e
1
e4x
C
8 arctg 2 +
√
√
+C
+C
+C
arctg(2t − 3) + arctg(2t + 3) + C, kde t = ex
√ 2
ln 1− x1−x − x1 arcsin x + C
√
√
1 + x2 · arctg x − ln x + 1 + x2 + C
p
p
√
√
4
(1 − x)5 − 43 (1 − x)3 + C
5 p
2
− 43 3 (1 + 2 cos
x)x + C
1
1
− 4 ln tg 2 + C
8 sin2 x
22+t+2√1+t+t2 + C, kde t = ex
326. − ln 2t
3
328. 23 (arctg ex ) + C
p
2
330. − 3 (arccos x)3 + C
Chapter 4. (Differential equations)
Aim
Introduce the students to the analytical solution of some types of ordinary differential equations and describe the general solution and particular solution.
Objectives
1. Determine the order of a differential equation, specify the number and type of initial conditions.
2. Classify a first order differential equation, learn to calculate some types of such equations.
3. A work of integration constants, determination of integration constants according to the initial conditions.
4. Analyze differential equation with constant coefficients of the second or higher order and understand various methods of their solution.
Prerequisities
function; graph of a function; curve; derivative; integral
Chapter 4. (Differential equations) – 1
Introduction
In the chapter, special kinds of ordinary differential equations are discussed, the solution is search either in general form or in particular form given by an
initial conditions. As long as a lot of physical processes or geometrical dependencies are expressed by differential equations, the equations are frequently
used in solving civil engineering problems. They can be found in mechanical analysis of constructions or their parts, in heat transfer problems or particle
transfer problems in buildings, or in optimizing and controlling of the construction management. Many differential equations do not provide an analytical
solution in a closed form, nevertheless, in several basic problems such a solution can be found which is the topic of the present chapter.
First, we will see which characteristics are important for taking into account an appropriate method for solving particular equation. Basic methodologies
of first order differential equation solutions cover the separation of variables and variation of constants. The second tool is especially useful for solving
linear differential equations. In addition to these basic types, also another types of differential equations will be shown which are usually solved by a
substitution to transform a more complicated equation to a simpler, in fact separable or linear.
The general solution will be looked for in all cases, however a special attention will also be paid to determination of particular solution based on prescribed
initial conditions. From the physical point of view or its technical implementation, the solution of a problem with initial conditions is more important.
The next part is focused on the solution of linear differential equation of the second or higher order with constant coefficients. Two commonly used
method used for their solution will be discussed. First, a method of special-form right-hand side uses a method of indeterminate coefficients for a series
of right-hand side functions. Second, if the function fails to be considered as a special form function, the method of constants variation is used as a
general tool for obtaining linear equations. Also in this case, appropriate initial conditions are required to correctly interpret the results in physical sense.
A few words will also be spent on solving the systems of first order differential equations which can be solved by substitutions to come to one higher
order equations.
The final part presents a series of word problems which can be converted to the solution of differential equations. The problems cover geometric
implementations and also physical ones.
Differential equations – Basic notions
Before we start to solve the differential equations, we present some basic differential equations terminology.
The n-th order ordinary differential equation is called the equation of the form
F (x, y, y 0 , y 00 , . . . , y (n) ) = 0,
where y=y(x) is an unknown function of one real variable x and F is an arbitrary function of n+2 unknowns. The number n is calld the order of the
differential equation. If the function F is a degree m polynomial with respect to variables y, y 0 , y 00 , . . . , y (n) , the number m is called the degree of the
differential equation. The degree one differential equation is called linear.
The solution or the integral of the differential equation is any function y=ϕ(x) satisfying the relation
F (x, ϕ(x), ϕ0 (x), ϕ00 (x), . . . , ϕ(n) (x)) = 0.
The graph of the function is called integral curve of the differential equation.
The integral
φ(x, y, c1 , . . . , cn ) = 0
of the aforementioned differential equation containing n arbitrary constants is called a general integral, or a general solution of the differential equation.
The constants c1 , . . . , cn are called integration constants. If the integration constants are substituted by particular values, the particular solution is
obtained.
Let for the solution y=ϕ(x) of the aforementioned differential equation following conditions hold at x=x0 :
y(x0 ) = b0 ,
y 0 (x0 ) = b1 ,
...,
y (n−1) (x0 ) = bn−1 ,
where b0 , b1 , . . . , bn−1 are arbitrary constants. Then these conditions are called Cauchy initial conditions.
The solution of a differential equations given together with initial conditions is called also a Cauchy problem.
Example 1 Determine the order and the degree of the differential equation
y 2 y 00 +y sin x y 0 =x4 .
Then specify required number of initial conditions for a unique determination of integration constants in general solution and then choose them arbitrarily.
Solution. The equation can be written, using the function F as follows:
F (x, y, y 0 , y 00 ) = y 2 y 00 +y sin x y 0 − x4 = 0.
The differential equation contains the second derivative of the unknown function y as the highest, thus this is the second order equation. It implies
that there are two integration constants. Moreover, the function F (x, y0 , y1 , y2 )=y02 y2 + sin x y0 y1 −x4 is with respect to the variables y0 =y, y1 =y 0 and
y2 =y 00 the third degree polynomial (due to the term y02 y2 ), hence the degree of the differential equation is three.
We need two initial conditions to uniquely determine integration constants. They could be prescribed, for example as
y(1) = 1,
y 0 (1) = 2.
Differential equations – Basic first-order equations
Rate of change of a physical quantity is expressed by the first derivative so that finding a relation between such a quantity and its rate leads to differential
equations of the first order. We will se the principal methods for their solving.
A general for of a differential equation of the first order is
dy
) = 0,
dx
which is usually expressed in explicit form with respect to the derivative y 0 =f (x, y). The first step in solving the equation identification of variable
separation or of equation linearity.
F (x, y, y 0 ) = F (x, y,
If a differential equation can be expressed in the form
ϕ(y)dy = ψ(x)dx
for some continuous functions ϕ and ψ, the equation is called the equation with separated variables.
The differential equation of the form
ϕ1 (x)ψ1 (y)dx = ϕ2 (x)ψ2 (y)dy
is called the equation with separable variables. Being such an equation divided by the product ϕ2 (x)ψ1 (y), the equation with separated variables is
obtained. The general integral is given by the following expression:
Z
Z
ψ2 (y)
ϕ1 (x)
dx −
dy = C.
ϕ2 (x)
ψ1 (y)
The solution is often written in an implicit form Φ(x)−Ψ(y)−C=0, but if it is possible we prefer an explicit form y=Υ(x, C).
A linear first-order differential equation with non-zero right-hand side is usually written in the form
y 0 + p(x)y = q(x),
where p and q are continuous functions with the only variable x. In the case of q(x)≡0, the equation is called the equation with zero right-hand side
(frequently also without right-hand side). The linear differential equation can be solved in different ways:
A. It can be proved that the general solution of a linear differential equation has the form
Z
R
R
p(x)dx
y=
q(x)e
dx + C e− p(x)dx .
B. Applying themethod of variation of constants, the original problem is solved first with zero right-hand side by separating the variables. The solution
of the simplified equation y0 can be always written in the form
y0 (x) = ce−
R
p(x)dx
.
In the second step, it is considered that the constant c is a variable c = C(x) and the solution of the original problem y (non-zero right-hand side)
is looked for in the form
R
y(x) = C(x)e− p(x)dx ,
such that the unknown function C is determined, if y and its derivative y 0 are substituted into the original equation with non-zero right-hand side.
The Cauchy problem prescribes besides the differential equation also the initial condition
y(x0 ) = b0 .
This condition uniquely determines the integration constant C in general integral.
Example 2 Solve the differential equation
(1 + ex )yy 0 = ex .
dy
dy
, the equation is expressed as (1 + ex )y dx
=ex . The equation is multiplied by the expression
Solution. As long as y 0 = dx
last equation is integrated and the general solution is rendered
dx
1+ex
x
e
to obtain ydy= 1+e
x dx. The
y2
− ln(1 + ex ) = C.
2
Example 3 Find the particular solution of the equation y 0 cos2 x=y ln y which satisfies the initial condition y( π4 )=e.
Solution. The equation can be written as
provides the general solution
dy
dx
cos2 x=y ln y, where the variables are separated to obtain
dy
y ln y
=
dx
.
cos2 x
The integral of the equation
ln | ln y| = tg x + C.
The initial condition gives y=e for
the particular solution is
x= π4 ,
thus after substitution to the general solution, the integration constant reads C= − 1. The implicit form of
ln | ln y| − tg x + 1 = 0.
Example 4 Solve the equation
2
y 0 + 2xy = 2xe−x .
Solution. We use the variation of variables to solve the equation. First, the equation with zero right-hand side is solved
y 0 + 2xy = 0,
with subsequent modifications:
dy
=
y
− 2xdx and ln |y|= − x2 + ln c. The general solution of the equation is
2
y=ce−x .
Supposing the variation c=C(x), the derivative of the original solution is
2
2
y 0 = C 0 (x)e−x − C(x)2xe−x .
2
2
Both y and y 0 are substituted into the original equation. The terms with C are eliminated and remains only C 0 (x)e−x =2xe−x , which provides the varied
constant C(x)=x2 +c1 . The function is substituted for c in general solution of the equation with zero right-hand side to obtain the general solution of
the equation with non-zero right-hand side
2
y = (x2 + c1 )e−x .
Remark 1 If an initial condition, say y(0)=1 is added to the differential equation in Example 4, the integration constant c1 can be determined
2
by the condition 1=(0+c1 )e−0 . The pertinent particular solution has the form
2
y = (x2 + 1)e−x .
The graphs in Fig. 4.1 show particular solutions of the differential equation for various integration constants c1 for x≥0. The solution corresponding
to the given initial condition is drawn by thick red line.
Differential equations – Other types of first-order differential equations
If the differential equation does not belong to any of the basic types, the equation is modified in order to obtain either a linear equation or an equation
with separated variables.
y
c1 =3
c1 =2
c1 = 23
c1 =1
c1 = 21
x
O
Figure 4.1: Particular solution for various values of c1 .
A function f (x, y) is called degree n homogeneous with respect to variables x and y, if for any t holds
f (tx, ty) = tn f (x, y).
The differential equation y 0 =f (x, y) is called homogeneous, if f (x, y) is degreee zero homogeneous function, i.e. f (tx, ty)=f (x, y) for all t. It also
implies f (x, y)=f (1, xy ) and the homogeneous differential equation can be written in the form
y
.
y0 = ϕ
x
The substitution z= xy converts the homogeneous equation to an equation with separable variables, because
y = zx,
Hence
dz
x+z=ϕ(z),
dx
y 0 = z 0 x + z,
z0 =
dz
.
dx
which gives
dz
dx
=
.
ϕ(z) − z
x
Another type of an equation is the Bernoulli equation. It is an equation of the form
y 0 + p(x)y = q(x)y α .
Special case cover the linear equation with non-zero right-hand side for α=0, or the linear equation with zero right-hand side for α=1. The other values
of α can be treated by the substitution
z = y 1−α ,
z 0 = (1 − α)y −α y 0 .
It converts the Bernoulli equation to a linear one. Notice that for α>0 the zero function y≡0 is a particular solution of the equation.
A slightly different type of an equation is the following:
M (x, y) + N (x, y)y 0 = 0.
Such an equation is called exact differential equation, if it holds
∂M (x, y)
∂N (x, y)
=
.
∂y
∂x
The general solution of the exact differention equation has an implicit form f (x, y)=C, where the function f (x, y) satisfies the conditions
∂f (x, y)
= M (x, y),
∂x
∂f (x, y)
= N (x, y).
∂y
(x2 + y 2 )y 0 = 2xy.
Solution. The differential equation is homogeneous as it can be written as
y0 =
2xy
+ y2
x2
and the right-hand side function satisfies the following relations:
2 xy
2(tx)(ty)
2xyt2
2xy
= 2
= 2
=
2 .
(tx)2 + (ty)2
(x + y 2 ) t2
x + y2
1 + xy
Put y=xz with y 0 =z 0 x+z and substitute it to the modified equation rendering z 0 x + z =
2z
,
1+z 2
or equivalently
x(1 + z 2 )z 0 + z 3 − z = 0.
This is an equation with separable variables which provides after separation the integral:
Z
Z
1 + z2
1
dz =
dx.
3
z −z
x
Both sides of the equation are integrated (the left-hand side is decomposed to partial fractions) and the integral reads
2
z − 1
− ln |x| = ln C2 ,
ln z where C2 =eC1 >0. Trivial modifications provide the equation x|z 2 − 1|=C2 |z|. Adding negative values to integration constant, the absolute value can
be removed and the general solution reads simply as
xz 2 − x = C,
where C6=0. Utilizing the back substitution from y=xz, the general solution has the form
y 2 − x2 = Cy.
Remark 2 The solution of the Cauchy problem, where the initial condition is y(0)=2, the integration constant C is determined by the condition
22 + 00 =C · 2. The implicit function of the particular solution is
y 2 − x2 = 2y.
Realize that the solution initiates at given point (physically). Thus, talking about the solution for x<0 lose the sense as these xs do not belong
to the domain of the solution. Let us try to demonstrate it by the graph in Fig. 4.2. The general solution is a set of all functions with various
constants C, in the figure e.g. C∈{ 21 , 23 , 2, 3}. However, our particular solution (the thick red line) is only a part of a curve beginning at the point
[0, 1].
2
y 0 + y = x4 y 2 .
x
Solution. It is the Bernoulli differential equation with α=2. As 2>0 one particular solution is y=0. Let us multiply the equation by the term y −2 . We
obtain
2
y −2 y 0 + y −1 = x4 .
x
−1
0
−2 0
Now, put z=y with z = − y y and substitute it to the original equation rendering
2
−z 0 + z = x4 .
x
y
C= 32
C= 12
3
C=3
C=2
C=1
2
3
2
1
1
2
x
O
C=3
C=2
C= 23
C= 12
Figure 4.2: Particular solutions for various C.
This is a linear differential equation, whose solution
x5
+ C 1 x2
3
can be found e.g. by the method of variation of variable as used in Example 4. Using the back substitution for z, the solution of the Bernoulli differential
equation reads
3
, y = 0,
y=
2
Cx − x5
where C=3C1 .
z=−
Remark 3 The zero solution does not correspond to any value C, though it can be obtained by a limit with C→∞. For fixed x, the limit is
always zero:
3
y(x) = lim
= 0.
C→∞ Cx2 − x5
(3x2 + 6xy 2 ) + (6x2 y + 4y 3 )y 0 = 0.
Solution. First, let us show that the equation is an exact differential equation. The partial derivatives of the terms M (without y 0 ) and N (with y 0 ) with
respect to pertinent variables read
∂(3x2 + 6xy 2 )
∂M (x, y)
=
= 12y,
∂y
∂y
∂N (x, y)
∂(6x2 y + 4y 3 )
=
= 12y.
∂x
∂x
As the derivatives are equal to each other, the equation is exact and its integral f satisfies the relation
∂f (x, y)
= M (x, y) = 3x2 + 6xy 2 .
∂x
It implies
Z
f (x, y) =
(3x2 + 6xy 2 )dx = x3 + 3x2 y 2 + ϕ(y),
∂f (x, y)
= 6x2 y + ϕ0 (y).
∂y
(x,y)
Simultaneously, the other condition should be valid ∂f∂y
=N (x, y)=6x2 y + 4y 3 , thus 6x2 y+ϕ0 (y)=6x2 y+4y 3 . The last relation provides ϕ(y)=y 4 .
The general solution of the given equation is any function
x3 + 3x2 y 2 + y 4 = C
with an arbitrary integration constant C.
Differential equations – Linear differential equations of the second or higher order with constant coefficients
The equations of a higher order cannot be easily integrated in general case, nevertheless a linear equation with constant coefficients can be solve by a
similar scheme as the linear equations of the first order.
A linear equation of any order n always enables to obtain the solution in two steps. In the first step, an equation with zero right-hand side is solved. In
the second step a particular solution of the equation with non-zero right-hand side is found. The method of finding this particular solution depends on
the function which appears in the right-hand side of the equation. The general solution of a linear equation is a combination of the two components.
As an additional final step, if it is required by the problem definition, the integration constants are determined satisfying the initial conditions
y(x0 ) = b0 ,
y 0 (x0 ) = b1 ,
...,
y (n−1) (x0 ) = bn−1 .
An equation
y (n) + a1 y (n−1) + · · · + an−1 y 0 + an y = 0,
where ai , i=1, 2, . . . , n are real numbers, is called the linear differential equation with constant coefficients with zero right-hand side. The polynomial
equation with the unknown r and with the same coefficients of the form
rn + a1 rn−1 + a2 rn−2 + · · · + an−1 r + an = 0
is called the characteristic equation of the differential equation. The general solution of the equation with zero right-hand side has a form
y = c1 y 1 + c2 y 2 + · · · + cn y n ,
where y1 , y2 , . . . , yn are linearly independent solutions of the equation and c1 , c2 , . . . , cn are arbitrary (integration) constants.
If r1 is a multiple root of the characteristic equation with multiplicity k, then the functions er1 x , xer1 x , x2 er1 x , . . . , xk−1 er1 x are linearly independent
solutions of the equation with zero right-hand side.
If α + βi is a k-multiple imaginary root of the characteristic equation, then the functions
eαx cos βx, xeαx cos βx, x2 eαx cos βx, . . . , xk−1 eαx cos βx,
eαx sin βx, xeαx sin βx, x2 eαx sin βx, . . . , xk−1 eαx sin βx,
are linearly independent solutions of the equation with zero right-hand side.
An equation
y (n) + a1 y (n−1) + · · · + an−1 y 0 + an y = f (x),
is called linear differential equation with constant coefficients with non-zero right-hand side. Its general solution is obtained as a sum of the general
solution of the equation with zero right-hand side and an arbitrary particular solution of the equation with non-zero right-hand side. If the right-hand
side f (x) is given in any of the forms below, the equation is said to have special right-hand side. In these cases, the form of particular solution can be
guessed and unknown parameters in the form are calculated by the method of indeterminate coeficients. The method is used under following conditions,
for example:
A. f (x)=Pm (x)eαx , where Pm (x) is an m-degree polynomial and α is a real number: the particular solution can be written in the form
y = (b0 + b1 x + b2 x2 + · · · + bm xm )eαx xk ,
