1
$\begingroup$

I was thinking what if I had a differential equation of the form:

$$\frac{d^2u}{dx^2} + vu(x) = 0 $$

where $v(y(x))$, that is $y$ is a function of $x$. What are the possible restrictions that I can put on this differential equation so that it admits a solution? Has anyone come across any differential equations that contain an implicit function?

$\endgroup$
  • $\begingroup$ Do you mean that $T$ is a function of $x$, is it just a function of $v$? Do you mean $y''+f(x)y=0$? $\endgroup$ – Semsem Jan 30 '14 at 2:47
  • $\begingroup$ I meant to say $y$ is a function of $x$.. I corrected it. $\endgroup$ – Millardo Peacecraft Jan 30 '14 at 4:46
  • $\begingroup$ $y''+f(y)=0$ do you mean like this $\endgroup$ – Semsem Jan 30 '14 at 4:52
  • $\begingroup$ My Q is $T(v(y))$ is just a function of $y$ why did you use both $T$ and $v$? $\endgroup$ – Semsem Jan 30 '14 at 4:54
  • $\begingroup$ I'm sorry it seems I wasn't thinking straight when I wrote the question. The one above is the correct form it should be in. $\endgroup$ – Millardo Peacecraft Jan 30 '14 at 5:02
0
$\begingroup$

I'm a bit lost with your notation changes but if you mean $\frac{d^2y}{dx^2}+f(y(x))=0$, then if you multiply by $\frac{dy}{dx}$ you can integrate once (as long as f is nice enough. The hard bit is usually solving $\frac{1}{2}(\frac{dy}{dx})^2 = const + \hat f$, where $\hat f$ is the integral of f(). E.g if f(y)=cos(y) then you get $\frac{1}{2}(\frac{dy}{dx})^2 = const + sin(y)$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.