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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?

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  • $\begingroup$ Laplace transform only works on linear DEs. The best method to solve a DE is the one that works. When more than one approach works, obviously you'll just want to use the one that is easier to implement. Knowing a priori the method that will be the easiest to implement comes from experience. $\endgroup$ – Mattos Mar 29 '17 at 4:38
  • $\begingroup$ Ah! I think that was the part I missed: "Laplace transform only works on linear DEs". So, if I were to linearize this using the v substitution I mentioned, I'd be able to take a Laplace transform and solve it that way (albeit that method being extremely long-winded)? $\endgroup$ – Aommaster Mar 29 '17 at 4:47
  • $\begingroup$ Using the $v$ substitution you gave gives $$(vx)' = \frac{4v - 3}{2 - v}$$ which is still non-linear, so no, you can't use the Laplace transform. $\endgroup$ – Mattos Mar 29 '17 at 7:13
  • $\begingroup$ Ah yes, you are correct. I seem to have mixed up the fact that an equation can be separable, but still non-linear. Thanks! $\endgroup$ – Aommaster Mar 29 '17 at 7:48
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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 $y(t(x))$.

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  • $\begingroup$ Very creative solution. Thank you! I seem to have forgotten how to solve systems of DE's, but there are plenty of resources online that I can read up on. Thanks again! $\endgroup$ – Aommaster Mar 29 '17 at 11:21

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