if α is a k-multiple root of the characteristic equation (if α is not a root of the characteristic equation, k=0). The real numbers b0 , b1 , . . . , bm are
the unknown parameters calculated by the method of indeterminate coefficients.
B. f (x)=eαx (A cos βx+B sin βx), where α, β are real numbers: the particular solution can be written in the form
y = eαx (M cos βx + N sin βx)xk ,
if α+βi is a k-multiple root of the characteristic equation. The constants M and N are again calculated by the method of indeterminate
coefficients.
Another possibility for finding the solution of the equation with non-zero right-hand side is the Lagrange method of variation of constants. This method
is usually used only in the case, if the right-hand side cannot be expressed as a special form function or a sum of such functions. Let y1 , y2 , . . . , yn are
n linearly independent solutions of the equation with zero right-hand side. The general solution of the equation with non-zero right-hand side is of the
form
y = C1 (x)y1 + C2 (x)y2 + · · · + Cn (x)yn ,
where the functions C1 (x), C2 (x), . . . , Cn (x) are obtained as a solution of the following equation system:
C10 (x)y1
C10 (x)y10
C10 (x)y100
..
.
+C20 (x)y2
+C20 (x)y20
+C20 (x)y200
(n−1)
C10 (x)y1
(n−1)
+C20 (x)y2
+ · · ·+Cn0 (x)yn
+ · · ·+Cn0 (x)yn0
+ · · ·+Cn0 (x)yn00
=0,
=0,
=0,
..
.
+ · · ·+Cn0 (x)yn(n−1) =f (x).
The determinant of the left-hand side matrix in the system is call the Wronski determinant or Wronskian W (C1 , C2 , . . . , Cn ). It can be proved that
the functions f1 , f2 , . . . , fk are linearly dependent in the interval ha, bi, if W (f1 , f2 , . . . , fk )=0 for any x∈ha, bi. If W (f1 , f2 , . . . , fk )6=0 at least at one
point x∈ha, bi, the functions are linearly independent. This implies that the system can be solved by Cramer rule and the solution is written in the form
Z
Z
Z
W2 (x)
Wn (x)
W1 (x)
dx + y2
dx + · · · + yn
dx,
y = y1
W (x)
W (x)
W (x)
were W (x), W1 (x), W2 (x), . . . , Wn (x) are pertinent determinants necessary for solving the system by the Cramer rule.
Finally, recall that linearity of the equation enable to use the superposition principle to find the general solution. If the solution of the equation
with zero right-hand side is denoted by y0 , the right hand side is decomposed to a sum of s function (say, in special form, but not necessarily)
f (x)=f1 (x)+f2 (x)+ · · · +fs (x) and the solutions of the equation with the right-hand side fi (x) are respectively denoted as yi , then the solution reads
y = y0 + y1 + y2 + · · · + ys .
y 000 − 2y 00 − 3y 0 = 0.
Solution. An equation with zero right-hand side is solved, we need to find the characteristic equation. It reads
r3 − 2r2 − 3r = 0.
The roots are r1 =0, r2 =−1, r3 =3. All the roots are simple and real, thus three independent solutions of the differential equation are: e0x , e−x and e3x .
The general solution then reads
y = c1 + c2 e−x + c3 e3x .
y 000 + 2y 00 + 3y 0 = 0.
Solution. An equation with zero right-hand side is solved again. The characteristic equation reads.
r3 + 2r2 + r = 0.
The roots are r1 =0 and r2 =r3 =−1. As the root “−1” is double, the pertinent paier of linearly independent solutions is e−x and xe−x . The third linearly
independent solution is a constant function so that the general equation of the differential equation is
y = c1 + c2 e−x + c3 xe−x .
y 00 + 2y 0 + 5y = 0
with initial conditions y(0)=2, y 0 (0)=0.
Solution. The characteristic equation is
r2 + 2r + 5 = 0.
The roots of the equation are imaginary (negative discriminant of the quadratic trinomial): r1 =−1+2i and r2 =−1−2i. The functions e−x cos 2x and
e−x sin 2x are linearly independent solutions of the equation and the general forma reads
y = c1 e−x cos 2x + c2 e−x sin 2x.
The integration constants can be determined by the initial conditions which also require the derivative of the general solution:
y 0 = c1 −e−x cos 2x − 2e−x sin 2x + c2 −e−x sin 2x + 2e−x cos 2x .
The given initial conditions render
0 = c1
2 = c1 e−0 cos (2·0) + c2 e−0 sin (2·0) = c1 ,
−e−0 cos (2·0) − 2e−0 sin (2·0) + c2 −e−0 sin (2·0) + 2e−0 cos (2·0) = −c1 + 2c2 .
The solution of the two equation system is a couple c1 =2, c2 =1 which provides the solution of the Cauchy problem in the form
y = e−x (2 cos 2x + sin 2x) .
y 00 + y 0 = 4x2 ex .
Solution. The root of the characteristic equation r2 +r=0 are r1 =0, r2 =−1, thus the general solution of the equation with zero right-hand side is
y0 = c1 + c2 e−x .
As α=1 is not a root of the characteristic equation, the particular solution y ∗ of the equation with non-zero right-hand side can be guessed in the form
y ∗ = (Ax2 + Bx + C)ex .
This function, for appropriately chosen parameters A, B, C is a solution of the equation, so it can be substituted into the equation together with its
derivatives
y ∗ 00
y ∗ 0 = (2Ax + B)ex + (Ax2 + Bx + C)ex ,
= 2(2Ax + B)ex + 2Aex + (Ax2 + Bx + C)ex .
The substitution renders
2Ax2 + (6A + 2B)x + 2A + 3B + 2C = 4x2 .
Comparing the coefficients by the same power of x, we have
2A = 4,
6A + 2B = 0,
2A + 3B + 2C = 0,
satisfied for A=2, B=−6, C=7. The particular solution is then
y ∗ = (2x2 − 6x + 7)ex
and the general solution of the differential equation is a sum of y0 and y ∗
y = c1 + c2 e−x + (2x2 − 6x + 7)ex .
y 00 + 10y 0 + 25y = 4e−5x
Solution. The characteristic equation r2 +10r+25=0 has a double root r1 =r2 =−5, hence the general solution of the equation with zero right-hand side
is
y0 = c1 e−5x + c2 xe−5x .
Moreover, as α=−5 is a root of characteristic equation with multiplicity k=2, the particular solution of the equation with non-zero right-hand side can
be expressed as
y ∗ = Ae−5x x2 .
The solution y ∗ and its derivatives
y ∗ 00
y ∗ 0 = 2Axe−5x − 5Ax2 e−5x ,
= 2Ae−5x − 20Axe−5x + 25Ax2 e−5x .
are again substituted to the equation providing
2Ae−5x = 4e−5x ,
with the solution A=2. The particular solution reads
y ∗ = 2x2 e−5x
and the general solution is obtained by adding the solution y0
y = y0 + y ∗ = c1 e−5x + c2 xe−5x + 2x2 e−5x .
Now, the integration constants should be determined. To use the initial conditions, the solution derivative is required
y 0 = −5c1 e−5x + c2 e−5x − 5xe−5x + 2 2xe−5x − 5x2 e−5x .
The start point provides the relations
0 = c1 e−5·0 + c2 · 0e−5·0 + 2 · 02 e−5·0 = c1 ,
1 = −5c1 e−5·0 + c2 e−5·0 − 5 · 0e−5·0 + 2 2 · 0e−5·0 − 5 · 02 e−5·0 = −5c1 + c2 ,
whose solution is a constant couple c1 =0, c2 =1. The Cauchy problem solution then reads
y = xe−5x (1 + 2x) .
y 00 − 6y 0 + 9y = 25ex sin x.
Solution. Also in this case the characteristic equation r2 −6r+9=0 has a double root r1 =r2 =3, the solution y0 (zero right-hand side) can be expressed
as
y0 = c1 e3x + c2 xe3x .
As to the special right-hand side eαx sin βx=ex sin x pertains the number α+βi=1+i (or α−βi=1−i), which is not the root of the characteristic equation,
the particular solution can be guessed in the form
y ∗ = ex (M cos x + N sin x).
The derivatives of this functions are
y ∗ 00
y ∗ 0 = ex (M cos x + N sin x) + ex (−M sin x + N cos x),
= ex (M cos x + N sin x) + 2ex (−M sin x + N cos x) + ex (−M cos x − N sin x)
an their substitution to the given equation renders
(3M − 4N ) cos x + (4M + 3N ) sin x = 25 sin x.
Comparing the coefficients by the functions cos x and sin x, the system of linear equations is obtained
3M − 4N = 0,
4M + 3N = 25,
whose solution is M =4, N =3. The general solution of the differential equation with non-zero right-hand side is
y = c1 e3x + c2 xe3x + ex (4 cos x + 3 sin x).
y 00 + y 0 = 5x + 2ex
Solution. The roots of the characteristic equation r2 +r=0 are r1 =0, r2 =−1 which provides the solution y0 in the form
y0 = c1 + c2 e−x .
The particular solution of the equation with non-zero right-hand side can be obtained by the superposition principle as a sum of particular solutions y1
and y2 of the equations
y 00 + y 0 = 5x,
andy 00 + y 0 = 2ex ,
respectively. The form of the particular solution y1 is y1 =(Ax+B)x, the form of the particular solution y2 is y2 =Cex . Both solutions are differentiated
and substituted into pertinent equations. The indeterminate coefficients are calculated: A= 52 , B=−5, C=1. The general solutio of the original equation
is
5
y = c1 + c2 e−x + x2 − 5x + ex .
2
Finally, the integration constants will be determined. With
y 0 = −c2 e−x + 5x − 5 + ex ,
the initial conditions provide the constants c1 and c2 by the relations
5
0 = c1 + c2 e−0 + 02 − 5 · 0 + e0 = c1 + c2 + 1,
2
0 = −c2 e−0 + 5 · 0 − 5 + e0 = −c2 − 4.
The solution of the system c1 =3, c2 =−4 renders the solution of the Cauchy problem
5
y = x2 − 5x + 3 − 4e−x + ex .
2
Remark 4 Let us draw some particular solutions for various values of integration constants. The graphs in Fig. 4.3 present five such couples,
including the solution of the Cauchy problem in Example 14. This solution is drawn by the thick red line and naturally only from the point of
c1 =0, c2 =1
y
c1 =1, c2 =0
c1 =4, c2 =−1
c1 =1, c2 =−2
c1 =3, c2 =−4
O
y=x
y=3
3x
Figure 4.3: Particular solutions for various values of integration constants.
the initial conditions: the graph begins at the point O and its tangent is the line y=x which means unit derivative at this point. Let us notice
that the scales of the coordinate axes are different due to exponential function which increases (or decreases) too fast for x just a bit greater than
one.
The graphs also show that while for the negative x the particular solutions are rather distinct, for the positive x are similar. This is again caused
by the exponential function ex present in the particular solution. All other function are for sufficiently large x negligible with respect to this
exponential.
Example 15 Find the general solution of the equation
y 00 + y =
1
.
cos x
Solution. The roots of characteristic equation r2 +1=0 are r1 =i, r2 =−i, hence the solution with zero right-hand side is
y0 = c1 cos x + c2 sin x.
We put c1 = C1 (x) and c2 = C2 (x) and use the method of variation of constants in the next calculation. We obtain the following system of equations
for unknown functions C1 , C2 :
C10 (x) cos x + C20 (x) sin x = 0,
1
,
−C10 (x) sin x + C20 (x) cos x =
cos x
whose solution is C10 (x) = − tg x a C20 (x) = 1. The functions C1 , C2 are rendered by integration
C1 (x) = ln | cos x| + c3 ,
C2 (x) = x + c4 .
The general solution can be written by the expression
y = (ln | cos x| + c3 ) cos x + (x + c4 ) sin x.
Remark 5 Let us notice the last form of the solution. As the solution of a linear equation, it can be expressed as a sum of the solutions with
zero right-hand side and with non-zero right-hand side. Let us rewrite it in the form
y = (c3 cos x + c4 sin x) + (ln | cos x| cos x + x sin x) .
The term in the first parentheses is y0 , though the notation for integration constants is different then above. In the second parentheses, the
particular solution is written.
Differential equations – systems of first-order differential equations
Relations of various physical quantities are frequently more complicated and require to solve not only one equation by a whole system of differential
equations.
A set of first-order differential equations
x01 = f1 (t, x1 , x2 , . . . , xn ),
x02 = f2 (t, x1 , x2 , . . . , xn ),
..
.
0
xn = fn (t, x1 , x2 , . . . , xn ),
is call a system of differential equations in normal form, the number n is called the order of the system. The solution of the system is considered any
n-component vector of differentiable functions (x1 (t), x2 (t), . . ., xn (t)) such that substituting it to the system, n equalities are obtained.
The system can be solved e.g. by the elimination method which converts the system of n equations of the first order to one equation of the order n
with one unknown function, say x1 . The form of the resulting equation is
(n)
(n−1)
x1 = φ(t, x1 , x01 , . . . , x1
)
with the solution x1 =ϕ1 (t, c1 , c2 , . . ., cn ). Subsequently, other unknown function are calculated until all n components of the solution are rendered
x1 = ϕ1 (t, c1 , c2 , . . . , cn ),
x2 = ϕ2 (t, c1 , c2 , . . . , cn ),
..
.
xn = ϕn (t, c1 , c2 , . . . , cn ).
Remark 6 Physically, the variable t is often interpreted as time and pertinent time-derivative distinguished by a “dot” instead of a “dash”. This is
rather useful if we tend to differentiate with respect to x which in the present case is the vector of unknown functions to be found. Moreover, the
“dash” can be used in more complicated situations for differentiation with respect to space variables in time-space partial differential equations.
Thus, in what follows, the “dot” notation will be used.
A system of two equations can be written in normal form as follows:
ẋ1 = f1 (t, x1 , x2 ),
ẋ2 = f2 (t, x1 , x2 ).
If the integration constants should be determined uniquely, two initial conditions have to be prescribed, e.g.
x1 (0) = b0 ,
x2 (0) = b2 .
Example 16 Find the particular solution of the system
ẋ = 3x + 8y,
ẏ = −x − 3y,
which satisfies the initial conditions x(0)=6, y(0)=−2.
Solution. First, we find the general solution of the system. The first equation renders
1
3
y = ẋ − x.
8
8
Let us differentiate it (with respect to t)
1
3
ẏ = ẍ − ẋ
8
8
and let us substitute for y and ẏ into the second equation of the system. Small formatting provides a simple second-order differential equation
ẍ − x = 0
with the general solution x=c1 et +c2 e−t . The dependence of y on x provides the expression for the other component of the solution y=− 41 c1 et − 12 c2 e−t .
The initial conditions x(0)=6, y(0)= − 2 are used to determine the values of integration constants
6 = c1 + c2 ,
1
1
−2 = − c1 − c2 .
4
2
The solution of the algebraic system is c1 =4 and c2 =2. The particular solution to have been found is thus
x = 4et + 2e−t ,
y = −et − e−t .
y
y=− 14 x
O
6
x
−2 t=0 t=0.6
t=1.1
t=1.6
t=2.1
t=2.6
Figure 4.4: The solution of the differential equations system with initial conditions.
Remark 7 The found particular solution is draw in Fig. 4.4 by the red line. It can be considered as a graph of a function given in the parametric
form, where [x; y] is a point lying on the graph of the function y=y(x) an t is the parameter. Physically, it corresponds to a motion of a mass
point in a plane with respect to time t.
Although, the graph seems almost linear, definitely, it is not a straight line but a part of a curve, a small curvature can be observed in the
neighborhood of the initial point. The reason for the shape is that the exponential function e−t is for larger t negligible with respect to the other
.
exponential function et and for sufficiently large x (and t, too), there holds y =− 14 x (the blue line in the picture). Moreover, the “motion” is
evidently accelerated: the distance of time-equally-spaced points is increasing.
Differential equations – Word problems
A differential relation is frequently sought in the physical or geometrical formulation of the problem, the solution of a differential equation is then only
one step leading to the complete solution of the problem.
Example 17 Find a curve passing through the point A[2, 3] for which the segment of any tangent bounded by the intersection points of the tangent
with the coordinate axes is halved by the tangent point.
Solution. Let us choose on a curve, considered as a graph of a function f given as y=f (x), a point B[xB , yB ] as it is shown in Fig. 4.5. The equation
of the tangent t to the curve f at the point B reads
t : y − yB = f 0 (xB ) (x − xB ) .
y
t By
3
y=f (x)
A
yB
O
B
2
xB
Bx x
Figure 4.5: The segment of the tangent is halved by the tangent point.
The tangent intersects the coordinate axes at points Bx a By , whose coordinates are
yB
Bx ∈ t ∩ ox =
xB − 0
;0 ,
f (xB )
By ∈ t ∩ oy = 0; yB − f 0 (xB )xB .
As long as the point B lies on t at the midpoint between these intersection points, its coordinates are given by the following relations:
yB
1
yB − 0
,
xB =
2
f (xB )
1
yB = yB − f 0 (xB )xB .
2
The points Bx , By and B lie on one line (the tangent), thus the equations are after some modifications, the same. Let us consider the second equation
and ignore the subscript index B as the equation should be valid for any point of f . Writing y 0 instead of f 0 (x), a linear first-order differential equation
with an initial condition determined by the point A is obtained
y + xy 0 = 0,
y(2) = 3.
The solution of this Cauchy problem is y= x6 which is the equation of the sought curve (a hyperbola).
Example 18 A mass point M of total mass m is moving rectilinearly affected by a constant force G. Simultaneously, due to resistance of the
surroundings there appears the force Fv directly proportional to the velocity of the motion and oriented against the motion direction, the coefficient of
proportionality is β. Moreover, the point is constrained such that the reaction force R is proportional (with the coefficient k) to the deflection from
the equilibrium position O and oriented to this equilibrium position, see Fig. 4.6. Find the equation of motion x=x(t), if in the beginning the point is
O
ẋ(t)
Fv =β ẋ
R=k x MG
x(t)
h
H x
Figure 4.6: Forces acting on the mass point M .
moved from the equilibrium position O to the distance h (the point H) and then released.
Solution. This classical example from mechanics can be interpreted as a motion of a mass point fixed by a spring of stiffness k which is affected by
gravity G and resistance of air Fv . The equation of motion for unknown deflexion x is written in differential form
mẍ = G − R − Fv ,
where G=m g, R=k x, Fv =β ẋ and g is the acceleration due tu gravity. The initial condition are given as follows:
x(0) = h,
ẋ(0) = 0.
The equation can be rewritten in the form
mẍ + β ẋ + kx = mg.
The characteristic equation m r2 +β r+k=0 may have real or imaginary roots, depending on the discriminant D=β 2 −4mk of the equation:
A. Ak D>0, tak
r1 =
−β +
p
β 2 − 4mk
,
2m
r2 =
−β + i
p
4mk − β 2
,
2m
p
−β − i 4mk − β 2
r2 =
.
2m
−β −
p
β 2 − 4mk
.
2m
B. Ak D<0, tak
r1 =
C. Ak D=0, tak
r1 = r2 = −
β
2m
The particular solution x∗ of the equation with non-zero right-hand side is simple as long as the right-hand side is only a constant. It is also constant:
. The general solution depends on D and there are three choices for its expression:
x∗ = mg
k
A. If D>0, then
√
√
x(t) = c1 e
−β+
β 2 −4mk
t
2m
+ c2 e
−β−
β 2 −4mk
t
2m
+
mg
.
k
B. If D<0, then
"
β
− 2m
t
x(t) = e
#
p
p
4mk − β 2
4mk − β 2
mg
t + c2 sin
t +
.
c1 cos
2m
2m
k
C. If D=0, then
β
x(t) = e− 2m t (c1 + c2 t) +
mg
.
k
The derivatives of the solutions are necessary for calculation of the integration constants. The derivatives also represent the velocity of the motion. We
obtain
A. If D>0, then
ẋ(t) = c1
−β +
p
p
√
√
β 2 − 4mk −β+ β2 −4mk t
−β − β 2 − 4mk −β− β2 −4mk t
2m
2m
e
e
+ c2
.
2m
2m
B. If D<0, then
"
"
# p
#
p
p
p
p
4mk − β 2
4mk − β 2
4mk − β 2 − β t
4mk − β 2
4mk − β 2
β −βt
e 2m c1 cos
t + c2 sin
t +
e 2m −c1 sin
t + c2 cos
t .
ẋ(t) = −
2m
2m
2m
2m
2m
2m
C. If D=0, then
β
− 2m
t
ẋ(t) = e
β
−
(c1 + c2 t) + c2 .
2m
The initial conditions also provide three different algebraic systems for determination of integration constants:
A. If D>0, then
mg
,
k
p
p
−β + β 2 − 4mk
−β − β 2 − 4mk
c1 +
c2 = 0.
2m
2m
c1 + c2 = h −
B. If D<0, then
mg
,
c1 = h −
k
p
4mk − β 2
−β
c1 +
c2 = 0.
2m
2m
C. If D=0, then
c1 = h −
mg
,
k
−β
c1 + c2 = 0.
2m
The substitution of the algebraic system solution into the general solution renders
A. If D>0, then

