Problem with solving hyperbolic PDE using canonical form

I'm stuck when im trying to solve this equation: $$\frac{1-n}{2}xu_x+\frac{n-1}{2}u-2x^2u_{xx}+b n\cdot x\cdot y \cdot u_{xy}=0$$, where $$u=u(x,y)$$ and $$b$$ and $$n>0$$ are constant parameters.

Comparing this with general 2-variable PDE: $$A(x, y)u_{xx} + B(x, y)u_{xy} + C(x, y)u_{yy} = Φ(x, y,u,u_x,u_y)$$

I get: $$A=-2x^2$$ ; $$B=bnxy\\$$ and $$C=0$$.

This is hyperbolic equation because $$B^2-AC=(bnxy)^2>0$$

From $$\frac{dy}{dx} = \frac{B \pm\sqrt{B^2-4AC}}{2A} = \frac{+bnxy-bnxy}{4x^2}=0$$ or $$\frac{dy}{dx}=\frac{2bnxy}{-4x^2}=-\frac{1}{2}\frac{bny}{x}$$, characteristic curves are: $$y=const$$ and $$y=x^{\frac{-bnx}{2}}\cdot constant$$

Using integration constants as coordinates: $$\xi=y$$ and $$\eta=x^{\frac{bnx}{2}}y$$

Then:

$$ξ_x=0$$, $$ξ_y=1$$ and $$\eta_x=\frac{bn}{2}x^{\frac{bn}{2}-1}y$$, $$\eta_y=x^{\frac{bnx}{2}}$$

Using chain rule:

$$u_x = u_ξξ_x + u_ηη_x=0+u_\eta\frac{bn}{2}x^{\frac{bn}{2}-1}y=\frac{bn}{2}x^{\frac{bn}{2}-1}y\cdot u_\eta$$ $$u_{xx} = u_{ξξ}ξ^2_x + 2u_{ξη}ξ_xη_x + u_{ηη}η^2_x+ u_ξξ_{xx} + u_ηη_{xx}=0+0+\frac{(bn)^2}{4}x^{bn-2} y^2\cdot u_{\eta\eta}+0+\frac{bn}{2}(\frac{bn}{2}-1)x^{\frac{bn}{2}-2}y\cdot u_\eta=\frac{(bn)^2}{4}x^{bn-2} y^2\cdot u_{\eta\eta}+\frac{bn}{2}(\frac{bn}{2}-1)x^{\frac{bn}{2}-2}y\cdot u_\eta$$ $$u_{xy} = u_{ξξ}ξ_xξ_y + u_{ξη} (ξ_xη_y + ξ_yη_x) + u_{ηη}η_xη_y + u_ξξ_{xy} + u_ηη_{xy}=\frac{bn}{2}x^{\frac{bn}{2}-1}u_\eta+\frac{bn}{2}x^{bn-1}yu_{\eta\eta}+\frac{bn}{2}x^{\frac{bn}{2}-1}y u_{\eta\xi}$$

Plugging this to equation, and simplifying this with expressions for $$\eta$$ and $$\xi$$: $$\Big[\frac{1-n}{2}(\frac{bn}{2}\eta)+bn\cdot\eta \Big]u_\eta+\frac{(bn)^2}{2}\xi\eta \cdot u_{\xi\eta}+\frac{n-1}{2}u=0$$

After further rearrangements equation takes form:

$$a_1u_\eta+a_2\xi u_{\xi\eta}+a_3\frac{1}{\eta}u=0$$ with $$a_1=(\frac{1-n}{4}bn+bn)$$, $$a_2=bn^2/2$$, $$a_3=(n-1)/2$$

Now i don't know what to do - i tried use Mathematica software to solve this but Mathematica doesn't support 2nd order PDE's with lower order terms. Can i use separation of variables to solve it? If so - how to separate this equation? Im trying to follow simillar aproach as Canonical form of PDE

• Edit: there was additional symbol in $u_x$ - variable $x$ which simplified equation in canonical form (so i started with wrong equation). Now i corrected this but still i don't know how to solve it. – StarPlatinumZaWardo Apr 16 '19 at 12:39

After hours of intensified thinking - i found satisfying me solution. Assuming: $$u=f(\xi)g(\eta)$$, I recasted equation in form:

$$a_1+a_2\xi\frac{f'(\xi)}{f(\xi)}+a_3\frac{1}{\eta}\frac{g(\eta)}{g'(\eta)}=0$$

then i split it into 2 equations:

$$a_1+a_2\xi\frac{f'(\xi)}{f(\xi)}=-k$$ and $$a_3\frac{1}{\eta}\frac{g(\eta)}{g'(\eta)}=k$$.

From this ODE's functions $$f$$ and $$g$$ are given by:

$$f(\xi)=\xi^{-(k+a_1)/a_2}C_1$$ and $$g(\eta)=\eta^{a_3/k}C_2$$

Where $$C_i$$ are integration constants and $$k$$ is arbitrary constant different from $$0$$ (it comes from separation of variables). Pluging back $$\eta=x^{\frac{bnx}{2}}y$$ together with $$\xi=y$$ should solve this PDE.