Answer
See below
Work Step by Step
Given:
$y''+3y'+2y=12te^{2t}$
and $y(0)=0\\
y'(0)=1$
We take the Laplace transform of both sides of the differential
equation to obtain:
$[s^2Y(x)-sy(0)-y'(0)]+3[sY(x)-y(0)]+2Y(s)=\frac{12}{(s-2)^2}$
Substituting in the given initial values and rearranging terms yields
$Y(s)(s^2+3y+2)-1=\frac{12}{(s-2)^2}$
That is,
$Y(s)(s^2+3y+2)=\frac{12}{(s-2)^2}+1=\frac{s^2-4s+16}{(s-2)^2}$
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:
$Y(s)=\frac{s^2-4s+16}{(s-2)^2(s+1)(s+2)}=\frac{7}{3(s+1)}-\frac{7}{4(s+2)}-\frac{7}{12(s-2)}+\frac{1}{(s-2)^2}$
We recognize the terms on the right-hand side as being the Laplace transform of appropriate exponential functions. Taking the inverse Laplace transform yields:
$y(t)=\frac{7}{3}e^{-t}-\frac{7}{4}e^{-2t}-\frac{7}{12}e^{2t}+te^{2t}$
