Question

Write the linear system of differential equations in matrix form then solve the system.

Answer 1

In matrix form, the system is

$\dfrac{\mathrm d}{\mathrm dt}\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}1&1\\4&1\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}$

First find the eigenvalues of the coefficient matrix (call it $\mathbf A$).

$\det(\mathbf A-\lambda\mathbf I)=\begin{vmatrix}1-\lambda&1\\4&1-\lambda\end{vmatrix}=(1-\lambda)^2-4=0\implies\lambda^2-2\lambda-3=0$

$\implies\lambda_1=-1,\lambda_=3$

Find the corresponding eigenvector for each eigenvalue:

$\lambda_1=-1\implies(\mathbf A+\mathbf I)\vec\eta_1=\vec0\implies\begin{bmatrix}2&1\\4&2\end{bmatrix}\begin{bmatrix}\eta_{1,1}\\\eta_{1,2}\end{bmatrix}=\begin{bmatrix}0\\0\end{bmatrix}$

$\lambda_2=3\implies(\mathbf A-3\mathbf I)\vec\eta_2=\vec0\implies\begin{bmatrix}-2&1\\4&-2\end{bmatrix}\begin{bmatrix}\eta_{2,1}\\\eta_{2,2}\end{bmatrix}=\begin{bmatrix}0\\0\end{bmatrix}$

$\implies\vec\eta_1=\begin{bmatrix}1\\-2\end{bmatrix},\vec\eta_2=\begin{bmatrix}1\\2\end{bmatrix}$

Then the system has general solution

$\begin{bmatrix}x\\y\end{bmatrix}=C_1\vec\eta_1e^{\lambda_1t}+C_2\vec\eta_2e^{\lambda_2t}$

or

$\begin{cases}x(t)=C_1e^{-t}+C_2e^{3t}\\y(t)=-2C_1e^{-t}+2C_2e^{3t}\end{cases}$

Given that $x(0)=1$ and $y(0)=2$, we have

$\begin{cases}1=C_1+C_2\\2=-2C_1+2C_2\end{cases}\implies C_1=0,C_2=2$

so that the system has particular solution

$\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}e^{3t}\\2e^{3t}\end{bmatrix}$