Find the solution of the initial value problem y 2y 5y e3

Find the solution of the initial value problem y\" + 2y\' + 5y = e^-3t, y(0) = 3, y\'(0) = 3 with L(y) obtained from homework 11 on paper problem 2a with the following steps a) express L(y) in terms of partial fraction and b) find the inverse of L(y). (see examples 4 and 5 on p. 214-215; examples 1, 2, and 3 under non-narrated Lecture 11 and examples on D2L; and example on page X of Lecture 11)

Solution

Solution :

Apply L to both sides:
(s² L(y) - 3s - 3) + 2(s L(y) - 3) + 5 L(y) = 1/(s - (-3)).

Solve for L(y):
(s² + 2s + 5) L(y) - 3s - 9 = 1/(s + 3)
==> (s² + 2s + 5) L(y) = (3s + 9) + 1/(s + 3)
==> L(y) = (3s + 9)/(s² + 2s + 5) + 1/[(s + 3)(s² + 2s + 5)].
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a) Since s² + 2s + 5 is irreducible, we only need to apply partial fractions to the second term.

1/[(s + 3)(s² + 2s + 5)] = A/(s + 3) + (Bs+C)/(s² + 2s + 5)
==> 1 = A(s² + 2s + 5) + (Bs+C)(s+3)

Letting s = -3 ==> 1 = 8A ==> A = 1/8.
So, 1 = (1/8)(s² + 2s + 5) + (Bs+C)(s+3)
==> 8 = (s² + 2s + 5) + 8(Bs+C)(s+3)
==> 0 = (s+3)(s-1) + 8(Bs+C)(s+3)
==> 0 = (s-1) + 8(Bs+C)
==> B = -1/8 and C = 1/8.

Hence,
L(y) = (3s + 9)/(s² + 2s + 5) + (1/8) [1/(s + 3) + (-s+1)/(s² + 2s + 5)]
.......= (1/8) [(24s + 72)/(s² + 2s + 5) + 1/(s + 3) + (-s+1)/(s² + 2s + 5)]
.......= (1/8) [(23s + 73)/(s² + 2s + 5) + 1/(s + 3)].
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b) Completing the square,
L(y) = (1/8) [(23s + 73)/((s+1)² + 4) + 1/(s + 3)]
......= (1/8) [(23(s+1) + 50)/((s+1)² + 4) + 1/(s + 3)].

Inverting:
y(t) = (1/8) [23 e^-t cos(2t) + (50/2) e^-t sin(2t) + e^-3t]
.....= (1/8) [23 e^-t cos(2t) + 25 e^-t sin(2t) + e^-3t].