Math 257/316 Assignment 1 Solutions
Problem 1. Find the solution to the initial value problem for the ODE
dy
1+y
=
dx
1+x
for each of the initial conditions
(a) y(0) = 1,
(b) y(0) = −1,
(c) y(0) = −2
The ODE is both separable and linear, so we can solve it either way. Let’s use its
separability:
Z
Z
dx
dy
=
=⇒ ln |1+y| = ln |1+x|+C or, taking the exponential |y+1| = K(x+1)
1+y
1+x
(since all the initial values are posed at x = 0, we have x + 1 > 0 and we can safely remove
the absolute value from x + 1).
(a)
y(0) = 1 =⇒ 2 = K =⇒ y + 1 = 2(x + 1) =⇒ y = 2x + 1
(since y(0) = 1, we have y + 1 > 0 and we can remove the absolute value from y + 1).
(b)
y(0) = −1 =⇒ 0 = K =⇒ y + 1 = 0 =⇒ y = −1
(again we can remove the absolute value).
(c)
y(0) = −2 =⇒ 1 = K =⇒ |y + 1| = x + 1 =⇒ −(y + 1) = x + 1 =⇒ y = −x − 2
(since y(0) = −2, y+1 < 0 and we must introduce a minus sign when removing the absolute
value).
Problem 2. Find the general solutions to:
(a) y 00 − 2y = 0
This is second-order
constant-coefficient, whose characteristic equation r2 − 2 = 0 has
√
roots r = ± 2, so the general solution is
√
y = c1 e
2x
√
+ c2 e−
1
2x
.
(b) y 00 + y 0 − 6y = 0
Also second-order constant coefficient, with characteristic equation 0 = r2 + r − 6 =
(r + 3)(r − 2) having roots r = 2 and r = −3. So the general solution is
y = c1 e2x + c2 e−3x .
(c) y 00 + 2y 0 + y = 0
Again, second-order constant coefficient, with characteristic equation 0 = r2 + 2r + 1 =
(r + 1)2 having double root r = −1, so the general solution is
y = c1 e−x + c2 xe−x .
Problem 3. Find the solutions to the initial value problems:
(a) y 00 + 9y = 0,
y(0) = 1, y 0 (0) = 6
The characteristic equation r2 + 9 = 0 has roots r = ±3i, so the general solution is
y = c1 cos(3x) + c2 sin(3x), and y 0 = −3c1 sin(3x) + 3c2 cos(3x).
1 = y(0) = c1 ,
(b) y 00 − 2y 0 + 5y = 0,
6 = y 0 (0) = 3c2 =⇒ c2 = 2 =⇒ y = cos(3x) + 2 sin(3x).
y(0) = 1, y 0 (0) = 7
√
The characteristic equation 0 = r2 −2r+5 has roots r = 1± 1 − 5 = 1±2i so the general
solution is y = ex (c1 cos(2x) + c2 sin(2x)), and y 0 = ex ((c1 + 2c2 ) cos(2x) + (c2 − 2c1 ) sin(2x)).
1 = y(0) = c1 ,
7 = y 0 (0) = c1 +2c2 = 1+2c2 =⇒ c2 = 3 =⇒ y = ex (cos(2x) + 3 sin(2x)) .
(c) x2 y 00 + 6xy 0 + 6y = 0,
y(1) = 0, y 0 (1) = 1
This is an Euler ODE, with characteristic equation 0 = r(r − 1) + 6r + 6 = r2 + 5r + 6 =
(r + 3)(r + 2) having roots r = −2 and r = −3, so the general solution (for x > 0) is
y = c1 x−2 + c2 x−3 , and so y 0 = −2c1 x−3 − 3c2 x−4 .
0 = y(1) = c1 + c2 =⇒ c2 = −c1 ,
1 = y 0 (1) = −2c1 − 3c2 = c1 ,
2
=⇒ y = x−2 − x−3 .
Problem 4. Find the first six non-zero terms in the power series y =
general solution of the second-order, linear, homogeneous ODE:
P∞
n
n=0 an x
of the
(x2 − 1)y 00 + 3xy 0 + y = 0
P∞
P
n−2 and plug into the ODE to
n−1 , y 00 =
Compute y 0 = ∞
n=2 n(n − 1)an x
n=1 nan x
find
0 = (x2 − 1)
∞
X
n(n − 1)an xn−2 + 3x
n=2
=
∞
X
∞
X
nan xn−1 +
n=1
n
n(n − 1)an x −
n=2
∞
X
n(n − 1)an x
n−2
∞
X
an xn
n=0
+
n=2
∞
X
n
3nan x +
n=1
∞
X
an xn
n=0
and shifting the index in the second series,
0=
∞
X
n
n(n − 1)an x −
n=2
∞
X
n
(n + 2)(n + 1)an+2 x +
n=0
∞
X
n
3nan x +
n=1
∞
X
an xn
n=0
The easiest way to proceed now, is to recognize that in both the first and third series, we
can just change the starting point to n = 0 without changing the sums (because the terms
we introduce by doing this are all zero), so:
0=
=
=
∞
X
n=0
∞
X
n=0
∞
X
n(n − 1)an xn −
∞
X
(n + 2)(n + 1)an+2 xn +
n=0
∞
X
3nan xn +
n=0
∞
X
an xn
n=0
[(n(n − 1) + 3n + 1)an − (n + 2)(n + 1)an+2 ] xn
(n + 1)2 an − (n + 2)(n + 1)an+2 xn .
n=0
(Note: we could also have combined the sums starting at n = 2, and collected the remaining
n = 0 and n = 1 terms from the last three sums separately – it would work out the same.)
Thus the recurrence relation is
n+1
(n + 1)2 an − (n + 2)(n + 1)an+2 = 0
=⇒
an+2 =
an , n = 0, 1, 2, 3, . . . .
n+2
So a0 and a1 are free, while
1
2
3
3
4
4·2
a2 = a0 , a3 = a1 , a4 = a2 =
a0 , a5 = a3 =
a1 , · · ·
2
3
4
4·2
5
5·3
and
1
3 4
2
4·2 5
y = a0 1 + x2 +
x + · · · + a1 x + x3 +
x + ··· .
2
4·2
3
5·3
3