Math 215 March 4, 2015 SOME SOLUTIONS TO THE HOMEWORK

Math 215 March 4, 2015 SOME SOLUTIONS TO THE HOMEWORK

Math 215
March 4, 2015
SOME SOLUTIONS TO THE HOMEWORK
HW1
Note: All derivatives, y 0 , y 00 , u0 , u00 , are with respect to x.
(1) Suppose that y is a function of x. Express the following in terms of x, y 0 and y 00 .
d 2
(a)
(y ) Answer: 2yy 0
dx
d2 2
(y ) Answer: 2(y 0 )2 + 2yy 00
(b)
dx2
d 2
(c)
(x y) Answer: x2 y 0 + 2xy
dx
(2) Suppose that u is a function of x and y = ux2 . Express the following in terms of x, u0 and
u00 .
(a) y 0 Answer: x2 u0 + 2xu. This is Question 1(c) again.
(b) y 00 Answer: x2 u00 + 4xu0 + 2u
(3) Find all solutions of the following differential equations:
(a) y 0 = 0 Answer: y = C with C a constant.
(b) y 00 = 0 Answer: y = c0 + c1 x with c0 and c1 constants.
(c) y 0 = x Answer: y = x2 /2 + C with C a constant.
(d) y 00 = x Answer: y = x3 /6 + c0 + c1 x with c0 and c1 constants.
(4) Consider the differential equation xy 0 − 2y = 0 with x > 0.
(a) Suppose y = ux2 is a solution of the equation for some function u of x. Show that
u0 = 0.
Answer: From y = ux2 we get y 0 = x2 u0 +2xu. Hence xy 0 −2y = x(x2 u0 +2xu)−2x2 u =
x3 u0 . Since x > 0, y = ux2 is a solution if and only if u0 = 0.
(b) Using this fact, find all solutions of the differential equation.
Answer: If u0 = 0, then u = C for some constant C, and hence y = x2 u = Cx2 is the
general solution.
(5) For what integers n is y = xn a solution of x2 y 00 + xy 0 − y = 0?
Answer: If y = xn then y 0 = nxn−1 and y 00 = n(n − 1)xn−2 . Plugging these expressions into
the differential equation we get
x2 y 00 + xy 0 − y = x2 (n(n − 1)xn−2 ) + x(nxn−1 ) − y = (n2 − 1)xn .
This is the zero function if and only if n2 − 1 = 0, that is, if n = ±1.
HW2
(1) Verify that√the following functions
are solutions of the corresponding differential equations.
√
(a) y = 2 x + c, y 0 = 1/ x.
√
Answer: Taking the derivative of both sides of y = 2 x + c gives
y0 =
√
d
(2x1/2 + c) = x−1/2 = 1/ x.
dx
(b) y 2 = e2x + c, yy 0 = e2x .
Answer: Taking the derivative of both sides of y 2 = e2x + c using the chain rule we get
2yy 0 = 2e2x ,
which simplifies to the given differential equation.
1
2
p
d
(c) y = arcsin xy, xy 0 + y = y 0 1 − x2 y 2 . Hint: arcsin x = sin−1 x and
(sin−1 x) =
dx
1
√
.
1 − x2
Answer: Using the chain rule, we first calculate the derivative of arcsin(xy) with respect
to x:
d
dy
d
1
1
y+x
(arcsin(xy)) = p
(xy) = p
.
dx
dx
1 − (xy)2 dx
1 − (xy)2
Now suppose that y satisfies y = arcsin(xy). Taking the derivative of both sides of this
equation we get the differential equation
dy
dy
1
y+x
=p
,
dx
dx
1 − (xy)2
which simplifies to xy 0 +y = y 0
equation.
p
1 − x2 y 2 . Hence y is a solution of the given differential
d
(d) x + y = arctan y, 1 + y 2 + y 2 y 0 = 0. Hint: arctan x = tan−1 x and
(tan−1 x) =
dx
1
.
1 + x2
Answer: Suppose that y satisfies x + y = arctan y. Taking the derivative of both sides
of this equation with respect to x using the chain rule we get the differential equation
1+
dy
1 dy
=
dx
1 + y 2 dx
which simplifies to 1 + y 2 + y 2 y 0 = 0. Hence y is a solution of the given differential
equation.
(2) Use separation of variables to solve the following equations:
(a) x5 y 0 + y 5 = 0 Answer: 1/y 4 + 1/x4 = C
(b) y 0 − y tan x = 0 Answer: Rewrite as dy/y = tan x dx. Solution is y = C sec x.
(c) (y + yx2 + 2 + 2x2 ) dy = dx Answer: y 2 /2 + 2y = tan−1 x + C
(d) y 0 /(1 + x2 ) = x/y and y = 3 when x = 1. Answer: The general solution of the
differential equation is 2y 2 = 2x2 + x4 + C. The solution such that y = 3 when x = 1
is 2y 2 = 2x2 + x4 + 15.
(e) y 0 = x2 y 2 and the graph of the solution passes through (−1, 2).
Answer: −1/y = x3 /3 − 1/6
dy
(3) The differential equation 2y
= x2 + y 2 − 2x is not separable. Supposing that y is a soludx
tion of this equation, show that v = x2 + y 2 is a solution of a separable differential equation.
Use this fact to find all solutions of the original differential equation.
dy
dv
Answer: First we find the relationship between
and
by differentiating v = x2 + y 2 :
dx
dx
dv
d 2
dy
=
(x + y 2 ) = 2x + 2y .
dx
dx
dx
dv
= v. This separable equation
dx
x
2
2
for v has solutions
√ v = Ae where A is a constant. From this equation and v = x + y we
x
2
find that y = ± Ae − x , with A > 0, are all solutions of the original differential equation.
