Problem 1
Find the solution of
(2y sin x − cos3 x) dx + cos x dy = 0,
subject to the initial condition y(π/4) = π/4.
Hint:
d
dx
ln sec x = tan x.
Solution 1. This is an ODE of the form M dx + N dy = 0 where M = 2y sin x − cos3 x and
N = cos x. Is it exact?
My − Nx = 2 sin x − (−sin x) = 3 sin x ̸= 0
so My ̸= Nx . However, (My − Nx )/N = 3 tan x is a function of x alone, so an integrating factor is
µ=e
R
3 tan x
= e3 ln sec x = sec3 x = 1/ cos3 x
by the hint. The scaled ODE is
cos x
2y sin x − cos3 x
dx +
dy = 0
cos3 x
cos3 x
(2y tan x sec2 x − 1) dx + sec2 x dy = 0
which we know must be exact. So we look for a potential U such that
Ux = 2y tan x sec2 x − 1
and Uy = sec2 x.
The second equation is easier to integrate. Doing so gives U = y sec2 x + C(x), whence
Ux = 2y sec x(sec x)′ + C ′ (x) = 2y sec x(sec x tan x) + C ′ (x) = 2y tan x sec2 x + C ′ (x).
Comparing the two expressions for Ux shows that C ′ (x) = −1 so C(x) = −x. Thus U = y sec2 x − x
is a potential, and its level sets
y sec2 x − x = c
are solution curves. To find c we use the initial condition y(π/4) = π/4 which gives
√ 2
c = (π/4) sec2 (π/4) − π/4 = (π/4)( 2 − 1) = π/4
because π/4 is 45 degrees. Thus
y sec2 x − x = π/4
or y = (x + π/4) cos2 x.
Solution 2. Rearranging the ODE into the form y ′ = G(x, y) gives
y′ =
dy
2y sin x − cos3 x
=−
= cos2 x − 2y tan x
dx
cos x
so
y ′ + 2y tan x = cos2 x.
1
An integrating factor is
µ=e
R
2 tan x
= e2 ln sec x = sec2 x
by the hint. The scaled ODE is
y ′ sec2 x + 2y tan x sec2 x = 1.
But the l.h.s. is equal to (yµ)′ , so
(y sec2 x)′ = 1.
Integrating gives
y sec2 x = x + c
for some constant c. To find c we use the initial condition y(π/4) = π/4 which gives
√ 2
c = (π/4) sec2 (π/4) − π/4 = (π/4)( 2 − 1) = π/4
because π/4 is 45 degrees. Thus
y sec2 x − x = π/4
or y = (x + π/4) cos2 x.
2
Problem 2
Find the general solution to
y ′′ − 2y ′ + y =
ex
.
1 + x2
Solution. Note: This is problem 3.6.8 from the textbook. This is a nonhomogeneous linear ODE, so
the general solution is y = yc +yp . Since the ODE has constant coefficients, we use the characteristic
polynomial:
r2 − 2r + 1 = (r − 1)2 .
Thus r = 1 is a repeated real root, so the complementary solution is yc = c1 ex + c2 xex . By the
method of variation of parameters, a particular solution is given by
yp = Aex + Bxex
where
A′ =
−y2 g
W
and B ′ =
Here,
g=
y1 g
.
W
ex
1 + x2
and the Wronskian is
W =
Thus
A′ =
ex
ex
1
xex
= e2x
1
(x + 1)ex
−xex ex
−x
=
2
2x
(1 + x )e
1 + x2
x
= e2x (x + 1 − x) = e2x .
x+1
and B ′ =
ex ex
1
=
.
2
2x
(1 + x )e
1 + x2
This means B = arctan(x). As for A, notice that the numerator is basically the derivative of the
denominator: (1 + x2 )′ = 2x. Thus A = − 21 ln(1 + x2 ). (The absolute value is not required because
1 + x2 is always positive.) Putting these together gives
1
yp = − ex ln(1 + x2 ) + xex arctan(x)
2
so the general solution is
1
y = c1 ex + c2 xex − ex ln(1 + x2 ) + xex arctan(x).
2
3
Problem 3
Consider the functions
u(x) = cos(x2 )
and v(x) = sin(x2 ).
(a) Show that y = u(x) and y = v(x) are solutions to xy ′′ − y ′ + 4x3 y = 0.
(b) Show directly that u and v are linearly independent in every open interval.
(c) Show that the Wronskian of u and v on R is neither nowhere zero nor identically zero.
(d) Why does this not contradict Abel’s identity?
Solution.
(a) By the chain rule and product rule,
u(x) = cos(x2 )
v(x) = sin(x2 )
u′ (x) = −sin(x2 )(2x)
v ′ (x) = cos(x2 )(2x)
= −2x sin(x2 )
= 2x cos(x2 )
u′′ (x) = −2 sin(x2 ) − 2x cos(x2 )(2x)
v ′′ (x) = 2 cos(x2 ) + 2x(−sin(x2 )(2x))
= −2 sin(x2 ) − 4x2 cos(x2 )
= 2 cos(x2 ) − 4x2 sin(x2 )
Thus
xu′′ (x) − u′ (x) + 4x3 u(x) = −2x sin(x2 ) − 4x3 cos(x2 ) − −2x sin(x2 ) + 4x3 cos(x2 ) = 0
and similarly
xv ′′ (x) − v ′ (x) + 4x3 v(x) = 2x cos(x2 ) − 4x3 sin(x2 ) − 2x cos(x2 ) + 4x3 sin(x2 ) = 0.
