Correct. So your argument for $\displaystyle ku$ being a solution is complete, because all the steps are reversible.

Actually, you'd get $\displaystyle c(x,y)(u^{2}+2uv+v^{2})$. So the proof for the $\displaystyle u+v$ case won't end up with zero, because you can't get zero out of this messier expression. Make sense?Quote:

$\displaystyle c(x,y)u^2+c(x,y)v^2$