√
 −β+
x(t) = e
β 2 −4mk
t
2m
2
p
−β + β 2 − 4mk
−
4mk
√
e
−β−
β 2 −4mk
t
2m

mg mg

+
.
 h−
k
k
B. If D<0, then
"
β
− 2m
t
x(t) = e
#
p
p
4mk − β 2
4mk − β 2 β
mg mg
cos
sin
t+ p
t h−
+
.
2m
2m
k
k
4mk − β 2
C. If D=0, then
x(t) = e
β
− 2m
t
β
mg mg
+
1+
t h−
.
2m
k
k
The physical meaning of these options is left up to physicists.
Self-assessment questions and problems for solving
1. Why the number of constants in general solution of a differential equation is equal to the order of the equation?
2. Can you prove the formulas for solving linear differential equations of the first order?
3. Find a reason, why the linear equation with zero right-hand side can be solved by the characteristic equation!
4. Can you prove the formula for solving the higher order differential equation by variation of constants?
5. Boundary conditions are prescribed in the case when the equation is solved on a bounded interval. How would you define the boundary conditions
for a differential equation of the second order? Is the solution of the differential equation with boundary conditions always unique?
6. Try to find situations in the subjects of your specialization, when the boundary conditions are prescribed instead of initial ones!
7. Referring to your knowledge from other fields, classify and try to solve differential equations used in specialized subjects!
In the problems 1. to 18. solve the differential equations by the method of variable separation.
√
0
1. 2y x = y .
√
0
2
2
2. y 1 − x − y − 1 = 0 .
0
3. x + xy + y (y + xy) = 0 .
3
x
2x
0
5. e sin y + (1 + e ) cos yy = 0 .
y
2
y
4. e (1 + x )dy − 2x(1 + e )dx = 0 .
p
√
0
2
2
6. x 1 + y + yy 1 + x = 0 .
0
7. y =
0
2
8. xy = (1 + y ) arctg y .
9.
√1
x
1−2x
y2
+
y0
√
y
.
= 0.
2
3
0
10. x (y + 1) + (x − 1)(y − 1)y = 0 .
√
√
11. cos xdx − xdy = 0 .
12. (y + 4)dx + (x − 5)dy = 0 .
x+y
0
.
13. y = 10
0
2
14. xyy = 1 − x .
2
0
15. y − y + xy = 0 .
2
0
16. 1 + y + xyy = 0 .
3 0
17. 2y − x y = 0 .
0
2
2
18. y − xy − y − xy − y = 0 .
In the problems 19. to 28. find the solution of the differential equation, which satisfies the initial condition.
19.
x
1+y
−
yy 0
1+x
= 0, y(0) = 1 .
0
21. y ln y + xy = 0, y(1) = 1 .
√
0
2y
x = y, y(4) = 1 .
23.
x
0
x
25. (1 + e )yy = e , y(0) = 1 .
√
1
0
2
27. y 1 − x = xy, y(0) = e .
20.
1+y 2
1+x2
− y 0 = 0, y(0) = 1 .
π
0
22. y sin x = y ln y, y( 2 ) = 1 .
0
24. sin y cos xy = cos y sin x, y(0) =
π
4
.
π
0
26. y sin x − y cos x = 0, y( 2 ) = 1 .
28.
1
dy
2y+1
− cotg xdx = 0, y( π4 ) = 12 .
In the problems 29. to 46. solve the first order linear differential equations.
0
−x
29. y + 2y = e .
2
0
31. y + x y =
0
33. y −
2
y
x+1
e−x
x
3
0
30. y − x y = x .
2
.
= (x + 1)3 .
0
32. xy + y = ln x + 1 .
0
34. y −
y
x
= x.
0
35. y cos x − y sin x = sin 2x .
0
36. y cos x + y sin x = 1 .
0
2x
37. y + 2y = e .
0
3
38. xy + y = x .
0
3
39. xy − 2y = x cos x .
2 0
40. (1 + x )y + y = arctg x .
0
41. tg xy − y = 2 .
0
42. y −
2x
y
x2 +1
= x2 + 1 .
0
2
43. y − y tg x = 2 cos x .
0
44. y +
x
y
x2 +1
=
0
45. y +
0
46. y x ln x − 2y = ln x .
1
y
x+1
= sin x .
√sin x
x2 +1
.
In the problems 47. to 52. find the solution of the first order linear differential equations satisfying the initial conditions.
0
2
48. xy + y = 3x , y(1) = 1 .
√
y
0
x
50. y − 2(x+1) = e x + 1, y(0) = 0 .
1
0
47. y + 3y = x, y( 3 ) = 1 .
3
0
49. y + x y =
2
,
x3
y(1) = 1 .
π
0
51. y + y cotg x = sin x, y( 2 ) = 1 .
0
52. y −
y
1−x2
= x + 1, y(0) = 0 .
In the problems 53. to 67. solve the homogeneous differential equations.
0
2 y
53. xy = y + x cos x .
2
0 2
54. y + y (x − xy) = 0 .
y
0
56. xy = y ln x .
0
57. y =
y2
x2
0
61. y =
2x+y
x
0
64. y =
2xy
x2 −y 2
− 2.
y
0
55. xy − y = xe x .
2
2
0
58. x + y = 2xyy .
59. xy 0 cos xy = y cos xy − x .
0
60. xy − y =
0
63. y =
x−y
x−2y
0
66. y =
x
y
p
.
+ xy .
x2 + y 2 .
3 0
2
2
62. x y = y(y + x ) .
.
.
0
65. y −
y
x
= tg xy .
0
67. (x + y)y − y = 0 .
In the problems 68. to 73. find the solution of the first order homogeneous differential equations satisfying the initial conditions.
0
68. y =
xy
,
x2 +y 2
y(0) = 1 .
0
70. x + 2y − xy = 0, y(2) = 2 .
69. y +
p
x2 + y 2 − xy 0 = 0, y(1) = 0 .
y
0
71. (xy − y) arctg x = x, y(1) = 0 .
2
2 0
72. (y − x )y + xy = 0, y(1) = 1 .
2
2
2
2 0
73. x + 2xy − y = (x − 2xy − y )y , y(2) = 2 .
In the problems 74. to 85. find the solution of the Bernoulli differential equations.
0
2
74. y + 2xy = 2xy .
0
76. y +
y
x
= y 2 ln x .
0
78. 2y + y =
x
y
2
0
80. y + x y =
2 sin x √
y.
x
.
0
82. y +
y
2
= 12 (x − 1)y 3 .
0
84. y −
y
x
=
1
2y
.
0
3
75. y + xy = xy .
√
0
77. y + y = x y .
0
79. 2y ln x +
y
x
=
cos x
y
.
2
0
3
81. y 2 sin x + y cos x = y sin x .
4
0
83. y − 3 y =
x+1
3y 2
.
0
2
85. xy + y = y ln x .
In the problems 86. to 89. find the solution of the Bernoulli differential equations satisfying the initial conditions.
0
2
86. y + y = 2y , y(0) = 2 .
0
88. y +
y
x
= y 2 , y(1) = 1 .
0
87. 2y +
x
y
1−x2
= xy , y(0) = 1 .
0
89. y + 2y tg x =
√
y cos2 x, y( π4 ) = 0 .
In the problems 90. to 99. find the general solution of the exact differential equations.
90. (3x2 − 2x − y)dx + (2y − x + 3y 2 )dy = 0 .
91. (6xy + x2 + 3)dy + (3y 2 + 2xy + 2x)dx = 0 .
92. (1 +
y2
)dx
x2
94. (2x +
96.
x2 +y 2
)dx
x2 y
x
dy
x2 +y 2
2
2
2
2
93. x(2x + y )dx + y(x + 2y )dy = 0 .
− 2 xy dy = 0 .
−
x2 +y 2
dy
xy 2
x
x
x
95. (1 + e y )dx + e y (1 − y )dy = 0 .
= 0.
y
− ( x2 +y
2 − 1)dx = 0 .
x 2
2
x
97. e (x + y + 2x)dx + 2ye dy = 0 .
y
y
99. e dx + (xe − 2y)dy = 0 .
2
2
98. x(x + 2y)dx + (x − y )dy = 0 .
In the problems 100. to 119. solve the differential equations with zero right-hand side.
00
0
100. y − 5y = 0 .
00
101. y − 9y = 0 .
00
0
102. y + 3y − 4y = 0 .
00
0
103. y + 6y + 9y = 0 .
00
0
104. y − 4y + 13y = 0 .
00
105. y + 16y = 0 .
00
0
106. y + 2y + 5y = 0 .
00
0
107. y − 6y + 8y = 0 .
00
108. y + 25y = 0 .
x
x
109. y = c1 e 2 + c2 xe 2 0 .
00
0
110. 2y + 11y − 6y = 0 .
00
0
111. 9y + 18y + 13y = 0 .
00
112. y + y = 0 .
000
0
113. y − y = 0 .
000
00
114. y − y = 0 .
000
00
0
115. y − 3y + 3y − y = 0 .
000
0
116. y − 3y − 2y = 0 .
000
0
117. y − 4y = 0 .
000
118. y + 8y = 0 .
0000
119. y + 4y = 0 .
5
5
In the problems 120. to 127. solve the linear differential equations with special right-hand side by the method of indeterminate coefficients.
00
0
x
120. y + 3y = e .
00
0
2x
121. y − 8y + 16y = xe .
00
0
122. y − 2y + 3y = x + 1 .
00
x
123. y + y = 4xe .
00
3x
124. y − 4y = 10e .
00
0
x
125. y − 3y + 2y = e .
00
0
−2x
.
126. y + 4y + 4y = e
00
127. y + y = 2 sin x − cos x .
In the problems 128. to 133. solve the equation y 00 −7y 0 +10y=f (x) for given f (x).
128. 40 .
2
129. 20x − 28x + 14 .
3x
130. −12e .
2x
131. −e (6x + 7) .
2
5x
132. (9x + 6x − 3)e .
2x
133. 8e sin x .
In the problems 134. to 136. solve the equation 3y 00 −4y 0 =f (x) for given f (x).
3
134. −32x + 84x + 50 .
4x
135. 4e 3 .
136. 25 sin x .
In the problems 137. to 139. solve the equation 9y 00 −6y 0 +y=f (x) for given f (x).
x
−
137. 4e 3 .
x
138. sin 3 .
2
139. 9x − 6x + 1 .
In the problems 140. to 142. solve the equation y 00 +4y=f (x) for given f (x).
140. cos 2x .
141. cos 3x .
−2x
.
142. e
In the problems 143. to 145. solve the equation y 00 −4y 0 +5y=f (x) for given f (x).
2x
143. e .
144. sin x .
2x
145. e sin x .
In the problems 146. to 151. solve differential equations.
00
0
x
146. y − y − 2y = 4x − 2e .
00
0
147. y − 3y = 18x − 10 cos x .
00
0
x
148. y − 2y + y = 2 + e sin x .
00
0
x
−x
149. y + 2y + 2y = (5x + 4)e + e .
00
0
2x
150. y − 4y + 5y = 1 + 8 cos x + e .
00
0
−x
−2x
.
151. y + 5y + 6y = e + e
In the problems 152. to 157. find the solution of differential equations satisfying the initial conditions.
152. y 00 − 6y 0 + 9y = 9x2 − 12x + 2, y(0) = 1, y 0 (0) = 3 .
153. y 00 + y = 2 cos x, y(0) = 1, y 0 (0) = 0 .
154. 4y 00 + y = 0, y(π) = 2, y 0 (π) = 3 .
155. y 00 − y 0 = −5e−x (sin x + cos x), y(0) = −4, y 0 (0) = 5 .
156. y 000 − y 0 = −2x, y(0) = 0, y 0 (0) = 1, y 00 (0) = 2 .
157. y 00 − 4y 0 + 3y = e2x , y(0) = 0, y 0 (0) = 0 .
In the problems 158. to 165. solve the linear differential equations by the method of variation of constants.
00
0
158. y − 6y + 9y =
00
160. y + y =
1
sin x
9x2 +6x+2
x3
.
.
00
0
159. y − 2y + y =
ex
x
00
161. y + 4y =
.
1
sin 2x
00
2
162. y + y = − cotg x .
00
163. y + y =
00
164. y + y =
00
165. y + y = tg x .
2
sin3 x
.
1
cos3 x
.
.
In the problems 166. to 177. solve the systems by elemination.
ẋ = 2x + y,
166. ẏ = 3x + 4y .
ẋ = x − 3y,
167. ẏ = 3x + y .
ẋ = −x + y,
168. ẏ = 3x + y .
ẋ = 4x − y,
169. ẏ = x + 2y .
ẋ = y + t,
170. ẏ = x − t .
ẋ = −x + y + et ,
171. ẏ = x − y + et .
ẋ = −2x − y + sin t,
172. ẏ = 4x + 2y + cos t .
ẋ = 4x − y − 5t + 1,
173. ẏ = x + 2y + t − 1 .
ẋ = −5x + 2y + 40et ,
174. ẏ = x − 6y + 9e−t .
ẋ = y − cos t,
175. ẏ = −x + sin t .
ẋ = −5x + 2y + et ,
176. ẏ = x − 6y + e−2t .
ẋ = y,
177. ẏ = x + et + e−t .
178. Find a curve passing through the point A[2, 0] with the property that the length of the tangent segment between the tangent point and the axis
Oy is 2 .
179. A mass point is moving along a line such that its kinetic energy in time t is directly proportional to the average velocity in the interval from 0 to
t. At the time t=0 the path s=0. Show that the motion is uniform .
180. Determine the time required by a body heated up to 100◦ C to be cooled to 25◦ C in a room with temperature 20◦ C, if it is cooled to 60◦ C during
10 minutes. According to the Newton law the rate of cooling is proportional to the temperature difference .
181. Find all curves for which the tangent segment between the tangent point and the axis ox is split to halves by the axis oy .
182. A tank is filled by 100l of water solution containing 10kg od salt. Water flows into the tank at a rate 3l per 1 minute, the uniform concentration
is maintained by stirring the solution in tank. Calculate the amount of salt in tank after 1 hour .
183. The body of mass 1kg started to move at time t=0s due to action of a force which was directly proportional to the time and indirectly proportional
to the velocity of the body. At the time t=10s, the velocity was v=0.5ms−1 and the force was F =4×10−5 N. Determine the velocity of the body after
one minute .
184. Find the curves, whose tangent distance from the origin is equal to the x coordinate of the tangent point .
185. Find the curves, whose arbitrary tangent segment on the axis oy has length equal to the x coordinate of the tangent point .
186. Find the curves, whose arbitrary tangent intersects with the axis ox at a point equally distant both from the origin and from the tangent point .
187. Find the curves, whose tangent at an arbitrary point P forms together with the axis oy an the segment OP , where O is the origin of the coordinate
system, a triangle of the area a2 .
188. Find the curves, whose tangent at an arbitrary point P0 [x0 , y0 ] makes a segment on the axis oy of the length y02 .
189. Find a curve, which passes through the point O and for which the midpoint between arbitrary tangent point and intersection point of the pertinent
normal to the curve at the same point with the axis ox lies on the parabola y 2 =ax .
190. How much time needs a boat, moving only by own inertia, to slow down to 1cms−1 , if its initial velocity is 1.5ms−1 . The resistance of water is
proportional to the velocity, such that during the first four seconds the boat is slowed down to 1ms−1 .
191. A mass point X of mass m is attracted by each point S1 , S2 while the force is directly proportional to the distance from the pertinent point. Find
the equation of motion for the point X knowing that in the initial instant the point X lies on the segment S1 S2 at the distance x0 from its midpoint
and the velocity is zero .
192. A cylindrical wooden block with cross-sectional area S, height h and mass density ρ floats in water. Its axis is vertical. Find the period of the
oscillation, which is invoked by pushing the cylinder to the water and subsequently releasing it. Neglect any resistance of the surroundings .
193. A body of mass m moves straightforwardly under the acting constant force F0 – free fall. Simultaneously, the resistance of the surroundings is
considered which is proportional to the velocity of the motion with proportionality coefficient β. Find the equation of the motion x=x(t), if x(0)=0 and
also velocity is zero at the time t=0. Find also velocity as a function of time .
Conclusion
The solution of several types of ordinary differential equations was discussed in the chapter. We learned to solve the first order differential equations,
