\section{Question 2}
\subsection{a)}
\[
\begin{array}{l l}
	\frac{dy_{1}}{dt} = y_{1}(t) + 3y_{2}(t) \\
	\frac{dy_{2}}{dt} = 2y_{1}(t) + 2y_{2}(t)
\end{array}
\]
Substitute in
\[
	y_{1}(t) = 3e^{-t} + 2e^{4t}
\]
and
\[
	y_{2}(t) = -2e^{-t} + 2e^{3t}
\]
then...
\[
\begin{array}{l l}
	\frac{dy_{1}}{dt} = -3e^{-t} + 8e^{4t} \\
	\frac{dy_{2}}{dt} = 2e^{-t} + 8e^{4t}
\end{array}
\]
So...
First for $\frac{dy_{1}}{dt}$
\[
\begin{array}{l l}
	y_{1}(t) + 3y_{2}(t) & \implies (3e^{-t} + 2e^{4t}) + 3(-2e^{-t}) + 2e^{4t}) \\
		& ... \\
		& = e^{-t}(3 - 6) + e^{4t}(2+6) \\
		& = -3e^{-t} + 8e^{4t} \\
		& = \frac{dy_{1}}{dt}
\end{array}
\]
Secondly, for $\frac{dy_{2}}{dt}$
\[
\begin{array}{l l}
	2y_{1}(t) + 2y_{2}(t) & \implies 2(3e^{-t} + 2e^{4t}) + 2(-2e^{-t}+ 2e^{4t}) \\
		& ... \\
		& = e^{-t}(6-4) + e^{4t}(4+4) \\
		& = -2e^{-t} + 8e^{4t} \\
		& = \frac{dy_{2}}{dt}
\end{array}
\]

\[ y_{1}(0) = 3e^{0} + 2e^{0} = 3+2 = 5 \]
\[ y_{2}(0) = -2e^{0} + 2e^{0} = -2+2 = 0 \]

Therefore we know that both $y_{1}$ and $y_{2}$ are solutions to the system given