1. No category

MA 30300 SAMPLE MIDTERM EXAM #1 SOLUTIONS
Problem 1. Consider the differential equation
x3 − x2 y 00 + (x − 1) y 0 + y = 0.
a. Find the singular points of the differential equation.
b. Classify these singular points as regular or irregular. Explain your reasoning.
Solution: (a) We have the coefficients

P (x) = x3 − x2 



Q(x) = x − 1




R(x) = 1

Q(x)
1



 p(x) = P (x) = x2
=⇒


1
R(x)

 q(x) =
= 2
P (x)
x (x − 1)
These functions fail to be continuous at x0 = 0 and x0 = 1, so these are the singular points.
(b) We follow the discussion of section 5.4 in the text. We compute two pairs of limits:
lim (x − x0 ) · p(x)
=
lim (x − x0 )2 · q(x)
=
lim (x − x0 ) · p(x)
=
lim (x − x0 )2 · q(x)
=
x→x0
lim x ·
x→0
x0 = 0 :
x→x0
x→x0
1
x2
lim x2 ·
x→0
x2
lim (x − 1) ·
x→1
x0 = 1 :
x→x0
1
(x − 1)
∞
=
−1
1
x2
lim (x − 1)2 ·
x→1
=
1
x2 (x − 1)
=
0
=
0
Hence x0 = 0 is an irregular singular point while x0 = 1 is a regular singular point.
Problem 2. Consider the differential equation
x2 y 00 + 4 x y 0 + x2 + 2 y = 0.
a. Find the indicial equation F (r) = 0, as well as the exponents r at the singularity.
b. Use Frobenius’ Method to find a solution corresponding to the smaller exponent at the singularity.
List the first three terms of your answer via a series in the form
"
2 k #
∞
X
x
r
y(x) = x 1 +
a2k (r)
,
x > 0.
2
k=1
1
Solution: We follow the discussion in section 5.6 of the text. Using part (b), we will guess that the solution
is a series in the form
"
2 k # X
∞
∞
X
x
1
a2k x2k+r .
y(x) = xr 1 +
=
a2k (r)
2
2k
k=1
k=0
Then we have the following derivatives:
∞
∞
X
X
1
1
2k+r
2
a
x
a2k x2k+r+2
x2 + 2 y =
+
2k
2k
2k
"
4 x y0 =
k=0
∞
X
k=0
x2 y 00 =
=
k=0
1
4 (2 k + r) a2k x2k+r
2k
=
∞
X
1
(2 k + r) (2 k + r − 1) a2k x2k+r
2k
=
k=0
∞ X
#
2 k
x
xr 2 a0 +
2 a2k + 2 a2k−2
2
k=1
"
2 k #
∞ X
x
r
x 4 r a0 +
4 (2 k + r) a2k
2
k=1


r (r − 1) a0
2 k 

∞ xr  X
x

+
(2 k + r) (2 k + r − 1) a2k
2
k=1
Adding these together, we find the expression
"
2 k #
∞ X
x
2 00
0
2
r
x y + 4 x y + x + 2 y = x F (r) a0 +
F (2k + r) a2k + 2 a2k−2
2
k=1
in terms of
F (r) = r (r − 1) + 4 r + 2 = r2 + 3 r + 2 = (r + 1) (r + 2).
(a) As we found above, the indicial equation is F (r) = r2 + 3 r + 2 = 0. The exponents at the singularity
are just the roots of this equation, so they are r1 = −1 and r2 = −2.
(b) We must also have F (2k + r) a2k + 2 a2k−2 = 0 for k ≥ 1. As the smaller exponent is r = −2, we find
that
1
F (2k + r) = (2k + r) + 1 (2k + r) + 2 = 2 k − 1 2 k
=⇒
a2k = −
a2k−2 .
(2 k − 1) k
The first few terms are
−1
(−1)k
−1
+1
a0 ,
a6 =
a0 ,
···
a2k =
a0 .
a2 =
a0 ,
a4 =
1·1
1 · 3 · 2!
1 · 3 · 5 · 3!
1 · 3 · · · (2k − 1) · k!
Hence the solution is
"
#
2 2
2 3
2 k
x
x
x
1
x2
1
1
(−1)k
y(x) = 2 1 −
+
−
+ ··· +
+ ···
x
2
1 · 3 · 2! 2
1 · 3 · 5 · 3! 2
1 · 3 · · · (2k − 1) · k! 2
Problem 3. Consider the initial value problem
y 00 + 6 y 0 + 25 y = 0
such that
y(0) = 0
a. Find the Laplace transform L y (s) of the solution y(t).
b. Find the solution y(t) to the initial value problem.
and
y 0 (0) = 4.
Solution: (a) Corollary 6.2.2 asserts that
L y (s)
L y 0 (s)
L y 00 (s)
=
=
Y (s)
s Y (s) −
2



y(0)
= s Y (s) − s y(0)
−
=⇒


y (0)
0
2

0 = L y 00 + 6 y 0 + 25 y (s)





= L y 00 (s) + 6 L y 0 (s) + 25 L y (s)




 = s2 Y (s) − 4 + 6 s Y (s) + 25 Y (s)
Solving for Y (s), we find that
4
.
+ 6 s + 25
(b) We complete the square in the denominator above to see that
b
4
=
in terms of
Y (s) =
(s + 3)2 + 42
(s − a)2 + b2
Using Table 6.2.1, we know that
f (t) = eat sin bt
=⇒
L f (s) =
Y (s) =
s2
so we see that the solution is
y(t) = e−3t sin 4t.
3
a = −3,
b
(s − a)2 + b2
b = 4.