Answer
y = $C_1e^{3x}$ + $C_2e^{-4x}$
Work Step by Step
Question: y"+y'-12y=0
We know:
y = $e^{rx}$
y' = $re^{rx}$
y" = $r^{2}e^{rx}$
This results in $(r^{2}+r-12)e^{rx}$=0
The corresponding characteristic equation is:
$r^{2}+r-12=0$
Factoring gives:
$(r-3)(r+4) = 0$
r=3 and r=-4
Adding the constants gives the following general solution:
y = $C_1e^{3x}$ + $C_2e^{-4x}$
