\begin{equation}
y'' + 3y' = 0 , y(0) = -2, y'(0) = 3
\end{equation}we get
\begin{equation}
r^2 + 3r = 0
\end{equation} Solve for r, we get the solution for r.
\begin{equation}
r1= -3, r2 = -3
\end{equation} For the repeated roots, the solution for y is
\begin{equation}
y(t) = C1e^{rt} + C2te^{rt}
\end{equation}So we have
\begin{equation}
y(t) = C1e^{-3t}+tC2e^{-3t}
\end{equation} Since
\begin{equation}
y(0) = -2, y'(0) = 3
\end{equation} We will have two equations about C1 and C2
\begin{equation}
C1 = -2, -3C1 + C2(1+0) = 3
\end{equation}So we have
\begin{equation}
C1 = -2, C2 = -3
\end{equation} Then the solution is
\begin{equation}
y(t) = -2e^{-3t}-3te^{-3t}
\end{equation}
\begin{equation}
y(t) = - (2+3t)e^{-3t}
\end{equation}