where the variable can be separated. We also learned to solve the linear first order differential equation. Besides we demonstrated the solution of other
types of the first order equations: homogeneous, Bernoulli and exact differential equations. For all these equations, we are able to prescribe an initial
condition and to find the solution satisfying this condition. For the case of linear differential equations we discussed also higher order equations though
only with constant coefficients. Now, we also know the methods of unique determination of integration constants by the initial conditions. We also
solved systems of differential equations of the first order which we are able to transform to a sole equation of a higher order. In the final part of the
chapter, several word examples and problems were presented which lead to differential equations discussed in the previous parts. In practice, definition
of the problem by words instead of equations and transformation of the word expression to the differential equation with boundary or initial conditions
is exactly what can be applied in studying other disciplines. Of course, crucial is the ability to solve such equations.
References
[1] Šiagová, J.: Advanced Mathematics, Bratislava, STU, 2011.
[2] Krivá, Z.: Mathematics I, Bratislava, STU, 2007.
[3] Bubeník, F.: Problems to Mathematics for Engineers, Praha, ČVUT, 1999.
[5] Neustupa, J., Kračmar S.: Problems in Mathematics I, Praha, ČVUT, 1998.
[6] Neustupa, J.: Mathematics I, Praha, ČVUT, 2004.
Problem solutions
The answers to the self-assessment questions, if you are unable to formulate, and also a lot of other answers can be found in provided references.
p
√
3. ex+y = c|(y + 1)(x + 1)| 4. 1 + ey = c(1 + x2 ) 5. arctg ex = 2 sin1 2 y + c 6. 1 + x2 + 1 + y 2 = c
1
√
√
√
7. y 3 = 3x − 3x2 + c
8. y = tg (cx)
9. x + y = c
10. ey (x3 − 1) 3 = c(y + 1)2
11. y = 2 sin x + c
12. (x − 5)(y + 4) =
1
2
−
y
1
c
13. 10x + 10−y = c
14. x2 + y 2 = ln cx2
15. y = 1−cx
16. x2 (1 + y 2 ) = c
17. y = ce x2
| = x2 + x + c
18. ln | y+1
√
√
√
√
21. y = 1, x 6= 0 22. y = 1 23. y = e x−2 24. cos x = 2 cos y 25. 2 ey2 = e(1+ex )
19. 2(x3 −y 3 )+3(x2 −y 2 ) = −5 20. y = 1+x
1−x
√
1. y = ce
x
2. arctg y = arcsin x + c
√
2
−x2
26. y = sin x 27. y = e− 1−x
28. y = 2 sin2 x − 12
29. y = ce−2x + e−x 30. y = cx3 − x2 31. y = c−e
32. y = ln x + xc
33. y =
2x2
2x
3
c−cos 2x
e
x
c
2 1 2
2
−2x
2
35. y = 2 cos x
36. y = c cos x + sin x 37. y = 4 + ce
38. y = 4 + x
39. y = x sin x + cx2
(x + 1) ( 2 x + x + c 34. y = cx + x
√
c
40. y = ce− arctg x + arctg x − 1 41. y = c sin x − 2 42. y = (x2 + 1)(c + x) 43. y = 32 tg x(3 − sin2 x) + 3 cos
44. y = (c − cos x) x2 + 1
x
√
x
50. y = (ex − 1) x + 1
45. y = c+sin
− cos x
46. y = ln(c ln x − 1)
47. y = e1−3x + x3 − 91
48. y = x2
49. y = x22 − x13
x+1
q
y
y
2x+4−π
1
1+x
51. y = 2x−sin
52.
y
=
(x(x
+
1)
−
arcsin x)
53. tg xy = ln cx
54. y = ce x
55. e− x = − ln cx
56. y = xe1+cx
4 sin x
2
1−x
p
x2
y
3
2
2
2 y2
2
2
57. y − 2x = cx (y + x)
58. y = x(x − c)
59. sin x + ln |x| = c
60. y + x + y = cx
61. y = 2x ln |x| + cx
62. x e = c
x
2
y
2
2
2
2
2
2
2
2
y
63. x − 2xy + 2y = c
64. x + y = cy
65. sin x = cx
66. y = 2x ln |cx|
67. y = ce
68. 2y ln y − x = 0
69. y = x 2−1
p
√
y
y
2
2
1
70. y = x2 −x 71. e x arctg x = x2 + y 2 72. xy2 = 1−2 ln |y| 73. y = 1+ 1 + 2x − x2 74. y = 1+ce
, y = 0 75. y 2 (cex +1) = 1, y = 0
x2
√
√
76. xy(ln2 x + c) = −2, y = 0
77. y = (c e−x + x − 2)2 , y = 0
78. y = ( ce−x + x − 1
79. y 2 ln x = c + sin x, y = 0
80. y =
2
(c−cos x)
5
1
2
2
x
3
4x x
2
2
, y = 0 81. y (c−x) sin x = 1, y = 0 82. y (ce +x) = 1, y = 0 83. y = ce 4 − 16
84. y = cx −x 85. y = cx+ln x+1 , y = 0
x2
86. y =
2
4−3ex
2
87. y =
p √
2 1 − x2 + x2 − 1
88. y =
2
1
x(1−ln x)
2 2
4
89. y =
1
16
3
cos2 x(2 sin x −
√ 2
2)
90. x3 + y 3 − x2 − xy + y 2 = c
x
91. 3y 2 x + x + x2 y + 3y = c
92. x − yx = c
96. x + arctg xy = c
93. x4 + x y + y = c
94. x y + x2 − y 2 = cxy
95. x + ye y = c
98. x3 + 3x2 y − y 3 = c
99. xey − y 2 = c
100. y = c1 e5x + c2
101. y = c1 e3x + c2 e−3x
102. y = c1 ex + c2 e−4x
97. (x2 + y 2 )ex = c
103. y = c1 e−3x + c2 xe−3x
104. y = c1 e2x cos 3x + c2 e2x sin 3x
105. y = c1 cos 4x + c2 sin 4x
106. y = c1 e−x cos 2x + c2 e−x sin 2x
5
1
5
x
x
x
−6x
2x
4x
110. y = c1 e + c2 e 2
111. y = c1 e−x cos 23 x + c2 ex sin 23 x
107. y = c1 e + c2 e
108. y = c1 cos 5x + c2 sin 5x 109. y = c1 e 2 + c2 xe 2
112. y = c1 cos x+c2 sin x 113. y = c1 +c2 ex +c3 e−x √
114. y = c1 +c2√
x+c3 ex 115. y = c1 ex +c2 xex +c3 x2 ex 116. y = c1 e−x +c2 xe−x +c3 e2x
2x
−2x
−2x
x
x
x
−x
117. y = c1 +c2 e +c3 e
118. y = c1 e +c2 e cos( 3x)+c3 e sin( 3x) √ 119. y = (c√
120. y =
1 cos x+c2 sin x)e +(c3 cos x+c4 sin x)e
x
1 x
1
5
x
−3x
4x
4x
2x
x
c1 + c2 e + 4 e
121. y = c1 e + c2 xe + 4 (x + 1)e
122. y = c1 e cos 2x + c2 e sin 2x + 3 + 9
123. y = c1 cos x + c2 sin x + (2x − 2)ex
1
124. y = c1 e2x + c2 e−2x + 2e3x 125. y = c1 e2x + c2 ex − xex 126. y = c1 e−2x + c2 e−2x + 2 x2 e−2x 127. y = c1 cos x + c2 sin x − (cos x + 21 sin x)x
128. y = c1 e2x + c2 e5x + 4
129. y = c1 e2x + c2 e5x + 2x2 + 1
130. y = c1 e2x + c2 e5x + 6e3x
131. y = c1 e2x + c2 e5x + (x2 + 3x)e2x
4x
132. y = c1 e2x + c2 e5x + (x3 − x)e5x
133. y = c1 e2x + c2 e5x + (3 cos x − sin x)e2x
134. y = c1 + c2 e 3 + 2x4 + 6x3 + 3x2 − 8x
4x
4x
4x
x
x
x
x
x
135. y = c1 + c2 e 3 + xe 3
136. y = c1 + c2 e 3 + 4 cos x − 3 sin x
137. y = c1 e 3 + c2 xe 3 + e− 3
138. y = c1 e 3 + c2 xe 3 + 21 cos x3
x
x
140. y = c1 cos 2x + c2 sin 2x + 14 x sin 2x
139. y = c1 e 3 + c2 xe 3 + 9x2 + 102x + 451
141. y = c1 cos 2x + c2 sin 2x − 15 cos 3x
142. y =
143. y = e2x (c1 cos x + c2 sin x) + e2x
144. y = e2x (c1 cos x + c2 sin x) + 81 (cos x + sin x) 145. y = e2x (c1 cos x +
c1 cos 2x + c2 sin 2x + 81 e−2x
c2 sin x) − 21 e2x sin 2x 146. y = c1 e−x + c2 e2x − 2x + 1 + ex
147. y = c1 + c2 e3x − 3x2 − 2x + cos x + 3 sin x 148. y = c1 ex + c2 xex − sin xex + 2
−x
x
−x
149. y = (c1 cos x + c2 sin x)e + xe + e
150. y = (c1 cos x + c2 sin x)e2x + cos x − sin x + e2x + 51
151. y = c1 e−2x + c2 e−3x + 12 e−x + xe−2x
155. y = 2ex + (sin x − 2 cos x)e−x − 4 156. y = 12 e−x + 23 ex + x2 − 2
152. y = x2 + e3x
153. y = cos x + x sin x 154. y = 2 sin x2 − 6 cos x2
157. y = 12 ex − e2x + 21 e3x
158. y = c1 e3x + c2 xe3x + x1
159. y = (x ln |x| + c1 x + c2 )ex
160. y = (c1 + ln | sin x|) sin x + (c2 − x) cos x
1
2x
1
x
+
cos
x
ln
| tg x2 | + 2 163. y = c1 cos x + c2 sin x − 2cos
161. y = c1 cos 2x + c2 sin 2x + 4 cos 2x ln | cos 2x| + 2 sin 2x 162. y = c1 cos x + c2 sin
cos x
x+1 165. y = c1 cos x + c2 sin x + 21 cos x ln sin
166. x = c1 et + c2 e5t , y = −c1 et + 3c2 e5t
167. x =
164. y = c1 cos x + c2 sin x + 2cossin2xx
sin x−1
1
t
t
−2t
2t
−2t
2t
3t
3t
3t
e (c1 cos 3t + c2 sin 3t), y = e (c1 sin 3t − c2 cos 3t) 168. x = −c1 e + 3 c2 e , y = c1 e + c2 e
169. x = c1 e + c2 te , y = c1 e + c2 (t − 1)e3t
t
−t
t
−t
−2t
t
−2t
t
170. x = c1 e −c2 e +t−1, y = c1 e +c2 e −t+1 171. x = c1 +c2 e +e , y = c1 −c2 e +e
172. x = c1 +c2 t+2 sin t, y = −2c1 −c2 (2t+
3t
3t
3t
3t
−4t
−7t
1) − 3 sin t − 2 cos t 173. x = c1 e + c2 te + t, y = c1 e + c2 (t − 1)e − t 174. x = c1 e + c2 e + 7et + e−t , y = 12 c1 e−4t − c2 e−7t + et + 2e−t
7 t
1 t
3 −2t
175. x = c1 cos t + c2 sin t − t cos t, y = −c1 sin t + c2 cos t + t sin t 176. x = c1 e−4t + c2 e−7t + 10
e + 15 e−2t
, y = −c1 e−4t + c2 e−7t + 40
e + 10
e
√
√
2
1 t
1 −t
1
2− 4−x
t
−t
t
−t
t
−t
2
2
177. x = c1 e − c2 e + 2 e (t − 1) − 2 e (t + 1), y = c1 e − c2 e + 2 t(e + e )
178. 4 − x + 2 ln | x |
180. ≈ 40min
181. y =cx
x
−1
2
2
2
2
2
2
182. ≈ 3.9kg
183. ≈ 2.7ms
184. x + y = cx
185. y = cx − x ln |x|
186. x + y = cy
187. a + cy = xy
188. y = c+x
x
k
k
189. y 2 = 4ax + 4a2 (1 − e a )
190. ms̈ = −k ṡ; ṡ = v = v0 e− m
t; s = m
v (1 − e− m t); t = m
ln vv0 ≈ 50s
191. mẍ = k( d2 − x) − k( d2 + x); x
k 0
k
- deflection of the point X from the mid point of the segment S1 S2 inqthe direction of the initial deflection x0 ; d is length of the segment S1 S2 ;
2k
x
m
2k
t
m
192. Sρẍ = −Sρv xg, x is the deflection of the block from the
q
ρv
ρv
equilibrium downwards and ρv is the density of water; equation: ẍ + ρ gx = 0; solution: x = c1 cos ωt + c2 sin ωt, ω =
g; period T = 2π
ρ
ω
equation: ẍ +
193. ẍ +
β
ẋ
m
=
F0
,
m
= 0, x(0) = 0, ẋ(0) = 0; solution: x = x0 cos
x(0) = 0, ẋ(0) = 0; x =
F0
(t
β
−
m
(1
β
β
− e− m t)); v =
F0
(1
β
β
− e− m t)
Chapter 5. (Double integral)
Aim
Introduce the students to double integral, concentrating methods of calculation and its applications.
Objectives
1. Determine the type of a domain, defining it geometrically.
2. Understand the notion of double integral on a domain and its calculation using partial integration.
3. Simplify the integrand or integration domain by an appropriate substitution.
4. Learn to apply double integrals in calculating various geometrical and physical characteristics of bodies.
Prerequisities
function; graph of a function; curve; derivative; partial derivative; determinate and indeterminate integral
Chapter 5. (Double integral) – 1
Introduction
This chapter discusses calculation techniques for an integral of a function of two variables on a bounded domain. Many geometrical and physical
quantities which can be met solving engineering problems are defined by integrals. Therefore, the calculation techniques for integrals have to be known
to any engineer, this is also the case of multiple integrals as long as practically all bodies are three dimensional in reality. The physical quantities are
defined on such bodies: as a simple example density of an nonhomogeneous body case can be considered. Although, the case of triple integral is not
discussed directly, the method explained for the double integral, as a reasonable simplification, are directly applicable also for the spacial integral.
First, an efficient way for calculation of double integral includes an efficient way of the integration domain description. Such a description enable to
convert the double integral into two simple integrals calculated subsequently. It will be also shown that under some condition the order of the integration
can be interchanges to define a better way of integral calculation.
Second, geometric description of the integration domain is crucial. Frequently, it is advantageous to use a substitution to change such a description to
simplify the calculation. Similarly to the simple integral, the substitution can lead to a simpler integral. Unlike the simple integral, however, where the
integration domain is always an interval, here the substitution is often used to modify and simplify the domain.
Finally, several examples for double integral applications will be provided, such as area of a domain, volume of a body, centroid of a planar region etc..
Double integral – Elementary domain and subsequent integration
The description of an integration domain has to be given in a special from, as it has to be able to provide the bounds of the integrals for each variable.
An elementary domain type [x, y], Fig. 5.1(a), is a set D of points in R2 , for which there exist continuous functions g1 and g2 in the interval ha; bi such
that for any x∈(a; b) holds g1 (x)<g2 (x) and which is given as: D={(x, y) ∈ R2 : a ≤ x ≤ b ∧ g1 (x) ≤ y ≤ g2 (x)}.
An elementary domain type [y, x], Fig. 5.1(b), is a set D of points in R2 , for which there exist continuous functions h1 a h2 in the hc; di such that for
any y∈(c; d) holds h1 (y)<h2 (y) and which is given as: D={(x, y) ∈ R2 : h1 (y) ≤ x ≤ h2 (y) ∧ c ≤ y ≤ d}.
A special case of a domain which belongs to both types [x, y] and [y, x] is also a two dimensional interval or a rectangle ha; bi×hc; di. A domain which
is not an elementary domain has to be divided into several parts such that they are already elementary domains.
Let there is a function f continuous in the elementary domain D of the type [x, y] (pertinent bounding functions are denoted g1 (x) and g2 (x) and
points a and b). For each x∈ha;bi, there is a continuous function hx (y)=f (x, y) defined in the interval y∈hg1 (x);g2 (x)i. Its integral depending on
Z g2 (x)
the parameter x is
hx (y) dy, and it is a function F (x) continuous in the interval x∈ha;bi. The function F (x) is integrable in the interval ha;bi
g1 (x)
y
d
y
D
g1 (x)
O
h2 (y)
h1 (y)
g2 (x)
D
c
a
b
x
x
O
(a)
(b)
Figure 5.1: Elementary domains: (a) type [x, y], (b) type [y, x].
rendering
Zb
Zb
F (x) dx =
a