Plugging this into the original differential equation gives
3
HW3
(1) Solve the following differential equations. Some are separable, some are linear, some are
both. If possible, express y as a function of x.
dy
(a) x3 − 3y − x
=0
x>0
dx
3
dy
+ y = x2 . So P = 3/x and Q = x2 .
Answer: Linear when written in the form
dx
x
R
R
R
We have
P dx = 3 ln x = ln x3 , so H = e P dx = x3 . Finally y = H −1 HQ dx =
R
x−3 x5 dx = x−3 (x6 /6 + C) = x3 /6 + Cx−3
dy
= 2e2x
dx
Answer: linear: y = (C + e2x )/x
(b) y + x
dy
= y2
x>0
dx
√
√
Answer: separable: y = x/(6x2 + C x + 2)
(c) 9x2 y 2 + x3/2
(d)
(e)
dy
x
=
such that y = −1 when x = 0
dx
y−1
√
Answer: separable: y = 1 − x2 + 4
dy
− 2xy = x such that y = 1 when x = 0
dx
2
2
Answer: linear: integration factor= e−x ; y = (3ex − 1)/2.
(f) y dx + (xy 2 + x − y) dy = 0. Hint: Think of x as a function of y. Answer: The trick
here is to think of x as a function of y. The given equation can be written as
dx
1
+ y+
x=1
dy
y
so we set P (y) = y +
Z
1
P (y) dy = y 2 + ln y
2
1
and Q(y) = 1. Then
y
H(y) = e
The implicit solution is x =
Ce−y
2
/2
R
P dy
y 2 /2
= ye
Z
H(y)Q(y) dy = ey
2
/2
+ C.
1
2
(ey /2 + C) which can be simplified to xy = 1 +
yey2 /2
.
(2) Show that if y is a solution of
dy
√
+ P (x) y = Q(x) y
dx
then v =
√
y is a solution of a first order linear equation.
dv
1 dy
1 dy
√
Answer: From v = y, we get
= √
=
. Substituting this into the differential
dx
2 y dx
2v dx
dv
equation yields 2v
+ P (x) v 2 = Q(x)v, which becomes linear after dividing by 2v.
dx
4
HW4
(1) Solve the following differential equations. All are exact—perhaps after a bit of algebraic
manipulation.
dy
(a) (2xy 2 + 2y) + (2x2 y + 2x)
= 0 Answer: Exact: x2 y 2 + 2xy = C.
dx
y
+ 6x dx + (ln x − 2) dy with x > 0. Answer: Exact: y ln x + 3x2 − 2y = C.
(b)
x
(c)
ax + by
dy
= −
with a, b, c, d constants. Answer: This is exact if written as (bx +
dx
bx + cy
cy) dy + (ax + by) dx = 0. Solution is ax2 + 2bxy + cy 2 = C.
(d) (y 2 + xy) dx − x2 dy = 0. Hint: Multiply this equation by the integration factor
1/(xy2 ) to make
it exact.
Answer: Multiplying by 1/(xy 2 ) gives the exact equa
1
x
1
x
+
tion
dx − 2 dy = 0 with solution + ln |x| = C.
x y
y
y
dy
(e) cos y sin x
+ 2 sin y cos x = 0. Hint: Multiply this equation by the integration facdx
tor sin x to make it exact. Answer: Multiplying by sin x gives the exact equation
dy
sin2 x cos y
+ 2 sin x cos x sin y = 0 with solution sin2 x sin y = C.
dx
(2) Solve the following differential equations. Whenever possible, express y as a function of x.
dy
− y = 2e2x Answer: Linear: y = 2e2x + Cex
(a)
dx
√
dy
(b) y
+x=0
y(0) = −3 Answer: Separable: y = − 9 − x2
dx
(c)
(d)
dy
y
y2
y
= + 2 x>0
Hint: Find a differential equation satisfied by u = . Start
dx
x x
x
by differentiating xu = y to get an equation relating u0 to y 0 .
Answer: Differentiating xu = y, gives u + xu0 = y 0 . Plugging this into the differential
equation yields the separable equation xu0 = u2 , with solution u = −1/(ln x + C).
Finally y = xu = −x/(ln x + C).
dy
= 2xy 2
dx
y(0) = 1 Answer: Separable: y = 1/(1 − x2 ).
(e) (cos x)y 0 = (sin x)y Answer: Written as (cos x) dy − (sin x)y dx this equation is exact.
1
Solution is y cos x = C or y = C sec x. This equation is also separable, dy = tan x dx,
y
with the same solution.
dy
(3) The equation cos y sin x
+ 2 sin y cos x = 0 from Question (1e) is separable. Use this fact
dx
to solve the equation.
Answer: Separating the variables we get cot y dy = −2 cot x dx, and then integration gives
ln | sin y| = −2 ln | sin x| + C. This implies sin y = ±eC (sin x)−2 , or sin y sin2 x = C1 where
now C1 = ±eC is an arbitrary constant.
(4) Show that every separable first order differential equation is exact.
Answer: A separable equation is one that can be written in the form
M (x) dx = −N (y) dy
for certain functions M (x) and N (y). The important point is that M is a function of x only
and N is a function of y only. This equation can be written as M (x) dx + N (y) dy = 0 which
∂N
∂M
is exact because
= 0 and
= 0.
∂y
∂x
5
HW5
In a leaking water tank, the height of water h is a decreasing function of
time t. Using Torricelli’s Law, we derived in class the differential equation
satisfied by h:
p
dh
A
= −a 2gh
dt
2
where g is the acceleration of gravity (9.8 m/s ), a is the area of the hole
through which the water is leaking and A is the area of the the surface of
the water at height h. In class we assumed that the tank was a vertical
cylinder and so A is constant in h. Even so, the differential equation is
valid even if A is a function of h.
