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