Solving $yu_{xx}+(x+y)u_{xy}+xu_{yy}=0$ Let $yu_{xx}+(x+y)u_{xy}+xu_{yy}=0$, how would you go about solving this?
So far, I have show it is hyperbolic everywhere except for the line $y=x$ and have been attempting to find the characteristic variables by solving:
$$\frac{d\phi}{dx}+\frac{x}{y}\frac{d\phi}{dy}=0$$
and
$$\frac{d\psi}{dx}+\frac{d\psi}{dy}=0$$
I have $\psi=x-y$, but am stuck on the first pde. Once I have these I plan to $u_{\phi \psi}=0$, the canonical form, to find the general solution. I would appreciate letting me know i'm on the right lines and any help very much, thanks!
EDIT: $\phi=y^2-x^2$ is a solution, so by the chain rule I end up with $(x^2+y^2)u_{\phi \psi}=0$ is this correct? As the solution is meant to be
$$u=\frac{1}{y-x}f(y^2-x^2)+g(y-x)$$
this seems to give
$$u=f(y^2-x^2)+g(y-x)$$
 A: Try let $v=x+y$ , $w=x-y$ ,
Then $\dfrac{\partial u}{\partial x}=\dfrac{\partial u}{\partial v}\dfrac{\partial v}{\partial x}+\dfrac{\partial u}{\partial w}\dfrac{\partial w}{\partial x}=\dfrac{\partial u}{\partial v}+\dfrac{\partial u}{\partial w}$
$\dfrac{\partial^2u}{\partial x^2}=\dfrac{\partial}{\partial x}\left(\dfrac{\partial u}{\partial v}+\dfrac{\partial u}{\partial w}\right)=\dfrac{\partial}{\partial v}\left(\dfrac{\partial u}{\partial v}+\dfrac{\partial u}{\partial w}\right)\dfrac{\partial v}{\partial x}+\dfrac{\partial}{\partial w}\left(\dfrac{\partial u}{\partial v}+\dfrac{\partial u}{\partial w}\right)\dfrac{\partial w}{\partial x}=\dfrac{\partial^2u}{\partial v^2}+\dfrac{\partial^2u}{\partial vw}+\dfrac{\partial^2u}{\partial vw}+\dfrac{\partial^2u}{\partial w^2}=\dfrac{\partial^2u}{\partial v^2}+2\dfrac{\partial^2u}{\partial vw}+\dfrac{\partial^2u}{\partial w^2}$
$\dfrac{\partial u}{\partial y}=\dfrac{\partial u}{\partial v}\dfrac{\partial v}{\partial y}+\dfrac{\partial u}{\partial w}\dfrac{\partial w}{\partial y}=\dfrac{\partial u}{\partial v}-\dfrac{\partial u}{\partial w}$
$\dfrac{\partial^2u}{\partial xy}=\dfrac{\partial}{\partial x}\left(\dfrac{\partial u}{\partial v}-\dfrac{\partial u}{\partial w}\right)=\dfrac{\partial}{\partial v}\left(\dfrac{\partial u}{\partial v}-\dfrac{\partial u}{\partial w}\right)\dfrac{\partial v}{\partial x}+\dfrac{\partial}{\partial w}\left(\dfrac{\partial u}{\partial v}-\dfrac{\partial u}{\partial w}\right)\dfrac{\partial w}{\partial x}=\dfrac{\partial^2u}{\partial v^2}-\dfrac{\partial^2u}{\partial vw}+\dfrac{\partial^2u}{\partial vw}-\dfrac{\partial^2u}{\partial w^2}=\dfrac{\partial^2u}{\partial v^2}-\dfrac{\partial^2u}{\partial w^2}$
$\dfrac{\partial^2u}{\partial y^2}=\dfrac{\partial}{\partial y}\left(\dfrac{\partial u}{\partial v}-\dfrac{\partial u}{\partial w}\right)=\dfrac{\partial}{\partial v}\left(\dfrac{\partial u}{\partial v}-\dfrac{\partial u}{\partial w}\right)\dfrac{\partial v}{\partial y}+\dfrac{\partial}{\partial w}\left(\dfrac{\partial u}{\partial v}-\dfrac{\partial u}{\partial w}\right)\dfrac{\partial w}{\partial y}=\dfrac{\partial^2u}{\partial v^2}-\dfrac{\partial^2u}{\partial vw}-\dfrac{\partial^2u}{\partial vw}+\dfrac{\partial^2u}{\partial w^2}=\dfrac{\partial^2u}{\partial v^2}-2\dfrac{\partial^2u}{\partial vw}+\dfrac{\partial^2u}{\partial w^2}$