This happens, for example, if the tank has the form of a paraboloid with area A0 at the top and
height H. Then the surface area of the water A as a function of the height of the water in the tank
varies linearly from h = 0 (where A = 0) to h = H (where A = A0 ):
A0
h.
H
Suppose that the tank is full (h = H) at t = 0. At what time is the tank empty?
Answer: Combining the two equations above we get
p
dh
A0
h
= −a 2gh
H
dt
A=
and separating variables h and t,
√
h dh = −
aH p
2g dt.
A0
Integrating gives
2 3/2
aH p
2g t + C.
h
=−
3
A0
We choose C = 32 H 3/2 so that the tank is full (i.e. h = H) at t = 0:
2 3/2
aH p
2
h
=−
2g t + H 3/2 .
3
A0
3
This is the implicit solution of the differential equation satisfying the condition that h = H when
t = 0. Now we find the time at which the tank is empty by setting h = 0 and solving for t. This
gives
s
t=
2A0
3a
H
.
2g
HW6
Solve the following differential equations.
(1) y 00 = 3y 0 Answer: y = c1 + c2 e3x
(2) y 000 = y 0 Answer: Auxiliary equation: m3 − m = m(m − 1)(m + 1). General Solution:
y = c1 + c2 ex + c3 e−x
(3) 2y 00 + y 0 − 6y = 0 Answer: y = c1 e−2x + c2 e3x/2
(4) (4D2 − 4D + 1) y = 0 Answer: The auxiliary equation 4m2 − 4m + 1 = (2m − 1)2 = 0 has
the double root m = 1/2. Thus the solution is y = (c1 + c2 x)ex/2 .
(5) 4y 000 = 3y 0 + y Answer: The auxiliary equation is 4m3 − 3m − 1 = 0 which factors as
(m − 1)(2m + 1)2 = 0, so it has the double root m = −1/2, as well as the simple root m = 1.
Thus the solution is y = (c1 + c2 x)e−x/2 + c3 ex .
6
(6)
dy
d2 y
+4
+ 4y = 0 with initial conditions y(0) = 1 and y 0 (0) = 1. Answer: General
dx2
dx
Solution: y = (c0 + c1 x)e−2x . Particular Solution: y = (1 + 3x)e−2x .
(7) (D2 − 4)2 y = 0 Answer: Auxiliary equation: (m2 − 4)2 = 0. Roots: m = ±2 are double
roots. General Solution: y = (c1 + c2 x)e2x + (c3 + c4 x)e−2x
(8) D3 y = 0 y(0) = 0, y 0 (0) = 1. Answer: Auxiliary equation: m3 = 0. Roots: m = 0 is a
triple root. General Solution: y = c0 + c1 x + c2 x2 . The condition y(0) = 0 implies c0 = 0,
and then the condition y 0 (0) = 1 implies c1 = 1. The remaining constant c2 is still arbitrary.
Final answer y = x + c2 x2 .
HW7
(1) Solve the following differential equations.
(a) y (4) + y 00 = 0 Answer: The auxiliary equation factors as (m − i)(m + i)m2 and so
y = c1 + c2 x + c2 cos x + c4 sin x.
(b) (D3 −5D2 +9D−5)y = 0 Answer: Auxiliary equation: m3 −5m2 +9m−5 = (m−1)(m2 −
4m + 5); Roots: m = 1, 2 ± i. General Solution: y = c1 ex + c2 e2x cos x + c3 e2x sin x.
dy
d2 y
−2
+ 5y = 0 Answer: y = ex (c1 cos 2x + c2 sin 2x).
dx2
dx
(d) (D2 − 4D + 20) y = 0 with initial conditions y(π/2) = 0 and y 0 (π/2) = 1. Answer:
General solution y = e2x (c1 cos 4x + c2 sin 4x). Particular solution: y = 41 e2x−π sin 4x.
(c)
d2 y
dy
+ 32
+ 25y = 0 y(0) = 0, y 0 (0) = 1. Answer: Auxiliary equation: 16m2 +
2
dx
dx
32m + 25 = 0. Roots: m = −1 ± 43 i. General Solution: y = e−x (c1 cos 34 x + c2 sin 43 x).
The condition y(0) = 0 implies c1 = 0, and then the condition y 0 (0) = 1 implies c2 = 43 .
Final answer y = 43 e−x sin 34 x.
(2) Suppose the function y satisfies the differential equation
(e) 16
y 00 + Ky = 0
for some constant K.
dE
(a) Show that the function E = (y 0 )2 + Ky 2 is constant. Hint: Calculate
.
dx
dE
Answer:
= 2y 0 y 00 + 2Kyy 0 = 2y 0 (y 00 + Ky) = 0. Hence E is a constant function.
dx
(b) Derive the general solution for the differential equation and confirm that E is constant
by direct calculation from your solutions. Note that the general solutions will have a
different form depending on whether K > 0, K = 0 or K < 0.
Answer:
• If K = −r2 < 0, then y = c1 erx + c2 e−rx and E = −4c1 c2 K.
• If K = ω 2 > 0, then y = c1 cos ωx + c2 sin ωx and E = (c21 + c22 )K.
• If K = 0, then y = c1 + c2 x and E = c22 .
In each case E is a constant.
HW8
d2 Q
dQ
1
In class we discussed the differential equation L 2 + R
+ Q = 0. We set ρ = R/2L and
dt
dt
C
√
d2 Q
dQ
ω0 = 1/ LC which simplified the equation to
+ 2ρ
+ ω02 Q = 0, and then we found the
dt2
dt
general solution in the case that ρ < ω0 .
