Assignment 2 Solution

Assignment # 2 (solution) Exercise 4.4 37. The Laplace transform of the given equation is L { f } + L {t} L { f } = L {t} Solving for L { f } we obtain L { f } =

1 . s +1 2

Thus f (t) = sin t.

Exercise 4.5 2. The Laplace transform of the differential equation yields 2 e−s L {y} = + s +1 s +1 so that y = 2e − t + e − ( t −1)U (t − 1).

4. The Laplace transform of the given equation yields L {y} =

1 4 e − 2πs so that 2 4 s + 16

1 y = sin 4(t − 2π )U (t − 2π ) 4 y=1/4 sin4t U(t-2π)

1

Exercise 4.6

3. Taking the Laplace transform of the system gives sL {x} +1 = L {x} − 2 L {y} sL {y} − 2 = 5L {x} − L {y} so that L {x} =

− s −5 s 5 3 = − − 2 2 2 s +9 s +9 3 s +9

and x = − cos 3t − Then y =

5 sin 3t 3

1 1 7 x − x ′ = 2 cos 3t − sin 3t 2 2 3

2

11. Taking the Laplace transform of the system gives s2 L {x} + 3(s + 1) L{ y} = 2 s2 L{x} + 3 L {y}=

1 (s+1)2

so that L {x} = − 2s + 1 = 1 + 1 + 1 2 − 1 3 2 3

s ( s + 1) s 1 Then x = 1 + t + t 2 − e −t 2

s

2s

3

s +1

1 1 1 1 1 And y = te −t − x ′′ = te −t + e −t − . 3 3 3 3 3

15 (b).

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