solve y''+y'-6y=12(e^(3t))+12e^(-2t)

Question

anonymous · Answer

Homogeneous solution:
$$y''+y'-6y=0$$
yields the characteristic equation
$$r^2+r-6=0~~\Rightarrow~~r=-3,2$$
So homogeneous part is $y_c=C_1e^{-3t}+C_2e^{2t}$.

Non-homogeneous solution:
$$y''+y'-6y=12e^{3t}+12e^{-2t}$$
Use the method of undetermined coefficients. Suppose $y_p=Ae^{3t}+Be^{-2t}$ is a solution, so you have
$$\begin{cases}
y_p=Ae^{3t}+Be^{-2t}\
y_p'=3Ae^{3t}-2Be^{-2t}\
y_p'=9Ae^{3t}+4Be^{-2t}
\end{cases}$$
Substitute into the original equation:
$$\begin{align*}\left(9Ae^{3t}+4Be^{-2t}\right)+\left(3Ae^{3t}-2Be^{-2t}\right)-6\left(Ae^{3t}+Be^{-2t}\right)&=12e^{3t}+12e^{-2t}\
6Ae^{3t}-4Be^{-2t}&=12e^{3t}+12e^{-2t}
\end{align*}$$
This tells you that $A=2$ and $B=-3$, and so your non-homogeneous part is $y_p=2e^{3t}-3e^{-2t}$.

Your final solution would be the sum of the non/homogeneous parts, or
$$y=y_c+y_p$$