# A fairly difficult differential equation

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?

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