Answer
See below
Work Step by Step
Given:
$y''+y=5te^{-3t}$
and $y(0)=2\\
y'(0)=0$
We take the Laplace transform of both sides of the differential
equation to obtain:
$[s^2Y(x)-sy(0)-y'(0)]+Y(s)=\frac{5}{(s+3)^3}$
Substituting in the given initial values and rearranging terms yields
$Y(s)(s^2+1)-2s=\frac{5}{(s+3)^3}$
That is,
$Y(s)(s^2+1)=\frac{5}{(s+3)^3}+2s=\frac{2s^3+12s^2+18s+5}{(s+3)^3}$
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:
$Y(s)=\frac{2s^3+12s^2+18s+5}{(s+3)^3(s^2+1)}=\frac{-1}{5(s+3)}+\frac{1}{2(s+3)^2}+\frac{17s}{10(s^2+1)}+\frac{4}{10(s^2+1)}$
We recognize the terms on the right-hand side as being the Laplace transform of appropriate exponential functions. Taking the inverse Laplace transform yields:
$y(t)=\frac{-1}{5}e^{-3t}+\frac{1}{2}te^{-3t}+\frac{17}{10}\cos t+\frac{4}{10}\sin t$
