Solve $y'(x)=\begin{pmatrix}1 & 1 \\ 4 & 1\end{pmatrix}y(x)$ Find the general solution for $y'(x)=\begin{pmatrix}1 & 1 \\ 4 & 1\end{pmatrix}y(x)$, for $y:\mathbb{R}\to \mathbb{R}^2$.

I've tried to solve this component-wise, that is I've tried to solve $y_1'-y_1=y_2$ and $4y_1+y_2=y_2'$ by plugging the first equation into the second and then solving for $y_2$ but this, along with the other approaches I've tried, doesn't seem to work.
 A: We have the system
$$
\begin{cases}
 y'_1=y_1+y_2, \\
y'_2=4y_1+y_2.
\end{cases}
$$
It can be readily obtained that
$$
(2y_1+y_2)'=3(2y_1+y_2), \\
(2y_1-y_2)'=-(2y_1+y_2).
$$
Hence
$$
2y_1+y_2=c_1e^{3x}, \\
2y_1-y_2=c_2e^{-x}.
$$
Thus, general solution
$$
y_1=\frac{1}4c_1e^{3x}+\frac{1}{4}c_2e^{-x}, \\
y_2=\frac{1}2c_1e^{3x}-\frac{1}{2}c_2e^{-x}.
$$
A: $$
\begin{cases}
 y'_1=y_1+y_2 \\
y'_2=4y_1+y_2
\end{cases}
$$
Multiply by $2$ first DE:
$$
\begin{cases}
 2y'_1=2y_1+2y_2 \\
y'_2=4y_1+y_2
\end{cases}
$$
Substract both DE:
$$ \implies y'_2-2y'_1=2y_1-y_2$$
$$  y'_2+y_2=2(y_1+y'_1)$$
Multiply by $e^t$ both sides:
$$ (y_2e^t)'=2(y_1e^t)'$$
Integrate.
$$ y_2=2y_1+c_1e^{-t}$$
Plug this in the first DE and solve.
$$ y'_1=y_1+y_2 $$
$$ y'_1=3y_1+c_1e^{-t}$$
A: Use the fact that any functions $y_1$ and $y_2$ will be eigenfunctions of the derivative map (possibly generalized), and that their eigenvalues will correspond to those of the given matrix. Then, you can write a combination of eigenfunctions scaling the eigenvectors of the given matrix to form a solution set.
A: i don't know a lot about matrices but i know that the solution of a differential that is : $ y'(x) = a.y(x) + b $ is $y$
where :  $y(x) = C.e^{a.x} - \frac{b}{a}$ $$C\in\mathbb{R}$$
i hope this will help you solving your problem
A: Let $y=Pw$ for some suitable invertible matrix $P$, then $y'=Ay$ becomes $w'=P^{-1}APw$. Consider A in $M_2(\Bbb C)$, then it has a Jordan form J. Hence, we may write $w'=Jw$ by choosing a suitable $P$.
The characteristic polynomial of A is $\det(tI-A)=(t-1)^2-4=(t-3)(t+1)$, so we may write $J=\begin{pmatrix} -1 \ 0\\ 0  \  3\end{pmatrix}$. Therefore $w=\begin{pmatrix}c_1 e^{-x} \\  c_2e^{3x}\end{pmatrix}$, and since a suitable choice of P may formed from the eigenvectors of A (akin to the process of matrix diagonalization), $v=Pw$ may be readily evaluated.
