I am trying to solve this PDE using Method of characteristics:
$$(u+e^x)u_x+(u+e^y)u_y=u^2-e^{x+y}$$
I don't know how the next equation is called in English, but it is used to solve the PDE:
$$\frac{dx}{u+e^x}=\frac{dy}{u+e^y}=\frac{du}{u^2-e^{x+y}}$$
I attempted to find $f(x)$, $g(y)$ and $h(u)$ such that
$$f(x)(u+e^x)+g(y)(u+e^y)+h(u)(u^2-e^{x+y})=0$$
other attempt I tried was, given that $d(e^{-x})=-e^{-x}dx$, then I get a fourth equcation
$$\frac{e^{-x}dx-e^{-y}dy}{ue^{-x}+1-ue^{-y}-1}=\frac{d(e^{-y}-e^{-x})}{u(e^{-x}-e^{-y})}$$
I am not 100% sure about the last one.
Best Answer
I usually find it easier to write the ODEs for the parametrized characteristic curves $(x(t),y(t),u(t))$: $$ \dot x = u+e^x ,\qquad \dot y = u+e^y ,\qquad \dot u = u^2 - e^{x+y} . $$ Now after a bit of playing around with this, you might (if you're lucky) notice that $$ \tfrac{d}{dt} (e^{-x}) = - e^{-x} \dot x = - e^{-x} (u+e^x) = -u e^{-x} - 1 $$ and then that $$ \tfrac{d}{dt} (u e^{-x}) = \dot u \, e^{-x} + u (-u e^{-x} - 1) = (u^2 - e^{x+y}) \, e^{-x} - u^2 e^{-x} - u = - (u + e^y), $$ where suddenly a very familiar expression appeared on the right-hand side!
Do you think you can take it from there? (I don't want to spoil your teacher's challenge entirely...)