Answer
$$y''+y'-6y=0$$
Work Step by Step
$y=c_{1}e^{2t}+c_{2}e^{-3t}\equiv{c}_{1}e^{\lambda_{1}t}+c_{2}e^{\lambda_{2}t}$, which is the format for the general solution.
Thus $\lambda_{1,2}=2,-3\Rightarrow(\lambda-2)(\lambda+3)=\lambda^2+\lambda-6$.
Recall the method of substituting $y=e^{\lambda{t}}$ so that $(\ln{y})'=\lambda$. Replacing dummy variable $\ln{y}$ with $y$,
$$y''+y'-6y=0$$
