Originally Posted by

**shawsend** I'm a little unsure. Here goes though. You have:

$\displaystyle \frac{\partial^2 z}{\partial v \partial n}=\frac{1}{2n}\frac{\partial z}{\partial v}$

and if the function is well-behaved, then:

$\displaystyle \frac{\partial^2 z}{\partial n \partial v}=\frac{1}{2n}\frac{\partial z}{\partial v}=\frac{\partial}{\partial n}\frac{\partial z}{\partial v}=\frac{1}{2n}\frac{\partial z}{\partial v}$

Let $\displaystyle \frac{\partial z}{\partial v}=g(n,v)$ then we have $\displaystyle \frac{\partial g}{\partial n}=\frac{1}{2n}g$ which is an ordinary DE in the variable n with v kept constant. Then:

$\displaystyle \frac{dg}{dn}-\frac{1}{2n}g=0$ (**)

Solving via integrating factor and noting that the constant of integration is now an arbitrary function of v, I obtain:

$\displaystyle d\left[n^{1/2}g\right]=0$

and so:

$\displaystyle g(n,v)=n^{-1/2} H(v)$ with $\displaystyle H(v)$ arbitrary.

Now:

$\displaystyle \frac{\partial z}{\partial v}=n^{-1/2} H(v)$

so that:

$\displaystyle z(n,v)=n^{-1/2}\int H(v) dv+G(n)$