# Homework Help: General solution to PDE: f(x-y) = g(y-x) for arbitrary functions f and g?

1. Sep 14, 2009

### kingwinner

1. The problem statement, all variables and given/known data
Quote:
" PDE: ∂u/∂x + ∂u/∂y = 0
The general solution is u(x,y) = f(x-y) where f is an arbitrary function.
Alternatively, we can also say that the general solution is u(x,y) = g(y-x) where g is an arbitrary function. The two answers are equivalent since u(x,y) = g(y-x) = f[-(x-y)] "

I don't see why the two different representations above [u(x,y) = f(x-y) and u(x,y) = g(y-x)] would describe exactly the SAME general solution. Why can we freely switch the order of x and y?
Also, I don't understand why u(x,y) = g(y-x) = f[-(x-y)].

2. Relevant equations
N/A

3. The attempt at a solution
I was thinking of odd and even functions? But I don't think the quote is meant to restrict only to these special functions...

Thanks for any help!

2. Sep 15, 2009

### LCKurtz

Maybe looking at an example will help you. Say you have the general solution f(x-y) and I claim the general solution is g(y-x). So you say, look, I can get exp(x-y) by taking f(x-y)=exp(x-y).

I say, so can I. I will take g(y-x) = exp(-(y-x))

It's because f(x-y) = f(-(y-x)) which I can call g(y-x). Clear as mud?