I have a differential equation here:
$$\frac{dy}{dx} = \frac{4y-3x}{2x-y}$$
This equation can be solved relatively easily by dividing through by $x$, and then doing a $y=vx$ substitution (making the equation separable).
However, my question is: can this be solved by taking a Laplace transform instead?
When I try to do so, I get this: $$\mathscr{L}(2ty') - \mathscr{L}(yy') = \mathscr{L}(4y) - \mathscr{L}(3t)$$
The RHS is fairly simple when it comes to a laplace transform. On the LHS, I believe that the $2ty'$ term can be translated to something like $-F'(s)$. But that second term is causing me a bit of trouble.
Can a Laplace transform even be taken for such an equation? I feel that it should, I'm just not quite sure how, the integral ends up being quite messy. If so, could you show me how? (and if not, why not?)
Additionally, I'd like to know how to best select a method to solve a differential equation. In the case of this one, clearly, the easiest way would be doing a substitution. But how do I know that would be the most efficient way?
You can get a linear system out of this differential equation by introducing an artificial time parameter and separating the equation into \begin{align} \frac{dy(t)}{dt} &= 4y(t)−3x(t)\\ \frac{dx(t)}{dt} &= 2x(t)−y(t) \end{align} or $$ \begin{bmatrix}\dot x\\\dot y\end{bmatrix} = \begin{bmatrix}2&-1\\-3&4\end{bmatrix} · \begin{bmatrix} x\\y\end{bmatrix} $$ where you now can proceed by computing eigenvalues and eigenvectors. In the end you have to invert the function $x(t)$ to get the solution of the original problem as $y(t(x))$.