Answer
See below
Work Step by Step
Given:
$y''-2y'=30e^{-3t}$
and $y(0)=1\\
y'(0)=0$
We take the Laplace transform of both sides of the differential
equation to obtain:
$[s^2Y(x)-sy(0)-y'(0)]-2[sY(x)-y(0)]=\frac{30}{s+3}$
Substituting in the given initial values and rearranging terms yields
$Y(s)(s^2-2s)-s+2=\frac{30}{s+3}$
That is,
$Y(s)(s^2-2s)=\frac{30}{s+3}+s-2=\frac{s^2+s+24}{s+3}$
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:
$Y(s)=\frac{s^2+s+24}{(s+3)(s^2-2s)}=\frac{2}{s+3}-\frac{4}{s}+\frac{3}{s-2}$
We recognize the terms on the right-hand side as being the Laplace transform of appropriate exponential functions. Taking the inverse Laplace transform yields:
$y(t)=2e^{-3t}-4+3e^{2t}$
