Differential Equations APM2611 University of South Africa 2011–2014 Complete exam material with memoranda and assignment solutions

APM2611/202/2/2014




Tutorial Letter 202/2/2014


DIFFERENTIAL EQUATIONS

APM2611

Semester 2

Department of Mathematical Sciences


This tutorial letter contains solutions
for assignment 02.




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, P∞
1. Let y = j=0 aj xj . The initial value problem becomes
∞
X ∞
X ∞
X
2 j−2 j−1
(x + 1) j(j − 1)aj x − 8x jaj x + 20 aj xj = 0, a0 = 1, a1 = 1.
j=2 j=1 j=0

Thus we have
∞
X ∞
X ∞
X ∞
X
j j−2 j
j(j − 1)aj x + j(j − 1)aj x −8 jaj x + 20 aj xj = 0, a0 = 1, a1 = 1.
j=2 j=2 j=1 j=0

To convert the second sum to easily comparable powers, we substitute j −2 → j so that j → j +2
and j(j − 1) → (j + 2)(j + 1) (the lower bound on the sum becomes j + 2 = 2 so that j = 0):
∞
X ∞
X ∞
X ∞
X
j j j
j(j − 1)aj x + (j + 2)(j + 1)aj+2 x − 8 jaj x + 20 aj xj = 0, a0 = 1, a1 = 1.
j=2 j=0 j=1 j=0

Separating terms in the sums which are not common yields
∞
X
(2a + 20a ) + (6a − 8a1 + 20a1 )x + (j(j − 1)aj + (j + 2)(j + 1)aj+2 − 8jaj + 20aj ) xj = 0.
| 2 {z 0} | 3 {z }
j=0 j=1 j=2


Comparing coefficients of x we find

a0 =1
a1 =1
j =0 2a2 + 20a0 = 0 a2 = −10
j =1 6a3 − 8a1 + 20a1 = 0 a3 = −2
..
.
(j − 4)(j − 5)
j(j − 1)aj + (j + 2)(j + 1)aj+2 − 8jaj + 20aj = 0 aj+2 = − aj
(j + 2)(j + 1)
..
.

Thus we find a4 = 5, a6 = 0 = a8 = a10 = · · · , and a5 = 1/5, a7 = 0 = a9 = a11 = · · · . It follows
that ∞
X x5
y(x) = aj xj = 1 + x − 10x2 − 2x3 + 5x4 + .
j=0
5

2. We have
Z ∞ Z 1 Z ∞ Z 1
−st −st −st
L {f (t)} = e f (t)dt = e f (t)dt + e f (t)dt = e−st (3t − 1)dt
0 0 1 0
 1
3t −st 3 1
= − e − 2 e−st + e−st
s s s 0
2 3 3 1
= − e−s − 2 e−s + 2 − .
s s s s


2

, APM2611/202


3. We use the table of Laplace transforms in the textbook.

(i)

d
L te2t sin 6t = (−1)1 L e2t sin 6t
 
ds 
d
=− L {sin 6t}
ds s→s−2
 
d 6
=−
ds s2 + 36 s→s−2
 
d 6
=−
ds (s − 2)2 + 36
12(s − 2)
= .
((s − 2)2 + 36)2

(ii)

L {cos(2t)U (t − π)} = L {cos(2(t − π) + 2π)U (t − π)}
= L {cos(2(t − π))U (t − π)}
s
= e−πs L {cos(2t)} = e−πs 2 .
s +4

(iii)
Z t   Z t 
−τ −t
L e cos τ dτ =L e t−τ
e cos τ dτ
0 0
Z t 
=L t−τ
e cos τ dτ
0 s→s+1
= L e L {cos t} s→s+1
 t

1 s
= 2
s − 1 s + 1 s→s+1
s+1
= .
s((s + 1)2 + 1)

4. (i) The table of Laplace transforms provides directly
 −πs   
−1 e −1 1
L =L U (t − π)
s2 + 1 s2 + 1 t→t−π

= sin t U (t − π)
t→t−π
= sin(t − π)U (t − π).

(ii) First find the partial fraction decomposition:

2s + 4 A B C
= + +
(s − 2)(s + 1)(s + 3) s−2 s+1 s+3


3

, 2s + 4 = A(s + 1)(s + 3) + B(s − 2)(s + 3) + C(s − 2)(s + 1)
2s + 4 = (A + B + C)s2 + (4A + B − C)s + (3A − 6B − 2C)
which provides the equations
s2 : 0 = A + B + C
s : 2 = 4A + B − C
1 : 4 = 3A − 6B − 2C.
The equations can be solved using Gauss elimination. Subtracting the second equation
from the first provides 2 = 3A − 2C which when inserted into the third equation yields
2 = −6B so that B = −1/3. The first two equation become
1 7
= A + C, = 4A − C
3 3
so that A = 8/15 and C = −1/5. Thus
 
−1 2s + 4
L
(s − 2)(s2 + 4s + 3)
     
8 −1 1 1 −1 1 1 −1 1
= L − L − L
15 s−2 3 s+1 5 s+3
2t −t −3t
8e e e
= − − .
15 3 5
5.
L {y 00 − 4y + 4y} = L t3 e2t


6
L {y 00 } − 4L {y 0 } + 4L {y} =
(s − 2)4
6
s2 L {y} − sy(0) − y 0 (0) − 4(sL {y} − y(0)) + 4L {y} =
(s − 2)4
6
(s2 − 4s + 4)L {y} =
(s − 2)4
6
L {y} =
(s − 2)6
 
6 −1 5!
y= L
120 (s − 2)6
1
y = t5 e2t .
20
6. We find
1 π 1 π
Z Z
a0 = f (x)dx = dx = 1
π −π π 0
1 π 1 π
Z Z
an = f (x) cos nx dx = cos nx dx = 0
π −π π 0
1 π 1 π
Z Z
1 π 1
bn = f (x) sin nx dx = sin nx dx = − cos nx 0
= (1 − (−1)n ) .
π −π π 0 nπ nπ


4