(1) Find the general solution of this differential equation in the case that ρ = ω0 .
Answer: Setting ρ = ω0 , the differential equation becomes (D2 + 2ρD + ρ2 )Q = 0 with
auxiliary equation m2 + 2ρm + ρ2 = (m + ρ)2 = 0. So m = −ρ is a double root and the
7
general solution is
Q = (c1 + c2 t)e−ρt .
(2) Find the general solution of this differential equation in the case that ρ > ω0 .
Answer: The roots of the auxiliary equation are the (negative) real numbers
q
m = −ρ ± ρ2 − ω02
and the general solution is
√ 2 2
√ 2 2
Q = c1 e(−ρ+ ρ −ω0 )t + c2 e(−ρ− ρ −ω0 )t .
1 2
(3) The energy stored in the capacitor is EC =
Q . The energy stored in the inductor is
2C
dQ
1
. The resistor does not store energy — it absorbs electrical
EL = LI 2 . Reminder: I =
2
dt
energy and gives off heat. Show that the total energy, E = EC + EL . satisfies
dE
= −RI 2 .
dt
This means that the total energy is a decreasing function of time.
Answer:
2 !
dE
d
1 2 1
dQ
=
Q + L
dt
dt 2C
2
dt
dQ d2 Q
1 dQ
Q
+L
C dt
dt dt2
dQ 1
d2 Q
=
Q+L 2
dt C
dt
dQ
dQ
=
−R
= −RI 2
dt
dt
=
Here we are using, of course, L
dQ
1
d2 Q
+R
+ Q = 0.
dt2
dt
C
HW9
(1) Find the general solutions of the following differential equations.
(a) y 00 + y 0 − 2y = 2x. Answer: y = − 12 − x + c1 ex + c2 e−2x
(b) y 00 + 2y 0 = 3 + 4 sin 2x Answer: y = c1 + c2 e−2x + 21 (3x − sin 2x − cos 2x)
(c) (D − 1)y = sin2 x. Hint: Linear combinations of sin2 x and its derivatives look like
yp = A + B sin2 x + C sin x cos x (Check this). Alternatively, using a trig identity,
rewrite sin2 x as a function of cos 2x.
Answer: Suppose that yp = A + B sin2 x + C sin x cos x. Then
(D − 1)yp = −A + C cos2 x + (2B − C) cos x sin x − (B + C) sin2 x
= −A + C + (2B − C) cos x sin x − (B + 2C) sin2 x.
So that yp is a solution, we want −A + C = 0, 2B − C = 0 and −(B + 2C) = 1. These
equations imply B = −1/5, A = C = −2/5, and so the general solution is
2 1
2
y = − − sin2 x − sin x cos x + c1 ex .
5 5
5
OR
8
Since sin2 x = 12 (1 − cos 2x), we try yp = A + B cos 2x + C sin 2x. Then
(D − 1)yp = −A + (2C − B) cos 2x + (−2B − C) sin 2x.
So that (D−1)yp = 21 (1−cos 2x), we want −A = 1/2, 2C −B = −1/2 and −2B−C = 0.
These equations imply A = −1/2, B = 1/10 and C = −1/5, and so the general solution
is
1
1
1
y=− +
cos 2x − sin 2x + c1 ex .
2 10
5
(d) y 00 − y = ex . Answer: y = 12 xex + c1 ex + c2 e−x
(e) y 00 + ω02 y = cos ω0 x where ω0 6= 0 is constant. Answer: y = c1 cos ω0 x + c2 sin ω0 x +
1
x sin ω0 x
2ω0
(2) Let u and v be functions of x that satisfy u0 = xv and v 0 = xu. Find second order linear
differential equations satisfied by u and v separately.
Answer: Differentiating u0 = xv gives u00 = v + xv 0 . Then, since v 0 = xu and v = u0 /x, we
get u00 = u0 /x + x(xu). Clearing denominators, this can be written as xu00 − u0 − x3 u = 0.
By symmetry, v satisfies the same differential equation: xv 00 − v 0 − x3 v = 0.
(3) Suppose that yp is a solution of (D3 + 2D2 + D + 3)y = 0. Find a differential equation
satisfied by yp0 . Hint: Don’t try to solve the differential equation! Differentiate it.
Answer: Differentiating the given equation gives (D4 + 2D3 + D2 + 3D)y = 0 or (D3 +
2D2 + D + 3)(Dy) = 0, so, if yp satisfies the given differential equation, Dyp satisfies the
same differential equation.
d2 y
dy
dy
−x 2 =0
x > 0. Hint: Let u =
. What differential equation does
(4) Solve x3 − 3
dx
dx
dx
u satisfy? HW3(1a).
du
dy
Answer: u satisfies x3 − 3u − x
= 0, so from HW3(1a), u = x3 /6 + Cx−3 and
=
dx
dx
x3 /6 + Cx−3 . Integrating the last equation gives y = x4 /24 − Cx−2 /2 + B where C and B
are constants.
HW10
Find particular solutions of the following differential equations using the method of undetermined
coefficients.
1 2
x − x ex
(1) y 00 − y = xex Answer: yp =
4
1
(2) y 00 + y = 2 sin x cos x Hint: Use trigonometric identity first. Answer: yp = − sin 2x
6
1
2
(3) y 00 + y = 2x sin x cos x Hint: Ditto. Answer: yp = − x sin 2x − cos 2x
6
9
1
00
2
(4) y + y = 2x sin x Answer: yp =
−2x cos x + 2x sin x
4
1
(5) y 000 − y 00 = x3 Answer: yp = − x2 x2 + 12
4
HW11
(1) Find particular solutions of the following differential equations using the operator method.
