Initial value problem for a linear system. Consider the linear system
$$
\frac{dY}{dt} = 
\begin{pmatrix}
1 & -1 \\
1 & 3 \\
\end{pmatrix}
Y
$$
(a) Show that the function
$$
Y(t) =
\begin{pmatrix}
te^{2t} \\
-(t + 1)e^{2t}\\
\end{pmatrix}
$$
is a solution to the differential equation.
I verified this without trouble. The second part I am stuck on:
(b) Solve the initial-value problem
$$
\frac{dY}{dt} = 
\begin{pmatrix}
1 & -1 \\
1 & 3 \\
\end{pmatrix}
Y
\text{, where }
Y(0) = 
\begin{pmatrix}
0 \\
2 \\
\end{pmatrix}
$$

Consider the differential system
\begin{align}
\frac{dY}{dt} = 
\begin{pmatrix}
1 & -1 \\
1 & 3 \\
\end{pmatrix}
Y .
\end{align}
Let 
\begin{align}
Y(t) = \begin{pmatrix} a(t) \\ b(t) \end{pmatrix}
\end{align}
such that
\begin{align}
\begin{pmatrix} a^{'} \\ b^{'} \end{pmatrix} = \begin{pmatrix} 1 & -1 \\
1 & 3 \\ \end{pmatrix} \cdot \begin{pmatrix} a \\ b \end{pmatrix}.
\end{align}
The set of differential equations is then given by
\begin{align}
a' &= a - b \\
b' &= a + 3b
\end{align}
of which $a$ and $b$ satisfy the second order equation $w'' - 4 w' +4 w = 0$. The solution of this second order equation is $w(t) = (A + B t ) e^{2t}$ and lead to the form of $Y(t)$ being
\begin{align}
Y(t) = \begin{pmatrix} A + B t \\ C + D t \end{pmatrix} \, e^{2t}.
\end{align}
Now the initial condition is 
\begin{align}
Y(0) = \begin{pmatrix} 0 \\ 2 \end{pmatrix}
\end{align}
for which 
\begin{align}
\begin{pmatrix} 0 \\ 2 \end{pmatrix} = \begin{pmatrix} A \\ C \end{pmatrix}
\end{align}
which yields the solution being
\begin{align}
Y(t) = \begin{pmatrix} B t \\ 2 + D t \end{pmatrix} \, e^{2t}.
\end{align}
Using this result to take its derivative and apply it to the differential equation it is seen that the result equations amount to $(B+D)t = -(B+2)$ and $(B+D)t = D-2$. In order for there to be solutions it is required that $B = -2$ and $D = 2$. The final solution becomes
\begin{align}
Y(t) = 2 \, e^{2t} \, \begin{pmatrix} - t \\ 1 + t \end{pmatrix}
\end{align}   
