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[tex]\begin{cases}x'=-5x+3y\\y'=-18x+10y\end{cases}[/tex]

[tex]\begin{bmatrix}x\\y\end{bmatrix}'=\begin{bmatrix}-5&3\\-18&10\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}[/tex]

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

[tex]\begin{bmatrix}x\\y\end{bmatrix}=C_1e^t\begin{bmatrix}1\\2\end{bmatrix}+C_2e^{4t}\begin{bmatrix}1\\3\end{bmatrix}[/tex]

Given that [tex]x(0)=4[/tex] and [tex]y(0)=11[/tex], we get

[tex]\begin{cases}4=C_1+C_2\\11=2C_1+3C_2\end{cases}\implies C_1=1,C_2=3[/tex]

so that the particular solution to the system is

[tex]\begin{bmatrix}x\\y\end{bmatrix}=e^t\begin{bmatrix}1\\2\end{bmatrix}+3e^{4t}\begin{bmatrix}1\\3\end{bmatrix}[/tex]

or in equivalent terms,

[tex]\begin{cases}x=e^t+3e^{4t}\\y=2e^t+9e^{4t}\end{cases}[/tex]

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