$\therefore y\left(\dfrac{\partial^2u}{\partial v^2}+2\dfrac{\partial^2u}{\partial vw}+\dfrac{\partial^2u}{\partial w^2}\right)+(x+y)\left(\dfrac{\partial^2u}{\partial v^2}-\dfrac{\partial^2u}{\partial w^2}\right)+x\left(\dfrac{\partial^2u}{\partial v^2}-2\dfrac{\partial^2u}{\partial vw}+\dfrac{\partial^2u}{\partial w^2}\right)=0$
$2(x+y)\dfrac{\partial^2u}{\partial v^2}-2(x-y)\dfrac{\partial^2u}{\partial vw}=0$
$v\dfrac{\partial^2u}{\partial v^2}-w\dfrac{\partial^2u}{\partial vw}=0$
Let $z=\dfrac{\partial u}{\partial v}$ ,
Then $v\dfrac{\partial z}{\partial v}-w\dfrac{\partial z}{\partial w}=0$
This belongs to a PDE of the form http://eqworld.ipmnet.ru/en/solutions/fpde/fpde1104.pdf
$z=\dfrac{\partial u}{\partial v}=c\left(\int\dfrac{1}{v}~dv+\int\dfrac{1}{w}~dw\right)=c(\ln v+\ln w)=c(\ln vw)=C(vw)$
$u=\int C(vw)~dv=\int C(v)~d\left(\dfrac{v}{w}\right)=\int\dfrac{C(v)}{w}dv=\dfrac{C_1(v)+C_2(w)}{w}=\dfrac{C_1(x+y)+C_2(x-y)}{x-y}$
Note that when the independent variables in the arbitrary functions are more simple, the general solutions are more general. 
A: Characteristic equation
$A(\dfrac{dy}{dx})^2-B(\dfrac{dy}{dx})+C=0\Rightarrow y(\dfrac{dy}{dx})^2-(x+y)(\dfrac{dy}{dx})+x=0$
then
$\dfrac{dy}{dx}=\dfrac{(x+y)\pm\sqrt{(x-y)^2}}{2y}=\dfrac{x}{y},\ 1
$
Characteristic curves: $s(x,y)$ and $t(x,y)$
$\left\{\begin{array}{c}
     \displaystyle{\frac{dy}{dx}=\frac{x}{y}}  \\
     \displaystyle{\frac{dy}{dx}=1} 
\end{array}\right.\ \Rightarrow\ 
\left\{\begin{array}{l}
     s(x,y)=y^2-x^2=c_1\  \\
     t(x,y)=y-x=c_2
\end{array}\right.
$
Reduce to Standard Form by using characteristic curves with chain rules
(i) First order derivatives
$
\begin{align*}
    u_{x}=\frac{\partial u}{\partial x}&=\frac{\partial u}{\partial s}\frac{\partial s}{\partial x}+\frac{\partial u}{\partial t}\frac{\partial t}{\partial x}=s_x u_s+t_x u_t
    \\
    u_{y}=\frac{\partial u}{\partial y}&=\frac{\partial u}{\partial s}\frac{\partial s}{\partial y}+\frac{\partial u}{\partial t}\frac{\partial t}{\partial y}=s_y u_s+t_y u_t
\end{align*}$
(ii) Second order derivatives
$
\begin{align*}
    u_{xx}&=\frac{\partial }{\partial x}(s_x u_s+t_x u_t)=\Big(s_{xx}u_s+t_{xx}u_t\Big)+\Big(s_x\frac{\partial u_s}{\partial x}+t_x\frac{\partial u_t}{\partial x}\Big)
    \\
    &=\Big(s_{xx}u_s+t_{xx}u_t\Big)+s_x\Big(s_xu_{ss}+t_xu_{st}\Big)+t_x\Big(s_xu_{ts}+t_xu_{tt}\Big)\\
    &=s_{xx}u_s+t_{xx}u_t+s_x^2u_{ss}+2s_xt_xu_{st}+t_x^2u_{tt}
    \\
    u_{yy}&=s_{yy}u_s+t_{yy}u_t+s_y^2u_{ss}+2s_yt_yu_{st}+t_y^2u_{tt}
    \\
    u_{xy}&=\frac{\partial }{\partial y}(s_x u_s+t_x u_t)=\Big(s_{xy}u_s+t_{xy}u_t\Big)+\Big(s_x\frac{\partial u_s}{\partial y}+t_x\frac{\partial u_t}{\partial y}\Big)
    \\
    &=\Big(s_{xy}u_s+t_{xy}u_t\Big)+s_x\Big(s_yu_{ss}+t_yu_{st}\Big)+t_x\Big(s_yu_{ts}+t_yu_{tt}\Big)\\
    &=s_{xy}u_s+t_{xy}u_t+s_xs_yu_{ss}+(s_xt_y+s_yt_x)u_{st}+t_xt_yu_{tt}
\end{align*}
$
(iii) Substitute for $u_{xx}, u_{yy}$ and $u_{xy}$ in the PDE：$y u_{xx}+(x+y)u_{xy}+x u_{yy}=0$.
