Answer
See below
Work Step by Step
Given:
$y''+4y=9\sin t$
and $y(0)=1\\
y'(0)=-1$
We take the Laplace transform of both sides of the differential
equation to obtain:
$[s^2Y(x)-sy(0)-y'(0)]+4Y(s)=\frac{9}{s^2+1}$
Substituting in the given initial values and rearranging terms yields
$Y(s)(s^2+4)-s+1=\frac{9}{s^2+1}$
That is,
$Y(s)(s^2+4)=\frac{9}{s^2+1}+s-1=\frac{s^3-s^2+s+8}{s^2+1}$
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:
$Y(s)=\frac{s^3-s^2+s+8}{(s^2+1)(s^2+4)}=-\frac{3}{s^2+1}+\frac{s-4}{s^2+4}$
We recognize the terms on the right-hand side as being the Laplace transform of appropriate exponential functions. Taking the inverse Laplace transform yields:
$y(t)=3\sin t+\cos 2t-2\sin 2t$
