We have the following equation: $$u_{yy}-2u_{xy}+2u_x-u_y=4e^x$$ and we want to find its general solution. We have brought it to the canonical form: $$-2u_{\xi\eta}+u_\xi=2e^\xi$$ where $\xi=x, \eta=x+2y$.
Our professor suggested to integrate it w.r.t. $\xi$ but we're not sure, and we don't know how to find the general solution. Any help would be appreciated. Thanks very much.
Integrating $-2u_{\xi\eta}+ u_{\xi}= 2e^{\xi}$ with respect to $\xi$, as suggested, $-2u_{\eta}+ u= 2e^{\xi}$. Now that is the same as the ordinary differential equation, $-2u'+ u= const$. The associated homogeneous equation is $-2u'+ u= 0$ with characteristic equation $-2r+ 1= 0$ with characteristic root $r= 1/2$ so the general solution to the associated homogeneous equation is $u(\eta)= Ce^{\eta/2}$ where, since we are treating $\xi$ as a constant, C can be an arbitrary function of $\xi$.
Since the non-homogeneous part, $e^{\xi}$, is a constant with respect to $\eta$ we look for a constant solution, u= D. Then $-2u'+ u= D= e^{\xi}$. The general solution to the entire equation is $u(\xi,\eta)= C(\xi)e^{\eta/2}+ e^{\xi}$.