1
1
1
(a) y 00 − 4y = e3x Answer: yp = 2
e3x = 2
e3x = e3x
D −4
3 −4
5
(b) y (4) = e3x Answer: yp =
1 3x
1
1 3x
e = 4 e3x =
e
4
D
3
81
(c) y 00 − 4y = e2x
Answer: yp =
D2
1
1
1
1 1
1
1
1
e2x =
e2x =
e2x = e2x · 1 = xe2x .
−4
D−2D+2
4D−2
4
D
4
9
(d) y 0 − 2y = xe2x Answer: yp =
1
1
1
1
xe2x = e2x
x = e2x x = x2 e2x .
D−2
(D + 2) − 2
D
2
(e) y 00 − 4y 0 + 4y = xe2x
1
1
1
1
Answer: yp =
xe2x = e2x
x = e2x 2 x = x3 e2x .
2
2
(D − 2)
((D + 2) − 2)
D
6
(f) y 00 − 4y = cos x
Answer: yp =
1
1
1
cos x =
cos x = − cos x
D2 − 4
−1 − 4
5
(g) y 00 − 4y = e2x cos x
Answer:
1
1
1
yp = 2
e2x cos x = e2x
cos x = e2x
cos x
2
D −4
(D + 2) − 4
(D + 4)D
1
D−4
1
= e2x
sin x = e2x 2
sin x = e2x
(D − 4) sin x
D+4
D − 16
−17
1 2x
1
(cos x − 4 sin x) =
e (4 sin x − cos x)
= e2x
−17
17
(2) Consider the differential equation (D2 − 1)y = x4 .
(a) Find a particular solution of this equation using the method of undetermined coefficients.
Answer: The complementary solution is yc = c1 ex + c2 e−x so we are in Case 1. We
guess that yp = Ax4 + Bx3 + Cx2 + Ex + F for some constants A, B, C, E, F . (Don’t
want to use D to represent a constant!) Then
(D2 − 1)yp = −Ax4 − Bx3 + (12A − C)x2 + (6B − E)x + 2C − F.
So that (D2 − 1)yp = x4 holds we need that A = −1, B = 0, C = −12, E = 0, and
F = −24. Thus yp = −x4 − 12x2 − 24.
(b) Find a particular solution of this equation using operator methods. Hint: You will need
1
the infinite series expansion of 2
.
D −1
2
Answer: Plugging x = D in the expansion
1
= 1 + x + x2 + x3 + x4 + x5 + · · ·
1−x
we get
1
1
=−
= −1 − D2 − D4 − D6 − D8 − · · ·
D2 − 1
1 − D2
so
yp =
D2
1
x4 = (−1 − D2 − D4 − D6 − D8 − · · · )x4
−1
= (−1 − D2 − D4 )x4
= −x4 − 12x2 − 24.
HW12
(1) Find particular solutions of the following:
(a) (D − 1)y = ex ln x. Answer: Using operator methods we find
yp =
1
1
ex ln x = ex ln x = ex (x ln x − x).
D−1
D
OR
10
This is a linear first order equation with P (x) = −1, Q(x) = ex ln x. The integrating
factor is H(x) = e−x so the general solution is
Z
Z
1
x
yp =
H(x)Q(x) dx = e
ln x dx = ex (x ln x − x + C).
H(x)
For a particular solution, just pick some number for C—zero would be as good as any
other number.
(b) (D − 1)y = 2 sin2 x.
Answer: Using the trigonometric identity 2 sin2 x = 1 − cos 2x, we get
1
1
1
1
2 sin2 x =
(1 − cos 2x) =
·1−
cos 2x.
yp =
D−1
D−1
D−1
D−1
Now we split the calculation into two parts:
1
1
1
·1=
· e0x =
= −1
D−1
D−1
0−1
D+1
D+1
1
cos 2x =
cos 2x = 2
cos 2x =
D−1
(D − 1)(D + 1)
D −1
1
1
− (D + 1) cos 2x = − (−2 sin 2x + cos 2x)
5
5
1
Combining these equations gives the final result yp = −1 + (−2 sin 2x + cos 2x).
5
OR
This is a linear first order equation with P (x) = −1, Q(x) = 2 sin2 x. The integrating
factor is H(x) = e−x so the general solution is
Z
Z
1
yp =
H(x)Q(x) dx = ex 2e−x sin2 x dx
H(x)
1 −x
= ex
e (−2 sin 2x + cos 2x − 5) + C .
5
1
= −1 + (−2 sin 2x + cos 2x) + Cex
5
The integral is hard! For a particular solution, just pick some number for C—zero
would be as good as any other number.
(2) Using the reduction of order method, find the general solution of the following differential
equations given the particular solution y1 .
(a) (D − 3)2 y = 0, y1 = e3x .
R
Answer: y 00 − 6y 0 + 9y = 0, p(x) = −6, e− p dx = e6x .
Z − R p dx
Z
e
v=
dx =
dx = x
y12
General solution: y = c1 e3x + c2 xe3x .
(b) x2 y 00 + 2xy 0 = 0, y1 = 1.R
Answer: p(x) = 2/x, e− p dx = x−2
Z − R p dx
Z
e
v=
dx
=
x−2 dx = −x−1
y12
General Solution: y = c1 + c2 x−1 .
√
(c) 2x2 y 00 + 3xy 0 − y = 0, y1 =
x.
R
Answer: p(x) = 3/2x, e− p dx = x−3/2 ,
Z − R p dx
Z
e
dx
=
x−5/2 dx = −2x−3/2 /3
v=
y12
√
General Solution: y = c1 x + c2 x−1 .
11
(d) x2 y 00 + xy 0 + (x2 − 1/4)yR = 0, y1 = x−1/2 sin x.
