Answer
$$y''+\frac{5}{2}y'+y=0$$
Work Step by Step
$y=c_{1}e^{-\frac{t}{2}}+c_{2}e^{-2t}\equiv{c}_{1}e^{\lambda_{1}t}+c_{2}e^{\lambda_{2}t}$, which is the format for the general solution.
Thus $\lambda_{1,2}=-\frac{1}{2},-2\Rightarrow(\lambda+\frac{1}{2})(\lambda+2)=\lambda^2+\frac{5}{2}\lambda+1$.
Recall the method of substituting $y=e^{\lambda{t}}$ so that $(\ln{y})'=\lambda$. Replacing dummy variable $\ln{y}$ with $y$,
$$y''+\frac{5}{2}y'+y=0$$