a

gZ2 (x)
Zb

hx (y) dy  dx =


f (x, y) dy  dx.


a
g1 (x)

gZ2 (x)
g1 (x)
The last integral contains subsequent integration. Similarly, subsequent integration of a function f (x, y) on an elementary domain of the type [y, x] can
be introduced.
If the elementary domain D is simultaneously of both types [x, y] and [y, x] and the function f (x, y) is continuous in D, then the order of integration
can be interchanged, i.e.




gZ2 (x)
hZ2 (y)
Zb
Zd




f (x, y) dy  dx =
f (x, y) dx dy.


a
c
g1 (x)
ZZ
Example 1 Interchange the order of integration in the integral
h1 (y)
Z
f (x, y) dx dy=
1
2
Z
2x2
f (x, y) dy
!
dx considering f a continuous function in
1
D
the domain D.
Solution. The subsequent integration is calculated first for the variable y with fixed x (e.g. with x1 along the drawn segment) and only then for the
variable x, the domain D is thus defined as elementary, type [x, y]. It can be written as D = {(x, y) ∈ R2 : 1 ≤ x ≤ 2 ∧ 1 ≤ y ≤ 2x2 } and it is draw
in left graph of Fig. 5.2. The domain has to be rewritten as an elementary of the type [y, x], in order to interchange the integration order. The left
y
y=2x2
y
r
y
x=
2
8
8
y2
D2
D
2
2
1
y1
1
O
1
x1
2
x
O
D1
1
x
2
Figure 5.2: Interchanging the integration order.
graph of Fig. 5.2 confirms that the domain D is not elementary of the type [y, x]. The point of the domain have y coordinates in the interval
h1; 8i,
p
nevertheless for y ∈ h1; 2i the x coordinates of pertinent points lie in the interval h1; 2i (e.g. for y1 ), while for y ∈ h2; 8i x depends on y as y2 ≤ x ≤ 2.
The domainD is to be split
two subdomains D1 , D2 , Fig. 5.2, right. These domain are written as D1 = {(x, y) ∈ R2 : 1 ≤ x ≤ 2 ∧ 1 ≤ y ≤ 2}
pint
y
2
and D2 = {(x, y) ∈ R :
≤ x ≤ 2 ∧ 2 ≤ y ≤ 8}. They both sum to whole D D = D1 ∪ D2 and do not intersect in the interior. According to the
2
property V2 (see page Chapter 5. (Double integral) – 5) we obtain






