# Partial differential equation

Hi there. I have this partial differential equation that I have to solve, and I thought that perhaps there was an easy way of solving this, like finding an equivalent differential for the right hand side of the equation, on such a way that I could get a simple differential equation, and then just integrating I could solve this.

The partial differential equation that I have to solve is this:

$$d \left( \frac{\mu}{T} \right )=ud(A^{-1/2}u^{-3/4}v^{1/2})+vd(2A^{-1/2}v^{-1/2}u^{1/4})$$

Is there an easy way for solving this? the idea I had was to merge both differentials on the right side in only one differential, but I couldn't find the way.

Ok. I think I got it. This is what I have done:
$$d \left( \frac{\mu}{T} \right )=ud(A^{-1/2}u^{-3/4}v^{1/2})+vd(2A^{-1/2}v^{-1/2}u^{1/4})$$

So I took
$$d(A^{-1/2}u^{-3/4}v^{1/2})=A^{-1/2}(-\frac{3}{4}u^{-7/4}v^{1/2}du+\frac{1}{2}u^{-3/4}v^{-1/2}dv)$$
And in the other hand:
$$d(2A^{-1/2}u^{1/4}v^{-1/2})=A^{-1/2}(\frac{1}{2}u^{-7/4}v^{1/2}du+u^{1/4}v^{-3/2}dv)$$

Then
$$d \left( \frac{\mu}{T} \right )=ud(A^{-1/2}u^{-3/4}v^{1/2})+vd(2A^{-1/2}v^{-1/2}u^{1/4})=A^{-1/2}(-\frac{3}{4}u^{-7/4}v^{1/2}du+\frac{1}{2}u^{-3/4}v^{-1/2}dv+\frac{1}{2}u^{-7/4}v^{1/2}du+u^{1/4}v^{-3/2}dv)$$
$$d \left( \frac{\mu}{T} \right )=A^{-1/2}[ -\frac{1}{4}u^{-3/4}v^{1/2}du-\frac{1}{2}u^{1/4}v^{-1/2}dv]=-\frac{1}{4}[u^{-3/4}v^{1/2}du+2u^{1/4}v^{-1/2}dv]=-A^{-1/2}d(u^{1/4}v^{1/2})$$