Answer
See below
Work Step by Step
Given:
$y''+2y'-3y=26e^{2t}\cos t$
and $y(0)=1\\
y'(0)=0$
We take the Laplace transform of both sides of the differential
equation to obtain:
$[s^2Y(x)-sy(0)-y'(0)]+2[sY(x)-Y(0)]-3Y(s)=\frac{26(s-2)}{(s-2)^2+1}$
Substituting in the given initial values and rearranging terms yields
$Y(s)(s^2+2s-3)-s-2=\frac{26(s-2)}{(s-2)^2+1}$
That is,
$Y(s)(s^2+2s-3)=\frac{26(s-2)}{(s-2)^2+1}+s+2=\frac{s^3+27s-42}{(s-2)^2+1}$
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:
$Y(s)=\frac{s^3+27s-42}{[(s-2)^2+1](s-1)(s+3)}=\frac{3}{2(s+3)}-\frac{5}{2(s-1)}+\frac{2(s-2)+3}{[(s-2)^2+1]}$
We recognize the terms on the right-hand side as being the Laplace transform of appropriate exponential functions. Taking the inverse Laplace transform yields:
$y(t)=\frac{3}{2}e^{-3t}-\frac{5}{2}e^{t}+2e^{2t}\cos t+3e^{2t}\sin t$
