1. ## Hyperbolic second order

$u_{xx}-3u_{yy}+2u_x-u_y+u=0$

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

$\displaystyle\omega_{\xi\xi}(\cos^2{\alpha}-3\sin^2{\alpha})+\omega_{\eta\eta}(\sin^2{\alpha}-3\cos^s{\alpha})-8\omega_{\xi\eta}\sin{\alpha}\cos{\alpha}+\omega_{ \xi}(2\cos{\alpha}-\sin{\alpha})-\omega_{\eta}(2\sin{\alpha}+\cos{\alpha})+\omega=0$

$\cos{\alpha}=1 \ \ \ \sin{\alpha}=0$

$\omega_{\xi\xi}-3\omega_{\eta\eta}+2\omega_{\xi}-\omega_{\eta}+\omega=0$

$u_{xx}-3u_{yy}+2u_x-u_y+u=0$

$u=\exp{(\beta x)}\omega(x,y)$

$\exp{(\beta x)}[\omega_{xx}-3\omega_{yy}+\omega_x(2\beta-3)-\omega_y+\omega(\beta^2+\beta+1)]=0$

$\displaystyla\beta=\frac{3}{2}$

$\displaystyle\omega_{xx}-3\omega_{yy}-\omega_y+\frac{19}{4}\omega=0$

I wasn't able to eliminate omega y. If I do $u=\exp{(\beta y)}\omega(x,y)$, I won't be able to eliminate omega x.

Should I continue or how should I approach this problem?

2. For equation

$
u_{xx}-3u_{yy}+2u_x-u_y+u=0
$

you may try this

$
u=\exp{(\alpha x)} \exp{(\beta y)} \: \omega(x,y).
$

3. Originally Posted by zzzoak
For equation

$
u_{xx}-3u_{yy}+2u_x-u_y+u=0
$

you may try this

$
u=\exp{(\alpha x)} \exp{(\beta y)} \: \omega(x,y).
$
So then I am setting w_x and w_y = 0 to obtain the respected alpha and beta?

4. $\exp{(\alpha x+\beta y)}[\omega_{xx}-3\omega_{yy}+\omega_{x}(2\alpha+2)-\omega_y(6\beta+1)+\omega(\alpha^2-3\beta^2+2\alpha-\beta+1)]=0$

$\displaystyle\alpha=-1 \ \ \ \beta=-\frac{1}{6}$

$\displaystyle\omega_{xx}-3\omega_{yy}+\frac{1}{12}\omega=0$

$\displaystyle\frac{1}{\mu^2}\omega_{\xi\xi}-\frac{3}{\nu^2}\omega_{\eta\eta}+\frac{1}{12}\omeg a=0$

$\displaystyle\frac{1}{\mu^2}=\frac{3}{\nu^2}=\frac {1}{12} \ \ \ \ \mu=\pm\sqrt{12} \ \ \ \ \nu=\pm\sqrt{36}$

$\displaystyle u_{xx}-u_{yy}+u=0$

5. Seems good to me.