- #1
kingwinner
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Homework Statement
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)].
Homework Equations
N/A
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!