We can get
$
\begin{align*}
    y&\Big[(-2)u_{s}+(-2x)^2u_{ss}+2(-2x)(-1)u_{st}+(-1)^2u_{tt}\Big]\\
    &+(x+y)\Big[(-2x)(-2y)u_{ss}-2(x+y)u_{st}+(-1)(1)u_{tt}\Big]\\
    &+x\Big[(2)u_{s}+(2y)^2u_{ss}+2(2y)(1)u_{st}+(1)^2u_{tt}\Big]=0
\end{align*}$
rearrange the terms
$-2(y-x)u_{s}-2(y-x)u_{st}=0\Rightarrow u_{s}+tu_{st}=0
$
For $y\neq x\Rightarrow$ 
$\dfrac{1}{t}\dfrac{\partial u}{\partial s}+\dfrac{\partial}{\partial t}(\dfrac{\partial u}{\partial s})=0.
$
Let $z=\dfrac{\partial u}{\partial s}$, then
$\dfrac{1}{t}+\dfrac{1}{z}\dfrac{\partial z}{\partial t}=0
$
integrate
$\ln{t}+\ln{z}=\ln{f(s)}\Rightarrow z=\dfrac{\partial u}{\partial s}=\dfrac{1}{t}f(s)
$
integrate again
$u(x,y)=\dfrac{1}{t}F(s)+g(t)=\dfrac{1}{y-x}F(y^2-x^2)+g(y-x)
$

Another method
By observing, we can express the PDE in
$(y\dfrac{\partial}{\partial x}+x\dfrac{\partial}{\partial y})(\dfrac{\partial}{\partial x}+\dfrac{\partial}{\partial y})u=0
$
Let $z=(\dfrac{\partial}{\partial x}+\dfrac{\partial}{\partial y})u$, then $y\dfrac{\partial z}{\partial x}+x\dfrac{\partial z}{\partial y}=0
$
Characteristic equation
$\dfrac{dx}{y}=\dfrac{dy}{x}=\dfrac{dz}{0}\Rightarrow\left\{\begin{array}{c}
     \displaystyle{\frac{dy}{dx}=\frac{x}{y}}  \\
     \displaystyle{\frac{dz}{dx}=0} 
\end{array}\right.\Rightarrow\left\{\begin{array}{l}
     \displaystyle{y^2-x^2=c_1}  \\
     \displaystyle{z=c_2} 
\end{array}\right.$
$\because c_2=f(c_1), z=f(y^2-x^2)$
$\Rightarrow f(y^2-x^2)=\dfrac{\partial u}{\partial x}+\dfrac{\partial u}{\partial y}$
Characteristic equation again
$\dfrac{dx}{1}=\dfrac{dy}{1}=\dfrac{du}{f(y^2-x^2)}\Rightarrow\left\{\begin{array}{l}
     \displaystyle{\frac{dy}{dx}=1}  \\
     \displaystyle{\frac{du}{d(y+x)}=f(y^2-x^2)} 
\end{array}\right.\Rightarrow\left\{\begin{array}{l}
     \displaystyle{y-x=c_3}  \\
     \displaystyle{u=\frac{1}{y-x}F(y^2-x^2)+c_4} 
\end{array}\right.$
$\because c_4=f(c_3)$, we finally obtain the solution
$u(x,y)=\dfrac{1}{y-x}F(y^2-x^2)+g(y-x)
$