Answer: p(x) = 1/x, e− p dx = x−1 ,
Z − R p dx
Z
e
1
v=
dx
=
dx = − cot x
y12
sin2 x
General Solution: y = c1 x−1/2 sin x + c2 x−1/2 cos x.
HW13
(1) Using the variation of parameters method, find particular solutions of the following:
(a) y 00 − 4y = e3x . Note: You solved this already in HW11(1a) using operator methods
(which is easy). Now try the variation of parameters method.
Answer: y1 = e2x , y2 = e−2x , W = y1 y20 − y2 y10 = −4.
Z
Z
1
1 x
−y2 Q(x)
dx =
e dx = ex ,
v1 =
W
4
4
and
Z
Z
y1 Q(x)
1
1
dx = − e5x dx = − e5x .
W
4
20
1
Thus yp = v1 y1 + v2 y2 = e3x .
5
(b) (x2 + 1)y 00 − 2xy 0 + 2y = 6(x2 + 1)2 , y1 = x. Hint: Use the reduction of order method
to find y2 . Don’t forget to divide the equation through by x2 + 1 before identifying Q.
Answer: First we find y2 for the homogeneous
equation (x2 + 1)y 00 R− 2xy 0 + 2y = 0.
R
2
− p dx
Setting p(x) = −2x/(x + 1) we find e
= x2 + 1, and so y2 = x (x2 + 1)/x2 dx =
2
x − 1. (We did this part in class!)
Now to the variation of parameters method: We have Q(x) = 6(x2 + 1), and the
Wronskian is
v2 =
W = y1 y20 − y2 y10 = x2 + 1
so
Z
v1 =
−y2 Q(x)
dx =
W
Z
6(1 − x2 ) dx = 6x − 2x3 ,
and
Z
v2 =
y1 Q(x)
dx =
W
Z
6x dx = 3x2 .
Thus yp = v1 y1 + v2 y2 = 3x2 + x4 .
(c) x2 y 00 + xy 0 + (x2 − 1/4)y = 4x3/2 sin x. Hint: You found y1 and y2 in HW10(2d).
Answer: We have y1 = x−1/2 sin x and y2 = x−1/2 cos x so W = −1/x. Also Q(x) =
4x−1/2 sin x. Hence
Z
Z
−y2 Q(x)
v1 =
dx = 4 cos x sin x dx = −2 cos2 x,
W
and
Z
v2 =
y1 Q(x)
dx =
W
Z
−4 sin2 x dx = 2 sin x cos x − 2x.
Thus yp = v1 y1 + v2 y2 = −2x1/2 cos x.
d2 y
1 dy
1
(2) Consider the equation
−
+ 2 y = 0.
2
dx
x dx x
(a) Find n so that y1 = xn is a solution.
Answer: Plugging y1 = xn into the differential equation gives (n − 1)2 xn−2 = 0. So,
choosing n = 1 gives the solution y1 = x.
12
(b) Find the general solution of this equation.
R
Answer: We use reduction of order to find y2 . We have p(x) = −1/x, e− p dx = x,
Z − R p dx
Z
e
1
v=
dx =
dx = ln x
2
y1
x
General Solution: y = c1 x + c2 x ln x.
1 dy
d2 y
1
−
(c) Now solve
+ 2 y = 16x3 .
2
dx
x dx x
Answer: We use variation of parameters to find yp . We have y1 = x, y2 = x ln x,
Q = 16x3 , W = y1 y20 − y2 y10 = x.
Z
Z
−y2 Q(x)
v1 =
dx = −16x3 ln x dx = x4 (1 − 4 ln x),
W
(this integral is done by parts with u = ln x and dv = −16x3 dx), and
Z
Z
y1 Q(x)
dx = 16x3 dx = 4x4 .
v2 =
W
Thus yp = v1 y1 + v2 y2 = x5 . The general solution is
y = yp + yc = x5 + c1 x + c2 x ln x = x(x4 + c1 + c2 ln x).
HW14
(1) Find the general solution of (x2 + 2x)y 00 − 2(x + 1)y 0 + 2y = (x + 2)2 . Hint: One solution of
the associated homogeneous equation is a power of x.
Answer: We start by finding a solution of the homogeneous equation of the form y1 = xn
for some n. Plugging y1 = xn into the homogeneous equation gives
(x2 + 2x)y100 − 2(x + 1)y10 + 2y1 = (n − 2)((n − 1)x + 2n) xn−1 .
This is zero if n = 2 and so y1 = x2 is a solution of the homogeneous equation.
Next
of order
to find y2 . Setting p(x) = p = −2(x + 1)/(x2 + 2x) we
R
R we use reduction
2
− p dx
find p dx = ln(x + 2x), e
= x2 + 2x, and so
Z 2
1+x
x + 2x
v=
dx = − 2 ,
2
2
(x )
x
and y2 = vy1 = −(1 + x)
Finally, we use variation of parameters to find the particular solution. We have y1 = x2
and y2 = −(1 + x) so W = x(x + 2). Also Q(x) = (x + 2)2 /(x2 + 2x) = (x + 2)/x. Hence
Z
Z
−y2 Q(x)
1
1
1
v1 =
dx =
+ dx = ln x − ,
W
x2
x
x
and
Z
Z
y1 Q(x)
v2 =
dx = 1 dx = x.
W
Thus yp = v1 y1 + v2 y2 = x(x ln x − x − 2). This can be simplified a bit since x2 is in the
complementary solution and so we can add x2 to yp to get x(x ln x − 2) which is another
particular solution. (Whew! That was hard!)
(2) Find the power series solution for y 00 + y = 0. Compare your answer with the solutions
of this equation you know from before and the known power series for the sine and cosine
functions.
