Hi everyone for quiz 4 we need to get the original equation of
$y''+y'-6y=12e^{-2t}+12e^{3t}$
First, we get the homogeneous form:
$r^2+r-6=0$
$(r+3)(r-2)=0$
$r1=-3, r2=2$
Then we can get:
$y=C_1e^{-2t}+C_2e^{3t}$
$y'=-2C_1e^{-2t}+3C_2e^{3t}$
$y''=4C_1e^{-2t}+9C_2e^{3t}$
so this means:
$-6B-2B+4B=12$
$-6A+3A+9A=12$
Solve to get
$A=2, B=-3$
So we rewrite:
$2e^{3t}-3e^{-2t}$
Plug in to the solution:
$y=2e^{3t}-3e^{-2t}+C_1e^{2t}+C_2e^{-3t}$
Please help correct my answer! Have a good weekend!