2
2
2x
2
2
8
2
ZZ
Z
Z
Z
Z
Z
Z






 dy,

f
(x,
y)
dx
f (x, y) dx dy =
 f (x, y) dy  dx =
 f (x, y) dx dy +


√y
1
1
1
1
2
D
2
and the order of integration is interchanged (i.e. first x, second y).
Double integral – Double integral, the Fubini theorem
The character of physical and geometrical quantities is usually closely related with the domain, with no need to know about the type of the domain.
Thus the integral is refferd to as integral over a planar domain.
The double integral over a closed bounded domain D of a function f (x, y) which is bounded on D is a generalization of simple (definite) integral
Z b
ZZ
k(x) dx of a function k(x). The integral is denoted
f (x, y) dx dy. We will assume that D is closed bounded planar domain in what follows.
a
D
Calculation of the double integral is usually based on he method of subsequent integration using the following Fubini theorem: Let the function f is
continuous on the domain D, which is an elementary domain of the type [x, y] or [y, x]. Then, it holds respectively: platí


gZ2 (x)
ZZ
Zb


f (x, y) dx dy =
f (x, y) dy  dx

D
a
ZZ
Zd
or
f (x, y) dx dy =
g1 (x)



f (x, y) dx dy.


c
D
hZ2 (y)
h1 (y)
Basic properties additivity and homogeneity of the double integrals are the same as we learned for the simple integrals. Providing that pertinent integrals
in the following formulas, the properties are:
ZZ
V1 :
ZZ
(c1 f1 (x, y) + c2 f2 (x, y)) dx dy = c1
D
ZZ
f (x, y) dx dy =
D
f1 (x, y) dx dy + c2
D
ZZ
V2 :
ZZ
D
ZZ
f (x, y) dx dy +
D1
f2 (x, y) dx dy,
f (x, y) dx dy,
D = D1 ∪ D2 ,
D2
where the domains D1 and D2 do not have common inner points. The properties can be extended to a finite number of functions or domains.
Converting the double integral to two subsequently applied simple integrals, the method of definite integrals known from the previous course can be
applied, e. g. the method of substitution, integration by parts etc..
ZZ
Example 2 Calculate the integral
(2x + 3y 2 − 7xy) dx dy, where D = h0; 2i × h−1; 1i.
D
Solution. The function f (x, y) = 2x + 3y 2 − 7xy is continuous in the interval D, which is an elementary domain of the type [y, x], thus it is integrable
in this domain. The Fubini theorem renders


ZZ
Z1 Z2


2
(2x + 3y 2 − 7xy) dx dy =
 (2x + 3y − 7xy) dx dy =
−1
D
0
Z1 =
7
x + 3y x − x2 y
2
2
2
−1
Z1
x=2
dy =
x=0
1
(4 + 6y 2 − 14y) dy = 4y + 2y 3 − 7y 2 −1 = 12.
−1
Remark 1 As long as the domain D is also of the type [x, y], the same theorem provide alternative calculation


y=1
ZZ
Z2 Z1
Z2 Z2
2
7


2
(2x + 3y 2 − 7xy) dx dy =
dx = (4x + 2) dx = 2x2 + 2x 0 = 12.
2xy + y 3 − xy 2
 (2x + 3y − 7xy) dy  dx =
2
y=−1
0
D
−1
0
0
The result does not depend on the order of integration.
ZZ
Example 3 Calculate
(x2 + y) dx dy, where D = {(x, y) ∈ R2 : 0 ≤ x ∧ 2x2 ≤ y ≤
√
x}.
D
Solution. First, let us plot the domain D. Its boundary is formed by √parts
of two parabolas with the equations y = 2x2 and y =
√
3
3
intersection points of the curves are the points O = (0, 0) and P = ( 22 , 24 ), obtained by solving the system of equations
√
y = 2x2 ,
y = x.
√
x, Fig. 5.3. The
The domain D shown in the picture can be seen as an elementary domain of the type [x, y] defined by the inequalities
√
3
√
2
,
2x2 ≤ y ≤ x.
0≤x≤
2
y
√
3
4
2
P
√
y= x
D
y=2x2
√
3
O
2
2
x
Figure 5.3: The domain D = {(x, y) ∈ R2 : x ≥ 0 ∧ 2x2 ≤ y ≤
√
x}.
As the function f is continuous on D, hence integrable, the Fibini theorem enables the following calculation:
ZZ
√
32
2
2
√
Z Z
(x + y) dx dy =
D
√
32
x
2
2
Z (x + y) dy dx =
0
2x2
0
y
x y+
2
2
√
32
2
√
2 y= x
dx =
y=2x2
Z x
27
4
4
√ .
x + − (2x + 2x ) dx =
2
280 3 2
5
2
0
Remark 2 The domain D in the last example is also of the type [y, x]. The pertinent inequalities are then
√
r
3
4
y
2
0≤y≤
,
y ≤x≤
,
2
2
and according to the Fubini theorem we can alternatively treat it as:
 √y

√
√
34
34
2
x=√ y2
Z2
Z
Z2 3
ZZ


x

(x2 + y) dx
+ yx
dy =
(x2 + y) dx dy =

 dy =
3
x=y 2
D
0
0
y2
√
34
Z2
=
3
2
0
3
3
y2
2 3
+
y2
2
1
2
−
y6
+ y3
3
!
5
y7 y4
7
√ y2 −
dy =
−
21
4
15 2
√324
=
0
27
√ .
280 3 2
The result is the same.
Double integral – The method of substitution
Sometimes the integration domain cannot be easily described as an elementary domain or as a union of elementary domains. In such a case, the domain
has to be appropriately transformed.
The method of substitution is used, as well as for the simple definite integral, to simplify the calculation. Here, however, we have two possibilities: first,
to simplify the function, second, to simplify the domain.
Let there is an injective (possibly, besides the boundary) and regular mapping g=(gx , gy ) determined by the equations x=gx (ξ, η), y=gy (ξ, η) defined
on the domain D∗ ⊂R2 . Let the image of the domain D∗ is a domain D, i. e. g : D∗ 7→ D. Then for a continuous function f (x, y) on D holds
ZZ
ZZ
f (gx (ξ, η), gy (ξ, η)) |Jg (ξ, η)| dξ dη,
f (x, y) dx dy =
D
D∗
where Jg (ξ, η) is determinant
∂gx (ξ, η)
∂ξ
Jg (ξ, η) = ∂gy (ξ, η)
∂ξ
∂gx (ξ, η) ∂η ∂gy (ξ, η) ∂η called jacobian of the mapping g.
The most frequent substitution is the polar transformation, or sometimes generalized polar transformation. The relations required for calculation of the
integrals by this substitution are summarized in Tab. 5.1. The table also provides the types of domains which are transformed to the simplest possible
domain — interval. The pictures give us a guess of domains suitable for an appropriate transformation.
Polar coordinates (ξ, η) ↔ (ρ, ϕ)
y
ρ=co
nst.
x = ρ cos ϕ, y = ρ sin ϕ
gx (ρ, ϕ) = ρ cos ϕ, gy (ρ, ϕ) = ρ sin ϕ
ϕ=
(x, y) ∈ R , (ρ, ϕ) ∈ h0; +∞) × h−π; πi
co
n st
.
A domain D mapped to an interval
2
ρ=co
nst.
Jg (ρ, ϕ) = ρ
.
D
ϕ
(identifying the polar axis p with the axis x)
ZZ
f (ρ cos ϕ, ρ sin ϕ) ρ dρ dϕ
Generalized polar coordinates (ξ, η) ↔ (ρ̃, ϕ̃)
A domain D mapped to an interval
(x, y) ∈ R2 , (ρ̃, ϕ̃) ∈ h0; +∞) × h−π; πi
x = aρ̃ cos ϕ̃, y = bρ̃ sin ϕ̃,
y
a > 0, b > 0
ρ̃=cons
gx (ρ̃, ϕ̃) = aρ̃ cos ϕ̃, gy (ρ̃, ϕ̃) = bρ̃ sin ϕ̃
D
st.
con
D
ZZ
f (aρ̃ cos ϕ̃, bρ̃ sin ϕ̃) abρ̃ dρ̃ dϕ̃
D∗
ϕ̃
on
=c
ρ̃=
(identifying the polar axis p with the axis x)
f (x, y) dx dy =
t.
st.
Jg (ρ̃, ϕ̃) = abρ̃
ZZ
x≡ p
O
D∗
D
co
nst
.
f (x, y) dx dy =
ϕ̃=
ZZ
t
ons
c
=
O
x≡ p
Table 5.1: Substitutions for the double integrals.
ZZ
Example 4 Calculate
e−x
2 −y 2
dx dy, where D = {(x, y) ∈ R2 : 4 ≤ x2 + y 2 ≤ 9 ∧ |y| ≤ x}.
D
Solution. First, we draw the domain D such that we draw the boundary curves. They are:
circles:
x2 + y 2 = 4,
x2 + y 2 = 9,
lines:
x = y,
x = −y,
see Fig. 5.4(left). In this case, the polar transformation x = ρ cos ϕ, y = ρ sin ϕ is suitable, because the D is the image of an interval. We can verify
y
3
2
ρ=3
ϕ= π4
ρ=2
ϕ
π
4
D
O
2
3x
O
3ρ
2
D∗
g (ρ, ϕ)
−2
−3
ϕ=− π4
− π4
Figure 5.4: The domain D = {(x, y) ∈ R2 : 4 ≤ x2 + y 2 ≤ 9 ∧ |y| ≤ x} and the domain D∗ = h2; 3i × h− π4 ; π4 i, g : D∗ 7→ D.
this by substituting the transformation equations int the equations of the boundary curves. We obtain
x2 + y 2 = ρ2 cos2 ϕ + ρ2 sin2 ϕ = ρ2 ,
x = ±y ⇒ ρ cos ϕ = ±ρ sin ϕ ⇒ cos ϕ = ± sin ϕ ⇒ tg ϕ = ±1.
It implies that the boundary curves can be expressed in polar coordinates as ρ = 2, ρ = 3, ϕ = π4 , ϕ = − π4 . The domain D∗ bounded by these curves
is an interval, i. e. D∗ = {(ρ, ϕ) ∈ R2 : ρ ∈ h2; 3i ∧ ϕ ∈ h− π4 ; π4 i}. The mapping g(ρ, ϕ) = (ρ cos ϕ, ρ sin ϕ) maps the interval D∗ onto the domain
D. This mapping is also shown in Fig. 5.4, where the domain D∗ is drawn in a rectangle coordinate system Oρϕ such that x is replaced by ρ and y is
replaced by ϕ. In this way the domain D∗ is also optically represented by an interval in the coordinate system Oρϕ.
As the function f (x, y) = e−x
2 −y 2
is continuous on the domain D, the integral exists. Using the method of substitution, we obtain
π
π
ZZ
e
D
−x2 −y 2
ZZ
−ρ2 cos2 ϕ−ρ2 sin2 ϕ
e
dx dy =
Z4 Z3
e
ρ dρ dϕ =
− π4
D∗
−ρ2
Z3
Z4
ρ dρ dϕ =
− π4
2
e
dϕ
−ρ2
π
ρ dρ =
2
ρ2 =t
2
Z9
e−t
π −4
1
dt =
e − e−9 .
2
4
4
Remark 3 If the direct integration were used, the domain would be necessarily split int several parts in order to have been able to find the
integration bounds and to use the Fubini theorem.
ZZ
Example 5 Caculate by the method of substitution
xy 2 dx dy, where
D
D = {(x, y) ∈ R2 : x ≥ 0 ∧ y ≥ 0 ∧ 1 ≤ xy 2 ≤ 4 ∧ x ≤ y 3 ≤ 8x}.
Solution. Though, in the most of the problems we solve, the polar coordinates are used, there exist examples where other substitution is advantageous.
This is one of them.
The integration domain is drawn in Fig. 5.5(left). The domain look trivial even for direct integration and we would be able to find the integration
bounds for the subsequent integration in the Fubini theorem. Nevertheless, an appropriate transformation can simplify the calculation. Let us consider
the curves with the equations
y3
y3
= 1,
= 8,
xy 2 = 1, xy 2 = 4,
x
x
which form the contour of the domain D. In this case a power transformation can be effectively used:
(x, y) ∈ (0; +∞) × (0; +∞), (u, v) ∈ (0; +∞) × (0; +∞) : x = uα1 v β1 ,
where α1 , α2 , β1 , β2 ∈ R, and α1 β2 − α2 β1 = κ 6= 0,
β2
β1
with jacobian Jg (u, v) = κ uα1 +α2 −1 v β1 +β2 −1 and inverse mapping u = x κ y − κ ,
We define new variables u and v as
u = xy 2 ,
v=
y = uα2 v β2 ,
α2
v = x− κ y
α1
κ
.
y3
,
x
y
v
8
xy 2 =4
y3
=8
x
g (u, v)
y3
=1
x
D
1
xy 2 =1
x
O
T
O
1
4
u
Figure 5.5: The domain D = {(x, y) ∈ R2 : x ≥ 0 ∧ y ≥ 0 ∧ 1 ≤ xy 2 ≤ 4 ∧ x ≤ y 3 ≤ 8x} and the domain T = h1; 4i × h1; 8i, g : T 7→ D.
because either u or v are then constant on the boundary. The coefficients of the power transformation α1 , α2 , β1 , β2 can be determined by the conditions
β2
= 1,
κ
−
β1
= 2,
κ
−
α2
= −1,
κ
α1
= 3,
κ
which gives
β2 = κ, β1 = −2κ, α2 = κ, α1 = 3κ,
1
κ = α1 β2 − α2 β1 = 3κ2 + 2κ2 = 5κ2 ⇒ κ = ,
5
1
2
1
3
β2 = , β1 = − , α2 = , α1 = .
5
5
5
5
Our power transformation is in the present case defined by the relations
1 1 6
Jg (u, v) = u− 5 v − 5
5
and it maps the elementary domain T = h1; 4i × h1; 8i onto the domain D, see Fig. 5.5 — the figure also shows that replacing x by u and replacing y by
v, presents T as an interval in the coordinate system Ouv. The function f (x, y) = xy 2 is continuous on D which implies its integrability. The method
of substitution renders
" 9 #4 " 1 #8
√
√
ZZ
ZZ
Z8 Z4
5
4
6
1
8 858−1
1
u5
v− 5
2
−
xy dx dy =
u |Jg (u, v)| du dv =
u 5 v 5 du dv =
·
= ··· =
· √
.
5
5
5 95
9
− 15
8
−
1
1
1
3
2
x = u 5 v− 5 ,
D
T
1
1
y = u5 v 5 ,
1
1
Double integral – Double integral applications
Integral in general is used to calculate geometrical and physical quantities. There is a lot of such examples, only the most principal will be discussed.
The double integral can be used for various quantities which characterize a planar domain D, e.g.:
ZZ
Area of a domain D: A =
dx dy.
D
ZZ
s
Area of a curved surface P: A =
1+
∂f
∂x
2
+
∂f
∂y
2
dx dy, where D is an orthogonal projection of the surface P, given as a graph of a
D
finction z = f (x, y), to the plane Oxy.
ZZ
Volume of a cylindrical body T: V =
f (x, y) dx dy, where D is an orthogonal projection of the body T to the plane Oxy. The body T occupies
D
the space between the domain D and the graph of the function f (x, y) which is defined on D and satisfying condition f (x, y) ≥ 0 on D.
ZZ
Mass of a domain D: m =
σ dx dy, where σ(x, y) is the (surface) density of the body defined by the domain D.
D
Static moments of a domain D with respect to the coordinate axes x, y:
ZZ
Ux =
y σ(x, y) dx dy,
D
The centroid T of a domain D: T =
ZZ
Uy =
x σ(x, y) dx dy.
D
Uy Ux
,
.
m m
Moments of inertia of a domain D with respect to the coordinate axes x, y and the polar moment:
ZZ
ZZ
2
y σ(x, y) dx dy,
Jx =
Jy =
D
ZZ
2
x σ(x, y) dx dy,
Jr =
D
(x2 + y 2 ) σ(x, y) dx dy.
D
Example 6 Calculate area of a domain D which is bounded by a curve given by the equation (x2 + y 2 )3 = a2 (x4 + y 4 ) with a parameter a > 0.
ZZ
dx dy, the domain is transformed by
Solution. The planar body, whose area is to be calculated, is drawn in Fig. 5.6(left). To calculate the integral
D
y
ϕ
π
A
ω
ω
O D
x
O
−π
ρ
D∗
ρ=r(ϕ)
g (ρ, ϕ)
A
Figure 5.6: The domain D boundaed by the curve (x2 + y 2 )3 = a2 (x4 + y 4 ) and the domain D∗ , g : D∗ 7→ D.
polar coordinates. To this end, we substitute the relations x = ρ cos ϕ, y = ρ sin ϕ to the equation of the boundary curve for expressing it in a suitable
form. We obtain
(ρ2 cos2 ϕ + ρ2 sin2 ϕ)3 = a2 (ρ4 cos4 ϕ + ρ4 sin4 ϕ)
which makes
q
ρ = a (cos ϕ + sin ϕ) ⇒ ρ = a cos4 ϕ + sin4 ϕ,
2
2
4
4
since both ρ and a are nonnegative.
The polar coordinates has an explicit geometric meaning. The last relation for example gives the distance ρ of a point on the curve from the beginning
−→
of the coordinate system depending on the angle ϕ. Hence, if we choose ϕ = ω ∈ h−π; πi (the halfline OA), see Fig. 5.6(left), the points of the domain
D lying on the segment OA satisfy the relation
p
ρ ≤ a cos4 ω + sin4 ω.
We denote g the mapping which maps the domain D∗
q
∗
4
4
D = (ρ, ϕ) : ϕ ∈ h−π; πi ∧ ρ ∈ h0; r(ϕ)i, r(ϕ) = a cos ϕ + sin ϕ
to the domain D. Then Fig. 5.6 presents the domain D∗ in a rectangular coordinate system Oρϕ, where it is an elementary domain of the type [ϕ, ρ]
mapped by g onto D.
Using the Fubini theorem, the double integral can be calculated as follows:
ZZ
A=
Zπ
ZZ
dx dy =
ρ dρ dϕ =
D∗
D
a2
=
2
Zπ 

