Answer
See below
Work Step by Step
Given:
$y'-y=5\sin 2t$
and $y(0)=-1$
We take the Laplace transform of both sides of the differential
equation to obtain:
$[sY(x)-y(0)]-Y(s)=\frac{10}{s^2+4}$
Substituting in the given initial values and rearranging terms yields
$Y(s)(s-1)+1=\frac{10}{s^2+4}$
That is,
$Y(s)(s-1)=2+\frac{-s^2+6}{s^2+4}$
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:
$Y(s)=\frac{-s^2+6}{(s-1)(s^2+4)}=\frac{-2s}{s^2+4}-\frac{2}{s^2+4}+\frac{1}{s-1}$
We recognize the terms on the right-hand side as being the Laplace transform of appropriate exponential functions. Taking the inverse Laplace transform yields:
$y(t)=-2\cos 2t -\sin 2t+e^t$
