In matrix form, the system is
[tex]\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}[/tex]
First find the eigenvalues of the coefficient matrix (call it [tex]\mathbf A[/tex]).
[tex]\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[/tex]
[tex]\implies\lambda_1=-1,\lambda_=3[/tex]
Find the corresponding eigenvector for each eigenvalue:
[tex]\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}[/tex]
[tex]\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}[/tex]
[tex]\implies\vec\eta_1=\begin{bmatrix}1\\-2\end{bmatrix},\vec\eta_2=\begin{bmatrix}1\\2\end{bmatrix}[/tex]
Then the system has general solution
[tex]\begin{bmatrix}x\\y\end{bmatrix}=C_1\vec\eta_1e^{\lambda_1t}+C_2\vec\eta_2e^{\lambda_2t}[/tex]
or
[tex]\begin{cases}x(t)=C_1e^{-t}+C_2e^{3t}\\y(t)=-2C_1e^{-t}+2C_2e^{3t}\end{cases}[/tex]
Given that [tex]x(0)=1[/tex] and [tex]y(0)=2[/tex], we have
[tex]\begin{cases}1=C_1+C_2\\2=-2C_1+2C_2\end{cases}\implies C_1=0,C_2=2[/tex]
so that the system has particular solution
[tex]\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}e^{3t}\\2e^{3t}\end{bmatrix}[/tex]
