Differential equation system IVP appears to be wrong $Y(0) = (2,5)$ Let $A = \Bigg ( \begin{array}{cc} 1 & -1 \\ 1 & 3 \end{array} \Bigg )$
Need to solve the IVP: $(x(0),y(0)) = (2,5)$
Solving for the eigenvalues get the characteristic polynomial:
$\lambda^2 - 4 \lambda +4 \implies \lambda_1=\lambda_2 = 2$
for the eigenvector I get $(1,-1)$
so as a general solution I should have  $Y(t) = k_1e^{2t}(1,-1) + k_2e^{2t}(1,-1)$ but trying to solve for the initial condition given I am getting that it's not a possible solution. 
Am I missing something?
 A: I will give you a solution that uses matrix exponential of the jordan normal form. Generally, the solution of $y(t)=Ay(t), y(0)=(2,5)^{\top}=y_0$ is given by $y(t)=e^{At}y_0$, so we must focus on $e^{At}$. You already remarked that the characteristic equation has one root $\lambda=2$, but the corresponding eigenspace only dimension $\dim E_{\lambda=2}=1$ with
\begin{align}
E_{\lambda=2}=\ker(A-2I)=\ker\begin{pmatrix} -1 &-1 \\
1 &1
\end{pmatrix}=\left\langle \begin{pmatrix} 1\\ -1\end{pmatrix}\right\rangle
\end{align}
Let $v=(1,-1)^{\top}$ be the eigenvector of $A$ to $\lambda=2$. From $(A-2I)w=v$ we find a generalized eigenvector $w=(-1,0)^{\top}$. Since $A$ isn't diagonalizable, we know that the jordan normal form is
\begin{align}
J=
\begin{pmatrix}
2 & 1\\
0 & 2
\end{pmatrix}
\end{align}
Now we define
\begin{align}
P=\begin{pmatrix}
1 & -1\\
-1 & 0
\end{pmatrix}\qquad \Rightarrow\qquad
P^{-1}=
\begin{pmatrix}
0 & -1\\
-1 & -1
\end{pmatrix}
\end{align}
From that we get
\begin{align}
PJP^{-1}=\begin{pmatrix}
1 & -1\\
-1 & 0
\end{pmatrix}\begin{pmatrix}
2 & 1\\
0 & 2
\end{pmatrix}\begin{pmatrix}
0 & -1\\
-1 & -1
\end{pmatrix}=\begin{pmatrix}
1 & -1\\
1 & 3
\end{pmatrix}=A
\end{align}
This allows us to write
\begin{align}
\exp(At)=\exp(PJP^{-1}t)=P\exp(Jt)P^{-1}
\end{align}
The next step is to calculate $\exp(Jt)$. Maybe you find in some books or your lecture notes that this is
\begin{align}
\exp(Jt)=e^{2t}\begin{pmatrix}
1 & t\\
0 & 1
\end{pmatrix}=\begin{pmatrix}
e^{2t} & te^{2t}\\
0 & e^{2t}
\end{pmatrix}
\end{align}
You will then have
\begin{align}
y(t)=P\exp(Jt)P^{-1}y_0=
\begin{pmatrix}
(1-t)e^{2t} & -te^{2t}\\
te^{2t} & (1+t)e^{2t}
\end{pmatrix}\begin{pmatrix}
2\\ 5
\end{pmatrix}
=\begin{pmatrix}
2(1-t)e^{2t}-5te^{2t}\\ 2te^{2t}+5(t+1)e^{2t}
\end{pmatrix}
\end{align}

This was the first answer i gave...
The first solution is standard. An eigenvector of $\lambda=2$ is $v=(1,-1)^{\top}$, so $Y_1(t)=k_1e^{2t}v$. Since now the Eigenspace of $\lambda=2$ has dimension $\dim E_{\lambda}=1$, we need another linear independent solution. That is
\begin{align}
Y_2(t)=k_2e^{2t}(vt+w)
\end{align}
with a solution $w$ of $(A-2I)w=v$. (Take $Y_2$ into the differential equation an you will get this). That is $w=(-1,0)^{\top}$. The general solution is now
\begin{align}
Y(t)=k_1e^{2t}\binom{1}{-1}+k_2te^{2t}\binom{1}{-1}+k_2e^{2t}\binom{-1}{0}
\end{align}
The constants are determined by $Y(0)=(2,5)^{\top}$ and i hope you will get the same solution as above.
