Question:find a differential equation whose solution is $y=c_1e^{2t}+c_2e^{-3t}$
Solution:
Set
\begin{align*}
y_1 &=e^{2t} & y_2 &=e^{-3t}\\
y_1' &=2e^{2t} & y_2' &=-3e^{-3t}\\
r_1 &=2 & r_2 &=-3\\
\end{align*}
So the Wronskian
\begin{equation*}
\begin{split}
W[y_1,y_2](t) &= \begin{vmatrix}
y_1(t)&y_2(t)\\
y_1'(t)&y_2'(t)
\end{vmatrix} \\
&=y_1y_2'-y_2y_1' \\
&=-5e^{-t}\\
&\neq0
\end{split}
\end{equation*}
Then ${y_1,y_2}$forms a fundamental set of solution.We can write one characteristic equation as
\begin{align*}
(r-2)(r+3)=0\\
(r-2)(r+3)e^{rt}=0
\end{align*}
by multiplying $e^{rt}$on both sides,we get one equation for this general solution:
\begin{equation*}
y''+y'-6y=0
\end{equation*}
