## 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
$B$
.
$x(0)=1 \rightarrow 1=A+B$

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

$e$
$1+2e=Ae+Ae^{-4} \rightarrow A= \frac{1+2e}{1+e^{-4}}=\frac{e^4+2e^5}{e^4+1}$
$e^5$
$1-2e^5=B+Be^4 \rightarrow B=\frac{1-2e^5}{1+e^4}$
$x=\frac{e^4+2e^5}{e^4+1} e^{5t}+\frac{1-2e^5}{1+e^4}e^t, \: y=\frac{e^4+2e^5}{e^4+1} e^{5t}-\frac{1-2e^5}{1+e^4}e^t$