## Calculus 8th Edition

$y=c_{1}e^{3x}+c_{2}e^{-4x}$
$r^{2}+r-12=0$ $(r-3)(r+4)=0$ So general formula and general solution add up to make the auxiliary equation which is $ar^2+by+c=0$ therefore the final answer would be $y=c_{1}e^{3x}+c_{2}e^{-4x}$