Answer
$y=C_{1}e^{3x}+C_{2}e^{-2x}$
Work Step by Step
$y''-y'-6y=0$
Fill in $y=e^{rx}$
$y'=re^{rx}$
$y''=r^{2}e^{rx}$
Factorize
$r^{2}e^{rx}-re^{rx}-6e^{rx}=0$
$e^{rx}$ is always bigger than zero.
$(r^{2}-r-6)e^{rx}=0$
Factorize
$r^{2}-r-6=0$
$r_{1}=3$
$r_{2}=-2$
$(r-3)(r+2)=0$
$y_{1}=e^{3x}$
$y_{2}=e^{-2x}$
General solution is:
$y=C_{1}e^{3x}+C_{2}e^{-2x}$