Answer: If y = a0 + a1 x + a2 x2 + a3 x3 + . . ., then y 00 = 2 · 1a2 + 3 · 2a3 x + 4 · 3a4 x2 + · · · +
(n + 2)(n + 1)an+2 xn + . . .. If y 00 + y = 0, then matching coefficients we have 2 · 1a2 + a0 = 0,
3 · 2a3 + a1 = 0, 4 · 3a4 + a2 = 0,. . . , (n + 2)(n + 1)an+2 + an = 0. This gives the recurrence
relation
1
an+2 = −
an ,
(n + 2)(n + 1)
13
valid for n = 0, 1, 2, . . .. This relation determines an for n = 2, 3, .. once a0 and a1 are
chosen. For example,
a4 = −
1
1
a2 =
a0
4·3
4·3·2·1
a5 = −
1
1
a3 =
a1 ,
5·4
5·4·3·2
and, in general,
a2n = (−1)n
a0
(2n)!
a2n+1 = (−1)n
a1
.
(2n + 1)!
Putting these expressions into the power series for y gives
x2
x4
x6
x3
x5
x7
y = a0 1 −
+
−
+ · · · + a1 x −
+
−
+ ···
2!
4!
6!
3!
5!
7!
= a0 cos x + a1 sin x,
as expected.
(3) Find a recurrence relation for the power series solution of y 00 = xy. Express the solution of
the differential equation in the form y = a0 y0 + a1 y1 , giving the first three nonzero terms
of the power series for y0 and y1 . (This is called Airy’s equation and its solutions are Airy
functions.)
Answer: If y = a0 + a1 x + a2 x2 + a3 x3 + . . ., then the differential equation is
2 · 1a2 + 3 · 2a3 x + 4 · 3a4 x2 + 5 · 4a5 x3 + 6 · 5a6 x4 + . . .
= a0 x + a1 x2 + a2 x3 + a3 x4 + . . .
Matching coefficients gives 2a2 = 0, 3 · 2a3 = a0 , 4 · 3a4 = a1 , 5 · 4a5 = a2 and, in general,
(n + 2)(n + 1)an+2 = an−1
for n > 0. Using this recurrence relation we get
1
1
a0 a4 =
a1
3·2
4·3
and so y = a0 y0 + a1 y1 where
a2 = 0 a3 =
a5 = 0 a6 =
1
a0
6·5·3·2
a7 =
1
a1 ,
7·6·4·3
x3
x6
x4
x7
+
+ ···
y1 = x +
+
+ ··· .
3·2 6·5·3·2
4·3 7·6·4·3
Just for fun, here are the graphs of these functions:
y0 = 1 +
2.0
1.5
1.0
y0
!10
!8
!6
!4
0.5
!2
y1
2
!0.5
!1.0
The oscillatory form of these functions for x < 0 is no surprise. If x is near −a for some
large real number
a, then√the given differential equation is almost y 00 + ay = 0 with solutions
√
y = c1 cos( ax) + c2 sin( ax).
(4) In class we came across the function
2
22 5
23
24
y1 = x + x3 +
x +
x7 +
x9 + · · ·
3
3·5
3·5·7
3·5·7·9
14
Show that y1 satisfies the first order linear differential equation
dy
− 2xy = 1.
dx
Use techniques from the beginning of the quarter to find all solutions of this differential
equation. Your answer will be expressed in terms of an integral that cannot be evaluated in
closed form. Now find the solution of the differential equation that has the same value at
x = 0 as y1 . This solution must be y1 so you have expressed y1 inR closed form as a definite
x
integral. Hint: If h(x) is any function and a is a constant, then a h(t) dt is a function of
x andRan antiderivativeR of h. In fact, all antiderivatives of h have this form. Thus we can
x
write h(x) dx + C = a h(t) dt for the set of all antiderivatives of h.
Answer: Checking the differential equation is straightforward:
23 6
23
22
x +
x8 + · · ·
y10 (x) − 2xy1 = 1 + 2x2 + x4 +
3
3
·
5
3
·
5
·
7
2
22 5
23
x +
x7 + · · ·
− 2x x + x3 +
3
3·5
3·5·7
=1
The differential equation
y 0 − 2xy = 1 is first order linear with P (x) = −2x and Q(x) = 1.
R
2
P dx
Setting H(x) = e
= e−x , the general solution is
Z
Z x
Z
2
2
2
2
1
Q(x)H(x) dx = ex
e−x dx + C = ex
y=
e−t dt
H(x)
a
where a and C are constants. Since y1 is a solution of the differential equation, y1 must
have this form for some particular a:
Z x
2
x2
y1 = e
e−t dt.
a
Z
To determine a we evaluate both sides of this equation when x = 0 to give 0 =
0
2
e−t dt.
a
Thus a = 0 and
y1 = ex
2
Z
x
2
e−t dt.
0
HW15
(1) Use the Leibniz rule to find a recurrence relation for the power series solution of y 00 = xy.
Compare your answer with HW14(3).
Answer: By Leibniz, Dn (xy) = (Dn y)x + n(Dn−1 y)(Dx) + · · · = xy (n) + ny (n−1) . The nth
derivative of the differential equation is
y (n+2) = xy (n) + ny (n−1) .
Evaluating this at x = 0 gives y (n+2) (0) = ny (n−1) (0) and so (n + 2)!an+2 = n(n − 1)!an−1 .
Canceling n! from both sides of this gives (n + 2)(n + 1)an+2 = an−1 , just as in HW14(3).
(2) Find a recurrence relation for the differential equation (x2 + 2)y 00 + 2xy 0 + 3y = 0. Find the
first four nonzero terms of the solution of this equation that satisfies y(0) = 0 and y 0 (0) = 1.
