Answer
See below
Work Step by Step
Given:
$y''-y=0$
We take the Laplace transform of both sides of the differential
equation to obtain:
$[s^2Y(x)-sy(0)-y'(0)]-Y(s)=0$
Substituting in the given initial values and rearranging terms yields
$Y(s)(s^2-1)=0$
That is,
$Y(s)(s^2-1)=\frac{s}{s^2-1}y(0)+\frac{1}{s^2-1}y'(0)$
We must now determine the partial fractions decomposition of the right-hand side.
We have:
$\frac{s}{s^2-1}y(0)+\frac{1}{s^2-1}y'(0)=\frac{A}{s-1}+\frac{B}{s+1}$
for appropriate constants A, B.
Solving for A, and B, we obtain
$A=\frac{y(0)+y'(0)}{2}\\
B=\frac{y(0)-y'(0)}{2}$
Thus,
$y(s)=\frac{y(0)+y'(0)}{2(s-1)}+\frac{y(0)-y'(0)}{2(s+1)}\\
=\frac{c_1}{s-1}+\frac{c_2}{s+1}$
Taking the inverse Laplace transform of both sides of this equation yields
$y(t)=c_1e^t+c_2e^{-t}$
