$\displaystyle u_x-u_y+u=1$

Auxiliary condition $\displaystyle u(x,0)=\sin{x}$

$\displaystyle u(x,y)=u(\xi\cos{\alpha}-\eta\sin{\alpha},\xi\sin{\alpha}+\eta\cos{\alpha}) =\omega(\xi,\eta)$

$\displaystyle u_x=\omega_{\xi}\cos{\alpha}-\omega_{\eta}\sin{\alpha}$

$\displaystyle u_y=\omega_{\xi}\sin{\alpha}+\omega_{\eta}\cos{\al pha}$

$\displaystyle \omega_{\xi}\cos{\alpha}-\omega_{\eta}\sin{\alpha}-(\omega_{\xi}\sin{\alpha}+\omega_{\eta}\cos{\alpha })+\omega=1$

$\displaystyle \omega_{\xi}(\cos{\alpha}-\sin{\alpha})-\omega_{\eta}(\sin{\alpha}+\cos{\alpha})+\omega=1$

$\displaystyle \sin{\alpha}+\cos{\alpha}=0$

$\displaystyle \displaystyle\tan{\alpha}=-1\Rightarrow\alpha=\frac{3\pi}{4}+\pi k \ k\in\mathbb{Z}$

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

$\displaystyle \displaystyle -\sqrt{2}\omega_{\xi}+\omega=1\Rightarrow\omega_{\x i}-\frac{1}{\sqrt{2}}\omega=-\frac{1}{\sqrt{2}}$

$\displaystyle x=A\xi+B\eta$

$\displaystyle y=C\xi+D\eta$

$\displaystyle \omega_{\xi}=Au_x+Cu_y=u_x-u_y$

$\displaystyle A=1 \ \mbox{and} \ B=-1$

$\displaystyle x=\xi+B\eta$

$\displaystyle y=-\xi+D\eta$

$\displaystyle u_x-u_y+u=1\Rightarrow\omega_{\xi}+\omega=1$

$\displaystyle P(\xi)=1 \ \ Q(\xi)=1$

$\displaystyle \exp{\left(\int P(\xi)d\xi\right)}=\exp{\left(\int d\xi\right)}=\exp{\xi}$

$\displaystyle \displaystyle\int\frac{\partial}{\partial\xi}\left[\exp{\left(\int P(\xi)d\xi\right)}\omega\right]d\xi=\int\exp{\left(\int P(\xi)d\xi\right)}Q(\xi)d\xi$

$\displaystyle \displaystyle\int\frac{\partial}{\partial\xi}\left[\exp{(\xi)}\omega\right]d\xi=\int\exp{(\xi)}d\xi\Rightarrow\exp{(\xi)}\ome ga=\exp{(\xi)}+g(\eta)$

$\displaystyle \omega(\xi,\eta)=1+\exp{(-\xi)}g(\eta)$

How do I solve for the auxiliary equation?