Answer
See below
Work Step by Step
Given:
$y''+y'-2y=10e^{-t}$
and $y(0)=0\\
y'(0)=1$
We take the Laplace transform of both sides of the differential
equation to obtain:
$[s^2Y(x)-sy(0)-y'(0)]+[sY(x)-y(0)]-2Y(s)=\frac{10}{s+1}$
Substituting in the given initial values and rearranging terms yields
$Y(s)(s^2+s-2)-0-1=\frac{10}{s+1}$
That is,
$Y(s)(s^2+s-2)=\frac{10}{s+1}+1=\frac{s+11}{s+1}$
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:
$Y(s)=\frac{s+11}{(s+1)(s^2+s-2)}=\frac{s+11}{(s-1)(s+1)(s+2)}=\frac{2}{s-1}-\frac{6}{s+1}+\frac{3}{s+2}$
We recognize the terms on the right-hand side as being the Laplace transform of appropriate exponential functions. Taking the inverse Laplace transform yields:
$y(t)=2e^t-6e^{-t}+3e^{-2t}$
