Question

Find the solution to the linear system of differential equations {x′y′==−5x+3y−18x+10y satisfying the initial conditions x(0)=4 and y(0)=11

Answer 1

$\begin{cases}x'=-5x+3y\\y'=-18x+10y\end{cases}$

$\begin{bmatrix}x\\y\end{bmatrix}'=\begin{bmatrix}-5&3\\-18&10\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}$

The coefficient matrix has eigenvalues $\lambda=1,4$, with corresponding eigenvectors $\mathbf v=\begin{bmatrix}1\\2\end{bmatrix},\begin{bmatrix}1\\3\end{bmatrix}$. So the general solution is

$\begin{bmatrix}x\\y\end{bmatrix}=C_1e^t\begin{bmatrix}1\\2\end{bmatrix}+C_2e^{4t}\begin{bmatrix}1\\3\end{bmatrix}$

Given that $x(0)=4$ and $y(0)=11$, we get

$\begin{cases}4=C_1+C_2\\11=2C_1+3C_2\end{cases}\implies C_1=1,C_2=3$

so that the particular solution to the system is

$\begin{bmatrix}x\\y\end{bmatrix}=e^t\begin{bmatrix}1\\2\end{bmatrix}+3e^{4t}\begin{bmatrix}1\\3\end{bmatrix}$

or in equivalent terms,

$\begin{cases}x=e^t+3e^{4t}\\y=2e^t+9e^{4t}\end{cases}$