Is there an approach for solving ODEs of the following form?
$$\frac{ d}{dx}\begin{pmatrix} y_1 \\ y_2\end{pmatrix} = A(x) \begin{pmatrix} y_1 \\ y_2\end{pmatrix} $$
where $A(x)$ is a non-diagonal 2x2 matrix
I tried to do separation of variables, but I am not sure if the operation $$\frac{d \begin{pmatrix}y_1 \\ y_2\end{pmatrix}}{\begin{pmatrix}y_1 \\ y_2\end{pmatrix}}$$ has some meaning
Is it possible to get to an elegant generalized solution , say in the following form? $$\begin{pmatrix} y_1 \\ y_2\end{pmatrix} = C \exp \left( \int A(x)dx\right)$$
I will use the standard notation for my answer
$$\dot{\boldsymbol{x}}(t)=\boldsymbol{A}(t)\boldsymbol{x}(t).$$
If $\boldsymbol{A}(t)=\boldsymbol{A}_0=\boldsymbol{\text{const.}}$ (also called Linear time-invariant systems) then
$\boldsymbol{x}(t)=\exp\left[\boldsymbol{A}_0(t-t_0)\right]\boldsymbol{x}(t_0)$.
If $\boldsymbol{A}(t)$ (also called Linear time-variant systems), hence the coefficient function is only a function of time then you can express the solution by using the Peano-Baker-Series.
$${\mathbf {\Phi }}(t,t_0)={\mathbf {I}}+\int _{t_0 }^{t}{\mathbf {A}}(\sigma _{1})\,d\sigma _{1}$$ $$+\int _{t_0 }^{t}{\mathbf {A}}(\sigma _{1})\int _{t_0 }^{{\sigma _{1}}}{\mathbf {A}}(\sigma _{2})\,d\sigma _{2}\,d\sigma _{1}+\int _{t_0 }^{t}{\mathbf {A}}(\sigma _{1})\int _{t_0 }^{{\sigma _{1}}}{\mathbf {A}}(\sigma _{2})\int _{t_0 }^{{\sigma _{2}}}{\mathbf {A}}(\sigma _{3})\,d\sigma _{3}\,d\sigma _{2}\,d\sigma _{1}+...$$
and
$$\boldsymbol{x}(t) = \boldsymbol{\Phi}(t,t_0)\boldsymbol{x}(\tau).$$