[SOLVED] Second Order

$3u_{xx}+4u_{yy}-u=0$

$a=3, \ \ b=0, \ \ c=4$

$ac-b^2=3*4-0=12>0$

$\mbox{Elliptic} \ \ \omega_{\xi\xi}+\omega_{\eta\eta}+\gamma\omega=0$

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

$u_{xx}=(\omega_{\xi}\cos{\alpha}-\omega_{\eta}\sin{\alpha})^2=\omega_{\xi\xi}\cos^2 {\alpha}-2\omega_{\xi\eta}\sin{\alpha}\cos{\alpha}+\omega_{ \eta\eta}\sin^2{\alpha}$

$u_{yy}=(\omega_{\xi}\sin{\alpha}+\omega_{\eta}\cos {\alpha})^2=\omega_{\xi\xi}\sin^2{\alpha}+2\omega_ {\xi\eta}\sin{\alpha}\cos{\alpha}+\omega_{\eta\eta }\cos^2{\alpha}$

$\omega_{\xi\xi}(3\cos^2{\alpha}+4\sin^2{\alpha})+\ omega_{\eta\eta}(3\sin^2{\alpha}+4\cos{\alpha})+2\ sin{\alpha}\cos{\alpha} \ \omega_{\xi\eta}+\omega=0$

Since I am trying to eliminate $\omega_{\xi\eta}$ , does it matter what alpha I pick?

$2\sin{\alpha}\cos{\alpha}=0\Rightarrow \alpha=\frac{\pi}{2}+\pi k \ \mbox{or} \ \pi k \ \ \ k\in\mathbb{Z}$