r(ϕ)
π
2
Zπ q
Z
2 Z
1
a


4
ρ dρ dϕ =
a cos4 ϕ + sin ϕ dϕ =
cos4 ϕ + sin4 ϕ dϕ =

2
2
−π
1 + cos 2ϕ
2
2
+
−π
0
1 − cos 2ϕ
2
2
−π
a2
dϕ =
2
−π
Zπ 1
1 1
1
1 1
2
2
+ cos 2ϕ + cos 2ϕ +
− cos 2ϕ + cos 2ϕ dϕ =
4 2
4
4 2
4
−π
a2
=
2
Zπ
π
π
2 Z
2 Z
1 1
a
1
1
(1
+
cos
4ϕ)
a
3 1
3
2
+ cos 2ϕ dϕ =
+
dϕ =
+ cos 4ϕ dϕ = πa2 ,
2 2
2
2 2
2
2
4 4
4
−π
Z
−π
−π
π
because
cos 4ϕ dϕ = 0 due to appropriately oscillating character of the periodic function.
−π
Example 7 Calculate area of a part of a conical surface P, with P given as
P = {(x, y, z) ∈ R3 : z =
p
√ x
x2 + y 2 ∧ z ≤ 2
+ 1 }.
2
y
z
2
D
√
2−2 2 O
2
√
2+2 2 x
y
P
−2
x
Figure 5.7: The orthogonal projection of the surface P to the plane Oxy is the domain D.
p
Solution.
A
part
of
the
conical
surface
P
with
the
equation
z
=
f
(x,
y)
=
x2 + y 2 is drawn in Fig. 5.7(left) together with a part of the plane
√ x
z = 2 2 + 1 . The curved surface, whose area is to be calculated, lies below the given plane as shown in the picture. The orthogonal projection of
the surface to the plane Oxy has to be found first. Let us denote this projected domain D. We substituted z from the equation of the plane to the
equation of the cone. It renders
p
√ x
+ 1 = x2 + y 2 ,
2
2
which can be squared to come to
2
x
2
+ x + 1 = x2 + y 2 .
4
After small arrangements we obtain
x2
− 2x − 2 + y 2 = 0 ⇒ x2 − 4x − 4 + 2y 2 = 0,
2
2 x−2
y 2
2
2
√
= 1.
(x − 2) + 2y = 8 ⇒
+
2
2 2
x2 −
√
The last equation is an equation of a ellipse with the centre at the point (2, 0) and halfaxes of the lengths 2 2 and 2. The domain D lies in its interior.
This domain is drawn in Fig. 5.7(right). Calculating the partial derivatives of the function f (x, y)
s
2 2
√
∂f (x, y)
x
∂f (x, y)
y
∂f (x, y)
∂f (x, y)
=p
,
=p
⇒ 1+
+
= 2,
∂x
∂y
∂x
∂y
x2 + y 2
x2 + y 2
we come to a simple expression for the area A =
ZZ √
2 dx dy. The domain D is an ellipse and such a case is appropriate for using generalized polar
D
coordinates. This coordinates have to be shifted such that the origin is√at the point (2, 0). The substitution relations for the composed transformation
with parameters a and b from the polar transformation given by a = 2 2 and b = 2, respectively, read
√
√
x = gx (ρ̃, ϕ̃) = 2 + 2 2ρ̃ cos ϕ̃, y = gx (ρ̃, ϕ̃) = 2ρ̃ sin ϕ̃, Jg = 4 2ρ̃.
We substitute them to the equation of the ellipse with the result
!2 √
2
2 + 2 2ρ̃ cos ϕ̃ − 2
2ρ̃ sin ϕ̃
√
= 1 ⇒ ρ̃ = 1,
+
2
2 2
so that the domain D∗ which is mapped to the domain D is an interval D∗ = {(ρ̃, ϕ̃) ∈ R2 : ρ̃ ∈ (0; 1i ∧ ϕ̃ ∈ h−π; π)}. Now, the substitution method
can be used and the area is calculated
A=
ZZ √
ZZ √ √
Z1 Zπ
1
2 dx dy =
2 4 2 ρ̃ dρ̃ dϕ̃ = 8
ρ̃ dϕ̃ dρ̃ = 8 2π = 8π.
2
D∗
D
0 −π
Remark 4 The area could be calculated directly in rectangular coordinates which would save us from the substitution. Nevertheless, the
integration would be more complicated. We can compare both methods:
A=
ZZ √
M
q
2
2+2x− x2
√
2+2
Z 2
Z
2 dx dy = 2
√
2−2 2
0
√
2+2
Z 2
=2
√
2−2 2
√
√
2+2
Z 2
2 dy dx = 2
√
4 + 4x − x2 dx =
√
2−2 2
p
√
8 − (x − 2)2 dx = 2 8
√ s
2+2
Z 2
√
2−2 2
1−
x−2
√
2 2
2
1
√ Z √
√
dx = 2 8
1 − t2 8 d t =
√
t= x−2
2
2
−1
h√
i1
= 8 t 1 − t2 + arcsin t
= 8(arcsin 1 − arcsin(−1)) = 8π.
−1
Example 8 Calculate volume of a body T bounded by the paraboloid of rotation z = x2 + y 2 an the planes z = 0, y = 1, y = 2x and x + y = 6.
Solution. The given body T is bounded from below by the plane Oxy and from above by the given paraboloid. The rest of the boundary is formed by
three vertical planes, so that the orthogonal projection of the body T to the plane Oxy is the domain D drawn in Fig. 5.8. The volume V of the body
y
y=2x
6
4
y=6−x
D
1
O
2
1
2
y=1
x
56
Figure 5.8: The domain T for calculation of the body T volume.
T is in such a case given by simple integration of the upper bound
ZZ
V =
D
6−y
Z4 Z6−y
Z4 3
Z4 2511
x
15 3
2
2
2
2
(x + y ) dx dy =
(x + y ) dx dy =
+y x
dy =
.
− y + 12y − 36y + 72 dy =
y
3
8
32
2
2
1
y
2
1
2
1
Example 9 Find the coordinates of the centroid of a thin homogeneous plate, whose form is a circular sector of an angle α and a radius R.
Solution. The domain D of the thin plate is shown in Fig. 5.9. The plate is homogeneous so its density σ is a constant. We need to calculate the mass
m of the domain D and its static moments Sx and Sy with respect to coordinate axes. Let us choose the coordinate system Oxy as it is shown in the
picture. We have a free choice for this coordinate system so it is worth thinking about the most advantageous position utilizing the shape of the domain.
The mass of the domain is obtained by the substitution with the polar coordinates
α
m=
σ dx dy =
D
α
Z2 ZR
ZZ
σ ρ dρ dϕ = σ
−α
2
0
ZR
Z2
−α
2
dϕ
ρ dρ = α
R2
σ.
2
0
y
R sin α2
α
2
O
D
Rx
α
2
−R sin α2
Figure 5.9: The domain D for finding the centroid.
The static moment with respect to the y axis is
α
Sy =
α
Z2 ZR
ZZ
σ x dx dy =
σ ρ cos ϕ dρ dϕ = σ
−α
2
D
2
ZR
Z2
cos ϕ dϕ
−α
2
0
ρ2 dρ = sin
α 2R3
σ.
2 3
0
The homogeneous domain D is symmetric with respect to the axes x so the pertinent static moment should be zero. We can verify it by calculating the
integral
α
α
ZR
Z2 ZR
Z2
ZZ
Sx =
σ y dx dy =
σ ρ2 sin ϕ dρ dϕ = σ
sin ϕ dϕ ρ2 dρ = 0.
−α
2
D
0
−α
2
0
Hence, the coordinates of the thin plate T centroid are
3
sin α2 2R3 σ
4 sin α2
Sy
=
=
R
,
Tx =
2
m
3
α
α R2 σ
Ty =
Sx
= 0.
m
It means that the centroid lies on the axes of symmetry of the plate and its distance from the vertex of the sector is given by the value of Tx
Self-assessment questions and problems for solving
1. Describe common geometrical shapes as elementary domains, e.g. a circle, a circular segment, an ellipse, a cardioid etc.. How do you proceed in the
description?
2. Cut a rectangular hole in aforementioned shapes and split the resulting domain to elementary domains. How do you proceed in this case?
3. Could you integrate a discontinuous function on a given domain?
4. Can you imagine how a three dimensional integral (a triple integral) is calculated based on your experience with the double integral?
5. Are you able to integrate on a map with the system of contour and descent lines by the method of substitution?
6. Resuming your knowledge from other fields, explain which physical quantities have “integral” character and thus are calculated by the double integral?
In the problems 1. to 10. calculate subsequently the integrals.


Z1 Z1
x2


dx dy .
1.

2
1+y
0
Z2
3.


(1 + xy + x2 + y 2 ) dx dy .


0
Z2
6.
0


π
2Z
cos ϕ
Z2


(1 + ρ) dρ dϕ .
7.

0

(x2 + xy + x − y 2 ) dx dy .
Zx
dx
4.
1+cos
Z x

0
0
y 2 sin x dy .
Z1
Z1
0
dx
0

0
Z1
Zπ
5.
2.
0



1
Z2
1
x2


Zx2


 (x + y) dy  dx .
1
x
3Zsin ϕ
Zπ
ρ cos ϕ dρ .
dϕ
8.
0
xy 2 dy .
0
Z1
9.

Z1



π
v+1 
dv  du .
v
1
Z1
15.
17.
12.

√
Zx


 f (x, y) dy  dx .
x3

Z2−y

sin x
Z


f (x, y) dy  dx .

0
14.
√

Z1−x2


f (x, y) dy  dx .


0
 √

2
1−y
Z
Z1



f (x, y) dx
16.

 dy .
√
0
−


f (x, y) dx dy .


−6

−1

1 2
y −1
4

x4 dx dy .
cos y
Z1
2x

Z1


0

Z6−x


f (x, y) dy  dx .

0
Z2
Zπ
3

0

0
In the problems 11. to 18. change the order of integration.


Z2 Z4


11.
 f (x, y) dx dy .
13.
10.
x
1
2
Z2
Z2
Z2
18.

√
Z 2x
√


f (x, y) dy  dx .


