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Millersville University Name Answer Key
Mathematics Department
MATH 365, Ordinary Differential Equations, Test 3
November 24, 2008, 2:00-2:50PM
Please answer the following questions. Show all work and write neatly. Answers without
justifying work will receive no credit. Partial credit will be given as appropriate, do not leave
any problem blank. The point values of problems are indicated in parentheses.
- (10 points) The Chebyshev equation is
(1 − x
2 )y
′′ − xy
′
2 y = 0
where α is a constant. Show that x = 1 is a regular singular point and find the
exponents of singularity.
We will let P (x) = 1 − x
2 , Q(x) = −x, and R(x) = α
2
. Since P (1) = 0 then x = 1 is
a singular point. We also note that
lim x→ 1
(x − 1)
Q(x)
P (x)
= lim x→ 1
(x − 1)
−x
1 − x
2
= lim x→ 1
x
1 + x
= p 0 ,
lim x→ 1
(x − 1)
2
R(x)
P (x)
= lim x→ 1
(x − 1)
2
α
2
1 − x
2
= lim x→ 1
−α
2 (x − 1)
1 + x
= 0 = q 0.
Since the preceding two limits are finite, then x = 1 is a regular singular point. The
exponents of singularity are the roots of the indicial equation:
r(r − 1) + p 0 r + q 0 = 0
r(r − 1) +
r = 0
r
r −
r 1 =
and r 2 = 0.
- (15 points each) Solve the following ODEs/IVPs.
(a) t
2 y
′′
′
If z = ln t then
d
2 y
dz^2
dy
dz
and the characteristic equation is
r
2
r = − 1 ± 2 i
where i =
−1. Thus the general solution is
y(t) = t
− 1 (c 1 cos(2 ln t) + c 2 sin(2 ln t)).
(b) 2t
2 y
′′ − 6 ty
′
The Euler equation above is equivalent to
t
2 y
′′ − 3 ty
′
If z = ln t then
d
2 y
dz
2
dy
dz
and the characteristic equation is
r
2 − 4 r + 4y = 0
r = 2.
Thus the general solution is
y(t) = t
2 (c 1 + c 2 ln t).
- (15 points) Consider the ODE
xy
′′
which has a regular singular point at x = 0. Find the first three nonzero coefficients of
a solution to the ODE corresponding to the larger of the two exponents of singularity.
Assuming y(x) =
∑^ ∞
n=
anx
r+n .
0 = x
∞ ∑
n=
(r + n)(r + n − 1)anx
r+n− 2
n=
anx
r+n
∑^ ∞
n=
(r + n)(r + n − 1)anx
r+n− 1
n=
2 an− 1 x
r+n− 1
= r(r − 1)a 0 x
r− 1
∞ ∑
n=
(r + n)(r + n − 1)anx
r+n− 1
n=
2 an− 1 x
r+n− 1
= r(r − 1)a 0 x
r− 1
∑^ ∞
n=
[(r + n)(r + n − 1)an + 2an− 1 ] x
r+n− 1
Thus the exponents of singularity are r 1 = 1 and r 2 = 0.
Now we will look for the first three nonzero coefficients of the series solution corre-
sponding to r = r 1 = 1. The recurrence relation is
an = −
2 an− 1
n(n + 1)
If a 0 6 = 0 is arbitrary then
a 1 = −
a 0
a 2 = −
a 1 =
2
a 0
a 3 = −
a 2 = −
3
a 0
an =
n
(n + 1)! n!
a 0.
Thus one solution to the ODE has the form
y 1 (x) = a 0 x
[
∞ ∑
n=
n x
n
(n + 1)! n!
]
- (15 points) Use the definition of the Laplace transform to find L {t
2 e
at } , where a is a
constant.
L
t
2 e
at
0
e
−st t
2 e
at dt
= lim M →∞
M
0
t
2 e
−(s−a)t dt
= lim M →∞
[
t
2
s − a
e
−(s−a)t
M
0
2 t
(s − a)
2
e
−(s−a)t
M
0
(s − a)
3
e
−(s−a)t
M
0
]
= lim M →∞
[
M
2
s − a
e
−(s−a)M −
2 M
(s − a)^2
e
−(s−a)M −
(s − a)^3
e
−(s−a)M
(s − a)^3
]
(s − a)^3
for s > a.
