Answer
See below
Work Step by Step
Given:
$y''+5y'+4y=20\sin 2t$
and $y(0)=-1\\
y'(0)=2$
We take the Laplace transform of both sides of the differential
equation to obtain:
$[s^2Y(x)-sy(0)-y'(0)]+5[sY(s)-y(0)]+4Y(s)=\frac{20\times2}{s^2+4}=\frac{40}{s^2+4}$
Substituting in the given initial values and rearranging terms yields
$Y(s)(s^2+5s+4)+s-2+5=\frac{40}{s^2+4}$
That is,
$Y(s)(s^2+5s+4)=\frac{40}{s^2+4}-s-3=\frac{-s^3-3s^-4s+28}{s^2+4}$
Thus, we have solved for the Laplace transform of y(t). To find y itself, we must take the inverse Laplace transform. We first decompose the right-hand side into partial fractions to obtain:
$Y(s)=\frac{-s^3-3s^-4s+28}{(s^2+4)(s+4)(s+1)}=-\frac{1}{s+4}+\frac{2}{s+1}-\frac{2s}{s^2+4}$
We recognize the terms on the right-hand side as being the Laplace transform of appropriate exponential functions. Taking the inverse Laplace transform yields:
$y(t)=-e^{-t}+2e^{-t}-2\cos 2t$
