HW 8 due Oct 19, Friday 1. Find the inverse Laplace transform (a) F ...

1. Find the inverse Laplace transform
(a) F (s) = 3
s 2 + 4
3
s 2 + 4
(b) F (s) =
HW 8 due Oct 19, Friday
3
=
2 ·
2
s2 3
→
+ 22 2 sin(2t)
3s + 1
s 2 − s − 6
3s + 1
s 2 − s − 6 =
A = 7/5, B = 3/5
7/5
s − 3 → 7/5e3t ,
3s + 1
(s − 3)(s + 2)
3/5
s + 2
f(t) = 3/5
+ 3/5e−2t
s + 2
(c) F (s) =
2s + 1
s 2 − 2s + 2
2s + 1
s 2 − 2s + 2 =
2s + 1
(s − 1) 2 + 1
(d) F (s) = 8s2 − 4s + 12
s(s 2 + 4)
A B
= +
s − 3 s + 2
→ 3/5e−2t
8s2 − 4s + 12
s(s2 + 4) = 4 · 2s2 − s + 3
s(s2 + 4)
2s2 − s + 3
s(s2 A Bs + C
= +
+ 4) s s2 + 4
A = 3/4, B = 5/4, C = −1
8s2 − 4s + 12
s(s2 �
A Bs + C
= 4 +
+ 4) s s2 + 4
3
→ 3, or 3u(t)
s
5s − 4
s2 → 5 cos(2t) − 2 sin(2t)
+ 4
= 2(s − 1) + 3
(s − 1) 2 + 1 → 2 cos t · et + 3 sin t · e t
�
= 3 5s − 4
+
s s2 + 4
f(t) = 3 + 5 cos(2t) − 2 sin(2t) or 3u(t) + 5 cos(2t) − 2 sin(2t)
1

2. Solve the initial value problems.
(a) y ′′ + 3y ′ + 2y = 0; y(0) = 1, y ′ (0) = 0
L{y ′′ + 3y ′ + 2y} = L{0}
s 2 Y − sy(0) − y ′ (0) + 3[sY − y(0)] + 2Y = 0
[s 2 + 3s + 2]Y = s + 3
s + 3
Y =
s2 + 3s + 2
Method 1:
s + 3
Y =
s2 + 3s + 2 =
A = −1, B = 2
−1
→ −e−2t
s + 2
2
→ 2e−t
s + 1
y(t) = −e −2t + 2e −t
Method 2:
Y =
=
s + 3
s 2 + 3s + 2 =
s + 3
(s + 2)(s + 1)
s + 3
(s + 3/2) 2 − (1/2) 2
A B
= +
s + 2 s + 1
s + 3/2
(s + 3/2) 2 3/2
+
− (1/2) 2 (s + 3/2) 2 − (1/2) 2
s + 3/2
(s + 3/2) 2 − (1/2) 2 → e−3/2t cosh(1/2t) = e −3/2t · e1/2t + e−1/2t 2
3/2
(s + 3/2) 2 − (1/2) 2 → 3e−3/2t sinh(1/2t) = 3e −3/2t · e1/2t − e−1/2t 2
y(t) = e−t + e −2t
2
+ 3e−t − 3e −2t
2
(b) y ′′ − 4y ′ + 4y = 0; y(0) = 1, y ′ (0) = 1
= −e −2t + 2e −t
2
= e−t + e −2t
2
= 3e−t − 3e −2t
2

s 2 Y − s − 1 − 4[sY − 1] + 4Y = 0
(s 2 − 4s + 4)Y = s − 3
Y =
s − 3
s 2 − 4s + 4
1
→ e2t
s − 2
1
→ t · e2t
(s − 2) 2
y(t) = e 2t − te 2t
s − 3 (s − 2) − 1 1
= = =
(s − 2) 2 (s − 2) 2
(c) y ′′ − 2y ′ + 2y = cos t; y(0) = 1, y ′ (0) = 0
s − 2 −
s 2 Y − s − 2[sY − 1] + 2Y = s
s2 + 1
(s 2 − 2s + 2)Y = s
s2 + s − 2
+ 1
s
Y =
(s2 + 1)(s2 − 2s + 2) +
s − 2
s2 − 2s + 2
s
(s2 + 1)(s2 As + B
=
− 2s + 2) s2 Bs + D
+
+ 1 s2 − 2s + 2
A = 1/5, B = −2/5, C = −1/5, D = 4/5
s
(s2 + 1)(s2 �
1 s − 2
=
− 2s + 2) 5 s2 −s + 4
+
+ 1 s2 �
− 2s + 2
s − 2
s2 s
=
+ 1 s2 −2
+
+ 1 s2 → cos t − 2 sin t
+ 1
−s + 4
s2 −(s − 1) + 3
=
− 2s + 2 (s − 1) 2 + 1 → − cos t · et + 3 sin t · e t
y = 1
5 [cos t − 2 sin t − et cos t + 3e t sin t]
3
1
(s − 2) 2

(d) y ′′ − 2y ′ + 2y = e −t ; y(0) = 0, y ′ (0) = 1
s 2 Y − 1 − 2sY + 2Y = 1
s + 1
(s 2 − 2s + 2)Y = 1 s + 2
+ 1 =
s + 1 s + 1
s + 2
Y =
(s + 1)(s2 A Bs + C
= +
− 2s + 2) s + 1 s2 − 2s + 2
A = 1/5 B = −1/5 C = 8/5
Y = 1
�
1 −s + 8
+
5 s + 1 s2 �
− 2s + 2
1
→ e−t
s + 1
−s + 8
s2 −(s − 1) + 7
=
− 2s + 2 (s − 1) 2 + 1 → − cos t · et + 7 sin t · e t
y = 1
5 [e−t − e t cos t + 7e t sin t]
(e) y (4) − 4y ′′′ + 6y ′′ − 4y ′ + y = 0; y(0) = 0, y ′ (0) = 1, y ′′ (0) = 0, y ′′′ (0) = 1
(s 4 − 4s 3 + 6s 2 − 4s + 1)Y = s 2 − 4s + 7
s
Y =
2 − 4s + 7
s4 − 4s3 + 6s2 − 4s + 1 = s2 − 4s + 7
(s − 1) 4
= [s2 − 2s + 1] − 2s + 6
(s − 1) 4 = (s − 1)2 −2s + 6
+
(s − 1) 4 (s − 1) 4
1 −2(s − 1) + 4
= +
(s − 1) 2 (s − 1) 4
1 −2 4
= + +
(s − 1) 2 (s − 1) 3 (s − 1) 4
y = te t − 2 t2
2! et + 4 t3
3! et
4