Answer
See the solution.
Work Step by Step
We want to show $y_1(t)=e^{-3t}$ and $y_2(t)=e^t$ are solutions of $$y''+2y'-3y=0.$$
First, we show $y_1$ is a solution.
We take the first and second derivatives of $y_1$:
$$y_1'(t)=e^{-3t}(-3)=-3e^{-3t}$$ and
$$y_1''(t)=-3e^{-3t}(-3)=9e^{-3t}.$$
Now we substitute $y_1$ and its derivatives into our differential equation:
$$y_1''+2y_1'-3y_1\overset{?}{=}0$$
$$9e^{-3t}+2(-3e^{-3t})-3(e^{-3t})\overset{?}{=}0$$
$$9e^{-3t}-6e^{-3t}-3e^{-3t}\overset{?}{=}0$$
$$9e^{-3t}-9e^{-3t}\overset{?}{=}0$$
$$0=0.$$
So $y_1$ is a solution.
Next, we show $y_2$ is a solution.
We take the first and second derivatives of $y_2$:
$$y_2'(t)=e^t$$ and
$$y_2''(t)=e^t.$$
Now we substitute $y_2$ and its derivatives into our equation:
$$y_2''+2y_2'-3y_2\overset{?}{=}0$$
$$e^t+2e^t-3e^t \overset{?}{=}0$$
$$e^t-e^t \overset{?}{=}0$$
$$0=0.$$
Thus $y_2$ is also a solution.