Answer: By differentiating the equation n times using the Leibniz rule we get
[(x2 + 2)y (n+2) + 2nxy (n+1) + n(n − 1)y (n) ] + 2[xy (n+1) + ny (n) ] + 3y (n) = 0.
Plugging in x = 0 yields 2y (n+2) (0) + (n(n − 1) + 2n + 3)y (n) (0) = 0 or 2y (n+2) (0) + (n2 + n +
3)y (n) (0) = 0. Using the formula n!an = f (n) (0), we get 2(n + 2)!an+2 + (n2 + n + 3)n!an = 0
and after cancelling n!,
2(n + 2)(n + 1)an+2 + (n2 + n + 3)an = 0.
15
The power series solution such that y(0) = 0 and y 0 (0) = 1 results by setting a0 = 0 and
a1 = 1. By the recurrence relation an = 0 for all even n, and for odd n we have a3 = −5/12,
a5 = 5/32, a7 = −55/896 etc., and
y =x−
5 3
5
55 7
x + x5 −
x + ···
12
32
896
(3) Find a recurrence relation for the differential equation (1 − x2 )y 00 − 2xy 0 + 2y = 0. Find the
general infinite power series solution (not just the first few terms). Note: You will also find
finite power series (i.e. polynomial) solutions.
Answer: By differentiating the equation n times using the Leibniz rule we get
[(1 − x2 )y (n+2) − 2nxy (n+1) − n(n − 1)y (n) ] − 2[xy (n+1) + ny (n) ] + 2y (n) = 0.
Plugging in x = 0 yields y (n+2) (0) − (n(n − 1) + 2n − 2)y (n) (0) = 0 or
2y (n+2) (0) − (n + 2)(n − 1)y (n) (0) = 0.
Using the formula n!an = f (n) (0), we get (n + 2)!an+2 = (n + 2)(n − 1)n!an and after
canceling (n + 2)n!,
(n + 1)an+2 = (n − 1)an .
As usual, a0 and a1 are arbitrary, and the other coefficients are determined by the recurrence
relation. For example, a2 = −a0 , a4 = a2 /3 = −a0 /3, a6 = 3a4 /5 = −a0 /5, a8 = 5a6 /7 =
−a0 /7, etc. Plugging n = 1 into the recurrence relation gives a3 = 0, and so a5 = a7 = a9 =
· · · = 0. Thus
x6
x8
x4
−
−
− ···
y =a0 1 − x −
3
5
7
!
∞
X
xn
= a0 1 −
+ a1 x
n−1
n=2
2
+ a1 x
(4) BONUS QUESTION. Use the polynomial solution of (1 − x2 )y 00 − 2xy 0 + 2y = 0 you found
in the previous question and reduction of order to find the general solution in closed form.
HARD! Needs partial fraction expansions. Assume that −1 < x < 1 to avoid more complicated cases.
R
Answer: Set y1 = x and p(x) = −2x/(1 − x2 ). Then −p(x) dx = − ln(1 − x2 ) and so
Z
v=
e−
R
p dx
y12
Z
dx =
1
dx =
(1 − x2 )x2
=−
Z 1
1
1
+
−
x2
2(x + 1) 2(x − 1)
1 1
1
+ ln(1 + x) − ln(1 − x).
x 2
2
So the general solution is
y = c1 x + c2 (2 − x ln(1 + x) + x ln(1 − x)).
dx
16
Here’s another way of getting the second solution from the above power series solution:
x4
x6
x8
y2 = 1 − x2 −
−
−
− ···
3 3 5 5 7 7
x
x
x
=1−x x+
+
+
+ ···
3
5
7
Z
=1−x
1 + x2 + x4 + x6 + · · · dx
Z
1
dx
=1−x
1 − x2
x
= 1 − (ln(1 + x) − ln(1 − x))
2
1
= (2 − x ln(1 + x) + x ln(1 − x))
2
HW16
Solve the following systems:
(1) x0 = x + 3y + 3
y 0 = 3x + y − 1 − 8t
Answer: Eliminating y gives the equation (D2 − 2D − 8)x = −6 − 24t with solution x =
c1 e4t + c2 e−2t + 3t. Then y is obtained from y = (x0 − x − 3)/3: y = c1 e4t − c2 e−2t − t
(2) x0 = x + 3y − 11 sin t
y 0 = 3x + y − 6 sin t
Answer: Eliminating y gives the equation (D2 − 2D − 8)x = −11 cos t − 7 sin t with solution
x = c1 e4t + c2 e−2t + cos t + sin t. Then y is obtained from y = (x0 − x + 11 sin t)/3:
y = c1 e4t − c2 e−2t + 3 sin t
(3) x0 + y 0 + 2y = e2t
x0 − x + y 0 = 0
Answer: x = (1 − 2t − 2c1 )e2t , y = (t + c1 )e2t
Answer: Eliminating y gives the (first order!) equation (D − 2)x = −2e2t with solution
x = (C − 2t)e2t . Subtracting the two equations gives x + 2y = e2t and so y = (e2t − x)/2 =
(1 + 2t − C)e2t /2.
(4) x0 = y
y0 = z
z 0 = 2z − y
Here x, y and z are all functions of t.
Answer: We express everything in terms of x. First y = x0 . Taking the derivative of this
gives y 0 = x00 and so z = x00 . Since z 0 = x000 , z 0 = 2z − y becomes x000 = 2x00 − x0 .
This homogenous constant coefficient equation has solution x = c1 + (c2 + c3 t)et . Then
y = x0 = (c2 + c3 + c3 t)et and z = y 0 = (c2 + 2c3 + c3 t)et .