Answer
y = $C_{1}e^{x}$ + $C_{2}e^{-x}$
Work Step by Step
Question: y=y"
We write: y"-y=0
We know:
y = $e^{rx}$
y' = $re^{rx}$
y" = $r^{2}e^{rx}$
This results in $(r^{2}-1)e^{rx}=0$
The corresponding characteristic equation is:
$r^{2}-1=0$
Solving the equation gives:
$r^{2}=1$
$r=\frac{+}{-}1$
Adding the constants gives the following general solution:
y = $C_{1}e^{x}$ + $C_{2}e^{-x}$