0
1−y 2
2x−x2
In the problems 19. to 32. calculate the double integrals.
ZZ
1
19.
dx dy, where D = h0; 1i × h1; 2i .
x+y
D
ZZ
20.
x2 + y 2 − 2x − 2y + 4 dx dy, where D = h0; 2i × h0; 2i .
D
ZZ
21.
ln(1 + x)2y dx dy, where D = h0; 1i × h0; 1i .
D
ZZ
22.
π
x2 y sin(xy 2 ) dx dy, where D = h1; 2i × h0; i .
2
D
ZZ
23.
x2 − y 2
dx dy, where D = h0; 1i × h0; 1i .
(x2 + y 2 )2
D
ZZ
24.
(2x + y) dx dy, where D = {(x, y) ∈ R2 : x ≥ 0 ∧ y ≥ 0 ∧ x + y ≤ 3} .
D
ZZ
25.
xy dx dy, where D = {(x, y) ∈ R2 : x ≥ 0 ∧ y ≥ 0 ∧
√
x+
√
y ≤ 1} .
D
ZZ
26.
|x| dx dy, where D = {(x, y) ∈ R2 : x2 ≥ y ∧ 4x2 + y 2 ≤ 12} .
D
ZZ
27.
x2
√
2x
dx dy, where D = {(x, y) ∈ R2 : x ≥ 0 ∧ y ≤ 1 ∧ x ≤ y} .
2
+y
D
ZZ
28.
x2 y 2 dx dy, where D = {(x, y) ∈ R2 : 1 ≤ x2 + y 2 ≤ 4} .
D
ZZ
29.
arctg
√
x
x
dx dy, where D = {(x, y) ∈ R2 : 1 ≤ x2 + y 2 ≤ 9 ∧ √ ≤ y ≤ x 3} .
y
3
D
ZZ
30.
(x + y + xy) dx dy, where D is a domain bounded by the curve y = 4 − x2 and by the x axis .
D
ZZ
31.
x2
dx dy, where D is a domain bounded by the curve xy = 1 and by the lines x = 2 and y = x .
y2
D
ZZ
32.
x
e y dx dy, where D is a domain bounded by the parabola y 2 = x and by three lines, one of the being the y axis and the other two are parallel
D
to the x axis passing respectively through the points (0, 1) and (0, 2) .
In the problems 33. to 40. calculate the integrals by the substitution with polar coordinates.
ZZ
33.
(1 − 2x − 3y) dx dy, where D = {(x, y) ∈ R2 : x2 + y 2 ≤ 4 ∧ y ≤ x} .
D
ZZ p
x2 + y 2 dx dy, where D = {(x, y) ∈ R2 : 0 ≤ y ≤ x ∧ 0 ≤ x ≤ 1} .
34.
D
ZZ
35.
ln(x2 + y 2 )
dx dy, where D = {(x, y) ∈ R2 : 1 ≤ x2 + y 2 ≤ e ∧ 0 ≤ y} .
2
2
x +y
D
ZZ
36.
sin
p
x2 + y 2 dx dy, where D = {(x, y) ∈ R2 : π 2 ≤ x2 + y 2 ≤ 4π 2 } .
D
ZZ s
37.
1 − x2 − y 2
dx dy, where D = {(x, y) ∈ R2 : x2 + y 2 ≤ 1 ∧ 0 ≤ x} .
1 + x2 + y 2
D
ZZ
38.
(x + y) dx dy, where D = {(x, y) ∈ R2 : x2 + y 2 ≤ p(x + y)}, (p > 0) .
D
ZZ
39.
(xy)2 dx dy, where D = {(x, y) ∈ R2 : (x2 + y 2 )2 ≤ p2 (x2 − y 2 )} .
D
ZZ
40.
e−x
2 −y 2
cos(x2 + y 2 ) dx dy, where D = {(x, y) ∈ R2 : x2 + y 2 ≤ π4 } .
D
In the problems 41. to 44. calculate the integrals by the substitution with generalized polar coordinates.
ZZ
√
x2
2
+ y 2 ≤ 1 ∧ 0 ≤ x ≤ 12y} .
41.
(x − 2y) dx dy, where D = {(x, y) ∈ R :
4
D
ZZ p
42.
1 − x2 − 4y 2 dx dy, where D = {(x, y) ∈ R2 : x2 + 4y 2 ≤ 1} .
D
ZZ
43.
ln(1 + x2 + 9y 2 ) dx dy, where D = {(x, y) ∈ R2 : x2 + 9y 2 ≤ 36} .
D
ZZ
44.
(x2 + y 2 ) dx dy, where D = {(x, y) ∈ R2 : 3x2 + 12y 2 ≤ 1} .
D
In the problems 45. to 48. calculate the integrals by an appropriate substitution.
ZZ
45.
xy dx dy, where D = {(x, y) ∈ R2 : x3 ≤ y ≤ 2x3 ∧ 3x ≤ y 2 ≤ 4x} .
D
ZZ
46.
(2x − y) dx dy, where D = {(x, y) ∈ R2 : 1 ≤ x + y ≤ 2 ∧ 1 ≤ 2x − y ≤ 3} .
D
ZZ
47.
x2
dx dy, where D = {(x, y) ∈ R2 : 2x2 ≤ 6y ≤ 3x2 ∧ x ≤ y 2 ≤ 7x} .
y
D
ZZ
48.
x
dx dy, where D = {(x, y) ∈ R2 : x ≤ y ≤ 2x ∧ x2 ≤ y ≤ 2x2 } .
y
D
In the problems 49. to 57. calculate the area of a planar domain bounded by given curves.
49. 2x − y = 0, 2x − y = 7, x − 4y + 7 = 0, x − 4y + 14 = 0 .
50.
x2 y 2
x y
+
=
1,
+ = 1.
a2
b2
a b
2
51. y = x, y = x + 2, y = 2, y = −2 .
54. xy = 6, 3x − y = 7, x = 1 .
y 2 + b2
y 2 + a2
, x=
, a > b > 0.
2b
2a
√
√
√
55. x + y = a, x + y = a, a > 0 .
3
3
56. 6x = y − 16y, 24x = y − 16y .
57. xy = 2, xy = 8, y =
1
52. y = (x − a)2 , x2 + y 2 = a2 , a > 0 .
a
53. x =
x
, y = 2x .
2
In the problems 58. to 63. calculate the area of a planar domain bounded by given curves using polar coordinates.
2
2
2
2
59. (x + y ) = 2a xy, a > 0 .
2
2
5
60. (x + y ) = 2ax , a > 0 .
3
2
2
4
61. (x + y ) = 2axy , a > 0 .
3
62. (x + y) = xy .
2
2
3
2
63. (x + y ) = a(x − 3xy ), a > 0 .
2
2
2
2
58. (x + y ) = 4(x − y ) .
2
3
2
In the problems 64. to 67. calculate the area of a planar domain bounded by given curves using generalized polar coordinates.
2
2
2
2
x
y2
xy
x
y2
64.
65.
= 2 , a > 0, b > 0, c > 0 .
= x2 + y 2 , a > 0, b > 0 .
+ 2
+ 2
a2
b
c
a2
b
2
2
x2 y 2
y2
x2 y
x
66. 2 + 2 = x + y, a > 0, b > 0 .
67.
+
=
, a > 0, b > 0, c > 0 .
a
b
a2
b2
c3
In the problems 68. to 75. calculate the area of a curved interface given by the domain D.
68. D = {(x, y, z) ∈ R3 : 2x + 2y + z = 4 ∧ x2 + y 2 ≤ 9} .
69. D = {(x, y, z) ∈ R3 : 2z = x2 + y 2 ∧ x2 + y 2 ≤ 1} .
70. D = {(x, y, z) ∈ R3 : x2 + y 2 + z 2 = 27 ∧ x2 + y 2 ≤ 6z} .
√
y
71. D = {(x, y, z) ∈ R3 : x2 + y 2 = z 2 ∧ z ≥ 0 ∧ z ≤ √ + 2} .
2
72. D = {(x, y, z) ∈ R3 : z 2 = 2xy ∧ 0 ≤ x ≤ 2 ∧ 0 ≤ y ≤ 4} .
73. D = {(x, y, z) ∈ R3 : y 2 + z 2 = 4 ∧ x2 + y 2 ≤ 4} .
74. D = {(x, y, z) ∈ R3 : z 2 = x2 + y 2 ∧ z ≥ 0 ∧ x2 + y 2 ≤ 2y} .
75. D = {(x, y, z) ∈ R3 : 2z = x2 + y 2 ∧ x2 + y 2 + z 2 ≤ 3} .
In the problems 76. to 85. calculate the volume of a body by the double integral.
2
76. 2y = x, x + 2y + z = 4, z = 0 .
2
77. 2y = x , x + y + z = 4, z = 0, y = 1 .
2
2
2
2
78. x + y = z, x + y = 2x, z = 0 .
2
2
2
2
79. x + y = 1, y + z = 1 .
−x
80. z = e
2 −y 2
, x2 + y 2 = 1, z = 0 .
2
2
2
2
81. z = 2xy, (x + y ) = 2xy, z = 0 .
2
2
2
2
82. 2z = 2x + 3y , x + 4y = 1, z = 0 .
2
2
2
2
83. (x + 4y ) + 9z = 1 .
84. z = x2 + y 2 , y = 2x, y = 6 − x, z = 0, y = 1 .
85. z = x2 + y 2 , x2 + y 2 = x, x2 + y 2 = 2x, z = 0 .
86. Determine the mass of a planar domain bounded by an ellipse with the equation
to the distance r from the small halfaxis and for r=1 it is equal to two .
2
x2
+ y9 =1,
4
if the density is in each its point directly proportional
87. Determine the mass of a domain bounded by the parabolas y=2−x2 and y=x2 . The density at each point is directly proportional to the distance
of the point from the x axis and it is equal to four at the point (0, 2) .
88. Calculate the mass of a thin plate whose boundary is a curve given by the equation |x|+|y|= π2 , if the density σ of the plate is given by the relation
σ(x, y)= cos x cos y .
p
89. Calculate the mass of a thin plate bounded by the circle x2 +y 2 =2ax. The density σ of the circle is given by the relation σ(x, y)= x2 +y 2 .
√
90. Calculate the static moments of a thin plate bounded by the curves y=|x| and y= 2−x2 , if the density in each point is equal to four times the
distance of the point from the x axis .
2
91. Determine the coordinates of the centroid of a homogeneous domain bounded by the curves y=x2 and y= 1+x
2 .
92. Find the centroid coordinates of a homogeneous domain bounded by the curve y= cos x, x∈h−π; πi and by a line parallel to the x axis which
passes through the point (0, 1) .
93. Calculate the coordinates of the centroid of a planar domain bounded by the curve y=4−x2 and by the x axis. The density at each point is
directly proportional to the distance from the y axis .
94. Determine the centroid
coordinates of a quartercircle given by the relation x2 +y 2 ≤2 and lying in the first quadrant, if the density σ is given by
p
the relation σ(x, y)=2 2x2 +2y 2 .
95. The density of a equilateral rectangular triangle with the hypotenuse 2a is directly proportional to the distance from the hypotenuse. Determine
the height of the centroid above the hypotenuse of the triangle .
96. Calculate the coordinates of the centroid of a planar domain bounded by the curve
√
√ √
x+ y= 2 and by the coordinate axes .
97. Calculate the coordinates of the centroid of a planar domain bounded by the parabola y=x2 and by the line y=x+2, if the density in each point
is directly proportional to the distance of the point from the x axis .
98. Determine the moment of inertia of a homogeneous circular domain with the unit density and the radius r=4 with respect to its tangent .
99. A ring domain is bounded by concentric circles with radii r1 =1 and r2 =2. The density at each point of the domain is directly proportional to the
distance from the centre of the ring. Determine the moment of inertia of the domain with respect to a line passing through the centre of the ring .
100. The density at each point of an annular domain given by the inequalities 1≤x2 +y 2 ≤4 is indirectly proportional to the distance of the point from
the centre of the annular domain. Determine the moment of the inertia of the domain with respect to the x axis .
101. Determine the polar moment of inertia of an ellipse given by the inequality x2 +4y 2 ≤4, if the density at each point is directly proportional to its
distance from the small halfaxis .
102. Calculate the polar inertial moment of the part of the circle x2 +y 2 ≤2x+2y which lies in the first quadrant. Consider the circle to be homogeneous
with the unit density .
Conclusion
The chapter was intended to calculation of double integrals. First, we learned to describe the domain of integration such that a double integral is
converted to two subsequently integrated integrals. These simple definite integrals are then calculated by the methods which we learned in the previous
course of mathematics. We also saw that it is frequently useful to transform the domain of integration by a appropriate function so that the evaluation of
the integrals were simplified by the substitution method. The final part of the chapter was intended to application of this type of integral in calculation of
various geometrical and physical quantities. The acquired knowledge can be applied for understanding and studying other disciplines of our specialization.
References
[1] Šiagová, J.: Advanced Mathematics, Bratislava, STU, 2011.
[2] Krivá, Z.: Mathematics I, Bratislava, STU, 2007.
[3] Bubeník, F.: Problems to Mathematics for Engineers, Praha, ČVUT, 1999.
[5] Neustupa, J., Kračmar S.: Problems in Mathematics I, Praha, ČVUT, 1998.
[6] Neustupa, J.: Mathematics I, Praha, ČVUT, 2004.
Problem solutions
The answers to the self-assessment questions, if you are unable to formulate, and also a lot of other answers can be found in provided references.
1.
π
12
2. 0
Z
1
12.

3.
53
12
1

Z

0
6.
56
10
1
√
3
y
13.
Z
1
16.
Z
−1
1−
1 2
y
2
1−y 2
Z
f (x, y) dx dy +
− ln 2

Z
6
4
0
17.
Z
2
Z
√
1+
1−y 2
f (x, y) dx dy +
2
Z

1
2
1 2
y
2
Z

−1


√
2 x+1
√

Z
f (x, y) dy dx +
−2 x+1
f (x, y) dy  dx
1

Z √1−y2

√
−
0
Z

14.
0

1−x2
1
Z
8

Z


2

3

f (x, y) dx dy

Z
11.
6−y
Z
0
1
15π−16
150
10.

f (x, y) dy  dx


0
13
8
0
√


9.
f (x, y) dx dy +


8. 0
y
2
Z
0
f (x, y) dx dy
√

4
Z
Z
4+π
2
7.


y2


0
4
3
arcsin y
15.
18.
5.
f (x, y) dx dy
0
Z
1
40
π−arcsin y
Z

Z
4.

4
Z
1−y 2

f (x, y) dx dy

2−x
√
f (x, y) dy  dx
−2 x+1
0

f (x, y) dx dy
27
19. ln 16
20.
32
3
21. 2 ln 2 − 1
22.
3
2
√
√
1
10
π
21π
π2
256
9
1
8
23. neexistuje
24. 27
25.
26.
4
3
−
27.
ln
2
−
2
+
28.
29.
30.
31.
32.
33.
2π
+
2
20
3
2
8
6
15
2
3
√
√ 2
√ 4
1
π
π
1
1 6
π
2
π
π
2
3
34. 6 ( 2 + ln(1 + 2)) 35. 4
36. −6π
37. 4 (π − 2) 38. 2 πp
39. 90 p
40. 2
41. − 3 ( 3 − 1) 42. 3
43. 3 (37 ln 37 − 36)
√
8
8
2
5π
4
9
ab
π
40
a
− 65
44. 288
45. 5(2 − 1)(4 5 − 3 5 )
46. 3
47. 5
48. 16
49. 7
50. 2 ( 2 − 1)
51. 3
52. 12 (3π − 4)
53. 23 (a − b) ab
2 2
63
7
1
54. 6 ln 3 + 2 55. 13 a2
56. 16 57. 6 ln 2 58. 4 59. a2
60. 128
πa2
61. 128
πa2
62. 60
63. 14 πa2
64. a2cb2
65. π2 ab(a2 + b2 )
√
√
√
√
π a5 b3
67. 32
68. 27π
69. 32 π(2 2 − 1)
70. 18π(3 − 3)
71. 8π
72. 16
73. 32
74. 2π
75. 23 π(3 3 − 1)
c6
2
π
11
78. 32 π
79. 16
80. π(1 − 1e )
81. 12
82. 64
π
83. π12
84. 2511
85. 45
π
86. 0, 24−15π
87. 16
88. π
3
32
32
30π−20
3
3
3
90. 2+π, 0
91. 0, 24−15π
, √
95. a2
96. 25 , 25
, 235
92. 0, − 41
93. (0, 1)
94. π√
97. 25
98. 320π
30π−20
32 112
2 π 2
66. π4 ab(a2 + b2 )
77. 68
76. 81
5
15
89.
99.
32
|a|3
9
31
kπ
5
100. 37 kπ
101.
344
k
75
102. 3π+8

matematika ii (mathematics ii) - Stavebná fakulta TU v Košiciach

Transcription

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