# Partial Differentiation

1. Sep 21, 2009

### Gregg

1. The problem statement, all variables and given/known data

The function f(x,y) satisfies the d.e.

$$y{\partial f \over \partial x} + x{\partial f \over \partial y} = 0$$

By changing to new vars $$u = x^2-y^2$$ and $$v=2xy$$ show that f is a function of $$x^2-y^2$$ only.

2. Relevant equations

$$\frac{\partial }{\partial x}=\frac{\partial u}{\partial x}\frac{\partial }{\partial u}+\frac{\partial v}{\partial x}\frac{\partial }{\partial v}$$

$$\frac{\partial }{\partial y}=\frac{\partial u}{\partial y}\frac{\partial }{\partial u}+\frac{\partial v}{\partial y}\frac{\partial }{\partial v}$$

3. The attempt at a solution

$$y (\frac{\partial u}{\partial x}\frac{\partial }{\partial u}+\frac{\partial v}{\partial x}\frac{\partial }{\partial v}) g + x (\frac{\partial u}{\partial y}\frac{\partial }{\partial u}+\frac{\partial v}{\partial y}\frac{\partial }{\partial v}) g = y{\partial f \over \partial x} + x{\partial f \over \partial y} = 0 = 0$$

$$\left(2y^2+2x^2\right)\frac{\partial g}{\partial v}=x\frac{\partial f}{\partial y}+y\frac{\partial f}{\partial x}=0$$

I did that and so

$$\frac{\partial g}{\partial v}=0$$

How does this confirm g=g(u)? Is is because the slope of g with respect to v is 0? The so called function of v is just a constant or 0?

2. Sep 22, 2009

### LCKurtz

I didn't check your arithmetic, but to answer your question, if g is a function of u and v and gv = 0, then yes, that does show g is only a function of u.