$\displaystyle u_x-u_y+u=\exp{(x+2y)}$

$\displaystyle u(x,0)=0$

The solutions is $\displaystyle u(x,y)=-y\exp{(x+2y)}$

$\displaystyle \displaystyle\omega_{\xi}(\cos{\alpha}-\sin{\alpha})-\omega_{\eta}(\sin{\alpha}+\cos{\alpha})+\omega=ex p{[\xi(\cos{\alpha}+2\sin{\alpha})+\eta(2\cos{\alpha}-\sin{\alpha})]$

$\displaystyle \displaystyle\cos{\alpha}=-\frac{\sqrt{2}}{2} \ \ \mbox{and} \ \ \sin{\alpha}=\frac{\sqrt{2}}{2}$

$\displaystyle \displaystyle -\sqrt{2}\omega_{\xi}+\omega=\exp{\left(\frac{\sqrt {2}}{2}\xi-\frac{3\sqrt{2}}{2}\eta}\right)\Rightarrow \omega_{\xi}-\frac{1}{\sqrt{2}}\omega=\frac{\exp{\left(\frac{\s qrt{2}}{2}\xi-\frac{3\sqrt{2}}{2}\eta}\right)}{-\sqrt{2}}$

$\displaystyle x=A\xi+B\eta \ \ \mbox{and} \ \ y=C\xi+D\eta$

$\displaystyle \omega_{\xi}=Au_x+Cu_y=u_x-u_y\Rightarrow A=1 \ \ B=-1$

$\displaystyle x=\xi+B\eta \ \ \mbox{and} \ \ y=-\xi+D\eta$

$\displaystyle \omega_{\xi}+\omega=\mbox{?}$

Should this be set equal to $\displaystyle \displaystyle\exp{(x+2y)} \ \mbox{or} \ \exp{\left(\frac{\sqrt{2}}{2}\xi-\frac{3\sqrt{2}}{2}\eta}\right)$