Answer
See below
Work Step by Step
Given:
$y''-y'-6y=6(2-e^{t})$
and $y(0)=5\\
y'(0)=-3$
We take the Laplace transform of both sides of the differential
equation to obtain:
$[s^2Y(x)-sy(0)-y'(0)]-[sY(x)-y(0)]-6Y(s)=6(\frac{2}{s}-\frac{1}{s-1})$
Substituting in the given initial values and rearranging terms yields
$Y(s)(s^2-s-6)-5s+8=\frac{12}{s}-\frac{6}{s-1}=\frac{6s-12}{s(s-1)}$
That is,
$Y(s)(s^2-s-6)=\frac{6s-12}{s(s-1)}+5s-8$
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:
$Y(s)=\frac{5s^3-13s^2+14s-12}{s(s-1)(s+2)(s-3)}=\frac{3}{s-1}-\frac{2}{s}+\frac{22}{5(s+2)}+\frac{8}{5(s-3)}$
We recognize the terms on the right-hand side as being the Laplace transform of appropriate exponential functions. Taking the inverse Laplace transform yields:
$y(t)=3e^{t}-2t+\frac{22}{5}e^{-2t}+\frac{8}{5}e^{3t}$
