Finding solutions for system of ODE

Question

How does one find solutions for the system of differential equations of the form

$$2x'-5y'=4y-x \\ 3x'-4y'=2x-y$$ ? All I can think of, is finding $x'+y' = 3x-5y$ and then substituting $x'$ or $y'$ in one of the equations, but the result is not satisfactory (for instance I get $y'+2y-x=0$). This system is just one example, I found whole exercise with equations like these and I'm trying to learn how to tackle such problems. I know the other approach is to create and work with a matrix, but to do that the system should be somehow transformed to the form $x'=f(x,y)$ and $y'=g(x,y)$ and I don't see a way to do that.

Answer 1

Let $D$ stands for '. I mean $y'=Dy$ so the system can be written as follows. Note that after doing this job we treat the system as a linear system and you know that what I did in the third and the fourth systems:

Lets work on the last equation in which I omited the part $x(t)$'s. We get $$(ad-bc)y(t)=...=(D^2-1)y(t)=0\Longrightarrow y(t)=C_1e^{t}+C_2e^{-t}$$ wherein $C_1,~C_2$ are constants. Now work on $x(t)$ by omitting the $y(t)$'s parts from the third system.

Answer 2

The system can be written as \begin{equation*} \left( \begin{array}{cc} 2 & -5 \\ 3 & -4 \end{array} \right) \cdot \left( \begin{array}{c} x' \\ y' \end{array} \right) = \left( \begin{array}{cc} -1 & 4 \\ 2 & -1 \end{array} \right) \cdot \left( \begin{array}{c} x \\ y \end{array} \right) \end{equation*} and therefore \begin{equation} \left( \begin{array}{c} x' \\ y' \end{array} \right) = \left( \begin{array}{cc} 2 & -5 \\ 3 & -4 \end{array} \right)^{-1} \cdot \left( \begin{array}{cc} -1 & 4 \\ 2 & -1 \end{array} \right) \cdot \left( \begin{array}{c} x \\ y \end{array} \right) = \left( \begin{array}{cc} 2 & -3 \\ 1 & -2 \end{array} \right) \cdot \left( \begin{array}{c} x \\ y \end{array} \right) \end{equation} and the solution of the system is

\begin{equation} \left( \begin{array}{c} x \\ y \end{array} \right) = \exp( (t-t_0)\cdot \left( \begin{array}{cc} 2 & -3 \\ 1 & -2 \end{array} \right) )\cdot \left( \begin{array}{c} x_0 \\ y_0 \end{array} \right)= \ldots \end{equation}