Solving a System of Initial Value Coupled First Order Linear Differential Equations

Suppose we have the system of initial value coupled differential equations.
\[\dot{x}=3x+2y\]

\[\dot{y}=2x+3y\]

\[x(0)=1, \: y(1)=2\]
  To solve this system, we need to diagonalise the coefficient matrix  
\[M= \left( \begin{array}{cc} 3 & 2 & \\ 2 & 3 \end{array} \right)\]
. Write the system in matrix form as  
\[\begin{pmatrix}\dot{x}\\ \dot{y} \end{pmatrix}= \left( \begin{array}{cc} 3 & 2 \\ 2 & 3 \end{array} \right) \begin{pmatrix}x\\ y \end{pmatrix}\]

The eigenvalues of the matrix are the solutions to  
\[det(\left( \begin{array}{cc} 3 & 2 \\ 2 & 3 \end{array} \right) - \lambda \left( \begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array} \right))=0 \rightarrow det(\left( \begin{array}{cc} 3- \lambda & 2 \\ 2 & 3- \lambda \end{array} \right))=0 \rightarrow (3- \lambda)^2-4=\lambda^2-6 \lambda +5=0\]

This expression in  
\[\lambda\]
  factorises as  
\[(\lambda-5)(\lambda -1)=0\]
  and we solve the equation, obtaining  
\[\lambda=5, \: 1\]

Now find the eigenvectors for each eigenvalue.
For  
\[\lambda=5\]
, solve  
\[(\left( \begin{array}{cc} 3 & 2 \\ 2 & 3 \end{array} \right) - 5 \left( \begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array} \right)) \mathbf{v} =\mathbf{0}\]
  for  
\[\mathbf{v}=\begin{pmatrix}x\\y\end{pmatrix} \]
.
\[\left( \begin{array}{cc} -2 & 2 \\ 2 & -2 \end{array} \right)\begin{pmatrix}x\\y\end{pmatrix}=\begin{pmatrix}-2x+2y\\2x-2y\end{pmatrix}=\begin{pmatrix}0\\0\end{pmatrix}\]

Hence  
\[x=y\]
  and we can take the eigenvector corresponding to the eigenvector 5 as  
\[\begin{pmatrix}1\\1\end{pmatrix}\]
.
For  
\[\lambda=1\]
, solve  
\[(\left( \begin{array}{cc} 3 & 2 \\ 2 & 3 \end{array} \right) - 1 \left( \begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array} \right)) \mathbf{v} =\mathbf{0}\]
  for  
\[\mathbf{v}=\begin{pmatrix}x\\y\end{pmatrix} \]
.
\[\left( \begin{array}{cc} 2 & 2 \\ 2 & 2 \end{array} \right)\begin{pmatrix}x\\y\end{pmatrix}=\begin{pmatrix}2x+2y\\2x+2y\end{pmatrix}=\begin{pmatrix}0\\0\end{pmatrix}\]

Hence  
\[x=-y\]
  and we can take the eigenvector corresponding to the eigenvector 5 as  
\[\begin{pmatrix}1\\-1\end{pmatrix}\]
.
The matrix of eigenvectors is
\[P= \left( \begin{array}{cc} 1 & 1 \\ 1 & -1 \end{array} \right) \]
.
At the moment our system takes the form  
\[\dot{\mathbf{v}}=M \mathbf{v}\]
.
Let  
\[\mathbf{w}=P^{-1} \mathbf{v}\]
  so that  
\[\mathbf{v}=P \mathbf{w}\]
&. The system becomes  
\[P \mathbf{v}=M P \mathbf{w} \rightarrow \mathbf{w}=P^{-1}MP \mathbf{w}\]
.
\[P^{-1}MP\]
  will be a diagonal matrix with entries equal to the eigenvalues of  
\[M\]
  and  
\[\mathbf{w}\]
  will be an elementary basis vector. The new system will be  
\[\begin{pmatrix}\dot{w_1} \\\dot{w_2} \end{pmatrix}=\left( \begin{array}{cc} 5 & 0 \\ 0 & 1 \end{array} \right) \begin{pmatrix}w_1\\w_2\end{pmatrix}\]
.
This is equivalent to the system
\[\dot{w_1}=5w_1 \]

\[\dot{w_1}=1w_1 \]

The solutions are
\[w_1=Ae^{5t}, w_2=B e^t\]

Then
\[\mathbf{v}'=P \mathbf{w} = \left( \begin{array}{cc} 1 & 1 \\ 1 & -1 \end{array} \right) \begin{pmatrix}Ae^{5t}\\Be^t\\2\end{pmatrix} =\begin{pmatrix}Ae^{5t}+Be^t\\Ae^{5t}-Be^t\end{pmatrix}\]

Now use the initial conditions to find  
\[A\]
  and  
\[B\]
.
\[x(0)=1 \rightarrow 1=A+B\]

\[y(1)=2 \rigthtarrow 2=Ae^{-5}+Be^{-1}\]

Add  
\[e\]
  times the second to the first.
\[1+2e=Ae+Ae^{-4} \rightarrow A= \frac{1+2e}{1+e^{-4}}=\frac{e^4+2e^5}{e^4+1}\]

Subtract  
\[e^5\]
  times the second from the first.
\[1-2e^5=B+Be^4 \rightarrow B=\frac{1-2e^5}{1+e^4}\]

Then  
\[x=\frac{e^4+2e^5}{e^4+1} e^{5t}+\frac{1-2e^5}{1+e^4}e^t, \: y=\frac{e^4+2e^5}{e^4+1} e^{5t}-\frac{1-2e^5}{1+e^4}e^t\]