MTHE 225 Homework 5 Solutions

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MTHE 225 Homework 5 Solutions




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Solutions #5
1. Use Laplace transforms to solve the following initial value problems:
(a) y000 + 3y00 + 3y0 + y = 2e−t , y(0) = 1, y0 (0) = 2, y00 (0) = −1
d4x d2x
(b) 4 − 2 2 = 3 + 2t + e2t , x(0) = 1, x0 (0) = 2, x00 (0) = x000 (0) = 0
dt dt

Solution. (a) Let Y (s) = L {y(t)}(s), then apply the Laplace transform to both sides of the equation:

L y000 + 3y00 + 3y0 + y (s) = L 2e−t (s)
 

2
L y000 (s) + 3L y00 (s) + 3L y0 (s) + L {y} (s) =
  
s+1
2
s3Y (s) − s2 y(0) − sy0 (0) − y00 (0) + 3(s2Y (s) − sy(0) − y0 (0)) + 3(sY (s) − y(0)) +Y (s) =
s+1
2
s3Y (s) − s2 − 2s + 1 + 3s2Y (s) − 3s − 6 + 3Y (s) − 3 +Y (s) =
s+1

s3 + 6s2 + 13s + 10
Y (s) s3 + 3s2 + 3s + 1 =
s+1
3 2
s + 6s + 13s + 10 s + 6s2 + 13s + 10
3
Y (s) = = .
(s + 1)(s3 + 3s2 + 3s + 1) (s + 1)4
Let us now find the inverse Laplace transform of Y (s) using partial fractions decomposition. We
have
s3 + 6s2 + 13s + 10 A B C D
4
= + 2
+ 3
+
(s + 1) s + 1 (s + 1) (s + 1) (s + 1)4
s3 + 6s2 + 13s + 10 A(s + 1)3 + B(s + 1)2 +C(s + 1) + D
=
(s + 1)4 (s + 1)4
s3 + 6s2 + 13s + 10 = A(s + 1)3 + B(s + 1)2 +C(s + 1) + D
s3 + 6s2 + 13s + 10 = A(s3 + 3s2 + 3s + 1) + B(s2 + 2s + 1) +Cs +C + D
s3 + 6s2 + 13s + 10 = As3 + 3As2 + 3As + A + Bs2 + 2Bs + B +Cs +C + D



 A=1

3A + B = 6
Equating coefficients gives: =⇒ A = 1, B = 3, C = 4 and D = 2.


 3A + 2B +C = 13

A + B +C + D = 10
1 3 4 2
Hence, Y (s) = + 2
+ 3
+ , and so
s + 1 (s + 1) (s + 1) (s + 1)4
1
y(t) = e−t + 3te−t + 2t 2 e−t + t 3 e−t .
3
(b) Let X(s) = L {x(t)}(s), then apply the Laplace transform to both sides of the equation:

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Solutions #5 MTHE 225: page 2 of 16

n o
L x(4) − 2x00 (s) = L 3 + 2t + e2t (s)


n o 3 2 1
L x(4) (s) − 2L x00 (s) = + 2 +

s s s−2
3 2 1
s4 X(s) − s3 x(0) − s2 x0 (0) − sx00 (0) − x000 (0) − 2(s2 X(s) − sx(0) − x0 (0)) = + 2+
s s s−2

3 2 1
X(s) s4 − 2s2 = + 2 + + s3 + 2s2 − 2s − 4
s s s−2
s6 − 6s4 + 12s2 − 4s − 4
X(s) =
s2 (s − 2)(s4 − 2s2 )
s6 − 6s4 + 12s2 − 4s − 4 s6 − 6s4 + 12s2 − 4s − 4
X(s) = = √ √ .
s2 (s − 2)s2 (s2 − 2) s4 (s − 2)(s − 2)(s + 2)
Let us now find the inverse Laplace transform of X(s) using partial fractions decomposition. We
have
s6 − 6s4 + 12s2 − 4s − 4 A B C D E F G
√ √ = + 2+ 3+ 4+ + √ + √
s4 (s − 2)(s − 2)(s + 2) s s s s s−2 s− 2 s+ 2
√ √
s6 − 6s4 + 12s2 − 4s − 4 As3 (s − 2)(s2 − 2) + Bs2 (s − 2)(s2 − 2) +Cs(s − 2)(s2 − 2) + D(s − 2)(s2 − 2) + Es4 (s2 − 2) + Fs4 (s − 2)(s + 2) + Gs4 (s − 2)(s −
√ √ = √ √
s4 (s − 2)(s − 2)(s + 2) s4 (s − 2)(s − 2)(s + 2)


Hence, we get
(1) √ √
s6 −6s4 +12s2 −4s−4 = As3 (s−2)(s2 −2)+Bs2 (s−2)(s2 −2)+Cs(s−2)(s2 −2)+D(s−2)(s2 −2)+Es4 (s2 −2)+Fs4 (s−2)(s+ 2)+Gs4 (s−2)(s− 2).
√ √
Setting s = 0 gives D = −1. Setting s = 2 gives E = 18 . Setting s = 2 gives F = 14 , and s = − 2
gives G = 14 . Next, differentiate both sides of equation (1) with respect to s, then set s = 0 to get
C = − 32 . Differentiate one more time with respect to s, then set s = 0 to get B = 74 . Differentiate
one last time with respect to s then set s = 0 to get A = 38 .
3 7 3 1 1 1
1
Hence, X(s) = 8 + 42 − 23 − 4 + 8 + 4√ + 4√ , and so
s s s s s−2 s− 2 s+ 2
3 7 3 1 1 1 √ 1 √
x(t) = + t − t 2 − t 3 + e2t + e 2t + e− 2t .
8 4 4 6 8 4 4

2. Compute: (a) t ∗ t ∗ t ∗ t (b) et ∗ t (c) sint ∗ sint (d) t 2 ∗ cost.
Solution. (a) Notice that t ∗ t ∗ t ∗ t = t ∗ (t ∗ (t ∗ t)). First,
 2 t
τ3
Z t
τ
t ∗t = τ(t − τ) dτ = t −
0 2 3 0
t3 t3 t3
= − = .
2 3 6




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