2 Use a Laplace Transform to solve the initial value problem

Solution

Taking the Laplace transform of both sides gives L{y\" 4y\' + 4y} = 0

L{y \"} 4L{y \' } + 4L{y} = 0

s ² L{y} s y(0) y \' (0) 4(sL{y} y(0)) + 4L{y} = 0

(s ² 4s + 4)L{y} s + 3 = 0

so that

L{y} = s 3 /s² 4s + 4 .

Expanding this last term in partial fractions gives

s 3 / s ² 4s + 4 = s 3 /(s 2)² = A/ s 2 + B/ (s 2)²= A/(s 2) + B/ (s 2)

so that

A(s 2) + B = s 3

Plugging in s = 2 gives B = 1 and taking a derivative gives immediately A = 1. Therefore

y = L ^-1 ( 1 / (s 2 ) 1 /(s 2)²) = e ^2t t e^2t