(b) Suppose a cos(x2 ) + b sin(x2 ) = 0 for all x, for some constants a and b. If b ̸= 0 then
tan(x2 ) =
−a
sin(x2 )
=
cos(x2 )
b
would be constant, which is nonsense. So b = 0. But then a cos(x2 ) = 0 for all x, so a = 0.
(c) By definition, the Wronskian is
W =
u
u′
cos(x2 )
v
′ =
−2x sin(x2 )
v
sin(x2 )
2 2
2 2
2 = 2x cos (x ) + sin (x ) = 2x
2x cos(x )
which is zero for x = 0 and nonzero for x ̸= 0.
(d) Abel’s identity implies that the Wronskian of any pair of linearly independent solutions to
y ′′ + p(x)y ′ + q(x) = 0
is either nowhere zero or identically zero on any interval where the coefficient functions p and
q are continuous. In our case,
1
y ′′ − y ′ + 4x2 y = 0
x
we see that p(x) = −1/x is not continuous at 0, so Abel’s identity does not apply on R.
4
Problem 4
Solve the initial value problem
xy ′′ + 2y ′ + 6x = 1
y(1) = 1, y ′ (1) = 0.
Solution. This ODE is missing the variable y, so we let v = y ′ . This reduces it to the 1st-order
equation
xv ′ + 2v = 1 − 6x
which has x as an integrating factor:
x2 v ′ + 2xv = x − 6x2
The l.h.s. is (x2 v)′ so, integrating once, we get
x2 v =
1 2
x − 2x3 + c1 .
2
Dividing through by x2 gives
1
c1
− 2x + 2 .
2
x
Plugging in the initial condition v(1) = y ′ (1) = 0 gives
v=
0=
1
− 2 + c1
2
so c1 = 3/2. Thus
3
1
− 2x + 2 .
2
2x
But v = y ′ so, integrating once more, we get
v=
y=
3
x
− x2 −
+ c2 .
2
2x
Plugging in the initial condition y(1) = 1 gives
1=
so c2 = 3. Therefore
y=
1
3
− 1 − + c2
2
2
x
3
− x2 −
+ 3.
2
2x
5
Problem 5
Imagine a rocket flying through empty space, propelling itself via the expulsion of burning fuel. In
the absence of external forces, the motion of the rocket is governed by the equation
m
dm
dv
+s
=0
dt
dt
where v and m are the velocity and total mass, respectively, of the rocket (together with all its
unburned fuel) at time t, and s is the constant exhaust speed (relative to the rocket). Assume the
fuel constitutes half the initial mass of the rocket and is expelled at a rate proportional to s. If the
rocket starts from rest, how far will it have travelled the moment it runs out of gas?
Hints:
• The answer is proportional to the initial mass.
Z
•
ln u du = u ln u − u.
Solution. Note: This is similar to Homework 2 Question 5. The distance travelled is the integral
of the speed (v) w.r.t. time (t) over the duration T of the trip. To get v, we rearrange the equation
and separate variables, which gives
dm
dv = −s
m
and then integrate from 0 to t, which gives
v − v0 = −s (ln m − ln m0 ) = s ln m0 − s ln m.
Note that we don’t use absolute values because mass is always positive. Since the rocket starts
from rest, v0 = 0. This gives
v = s ln m0 − s ln m
for the speed of the rocket. To integrate this w.r.t. t, we need to write the r.h.s. in terms of t. So,
we need to know more about m. The assumption that the mass is expelled at a rate proportional
to s means
dm
= −ks
dt
for some constant of proportionality k > 0, so
m = m0 − kst.
Therefore the distance travelled is
Z T
Z
v dt = s
0
T
Z
ln m0 dt − s
0
T
ln(m0 − kst) dt.
0
The first integral is sT ln m0 . Since T is the time it takes for all the fuel to burn up,
m0 − ksT =
m0
,
2
6
so T =
m0
2ks
(*)
and thus
m0
ln m0 .
(1)
2k
For the second integral, we use the hint. Since m = m0 − kst and dm = −ks dt we can integrate
w.r.t. m instead of w.r.t. t (by change of variables). This gives
sT ln m0 =
Z
T
ln(m0 − kst) dt = −
s
0
=
=
=
=
=
1
k
1
k
1
k
1
k
1
k
1
k
Z
Z
m0 /2
ln m dm
m0
m0
ln m dm
m0 /2
m0
(m ln m − m)
m0 /2
h
m
m0
m0 i
0
(m0 ln m0 − m0 ) −
ln
−
2
2
2
h
m0 i
m0
[ln m0 − ln 2] +
m0 ln m0 − m0 −
2
2
m0
m0
m0 ln 2
ln m0 −
+
.
2
2
2
Plugging (1) and (2) into (*) gives that the distance travelled is
m0
1 m0
m0
m0 ln 2
ln m0 −
ln m0 −
+
2k
k 2
2
2
m0 ln 2
m0
−
=
2k
2k
m0
=
(1 − ln 2).
2k
7
(2)