Homework solutions

Math 215
February 5, 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
D −4
(D + 2)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
= e2x
(cos x − 4 sin x) =
e (4 sin x − cos x)
−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 − · · ·
2
D −1
1 − D2
so
1
x4 = (−1 − D2 − D4 − D6 − D8 − · · · )x4
yp = 2
D −1
= (−1 − D2 − D4 )x4
= −x4 − 12x2 − 24.