Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
Do you mean that $T$ is a function of $x$, is it just a function of $v$? Do you mean $y''+f(x)y=0$? – Sameh Shenawy Jan 30 '14 at 2:47
I meant to say $y$ is a function of $x$.. I corrected it. – Millardo Peacecraft Jan 30 '14 at 4:46
$y''+f(y)=0$ do you mean like this – Sameh Shenawy Jan 30 '14 at 4:52
My Q is $T(v(y))$ is just a function of $y$ why did you use both $T$ and $v$? – Sameh Shenawy Jan 30 '14 at 4:54
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. – Millardo Peacecraft Jan 30 '14 at 5:02

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 $\frac{1}{2}(\frac{dy}{dx})^2 = const + sin(y)$

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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