General Solution for $x^2u_{xx}-y^2u_{yy}=0$ I tried finding the General solution of the PDE: $x^2u_{xx}-y^2u_{yy}=0$
I first tried reducing it to canonical form and then I got stuck.
Here was what I did:
I got the characteristic equation to be:
$$\frac{dy}{dx}=\pm\frac{y}{x}$$
Then solving further, I got:
$$\ln y = \ln x+C_1$$
and
$$\ln y=-\ln x+C_2$$
So in order to reduce the PDE into its canonical form, I introduced the new functions: $\xi, \eta$
Such that:
$$\xi=\ln y-\ln x$$
and
$$\eta=\ln y+\ln x$$
Thus the function becomes:
$$u=[\xi(x,y),\eta(x,y)]$$
So I got $u_{xx}$ & $u_{yy}$ in terms of $u_{\xi}, u_{\eta}, u_{\xi\eta}, u_{\xi\xi}, u_{\eta\eta}$and slotted it into the  PDE and I got this:
$$u_{\xi}-2u_{\xi\eta}=0$$
Is that the right canonical form for the PDE? And if it, how can I solve further to get the General Solution
 A: Let $u(x,y)=X(x)Y(y)$ ,
Then $x^2X''(x)Y(y)-y^2X(x)Y''(y)=0$
$x^2X''(x)Y(y)=y^2X(x)Y''(y)$
$\dfrac{x^2X''(x)}{X(x)}=\dfrac{y^2Y''(y)}{Y(y)}=\dfrac{4s^2-1}{4}$
$\begin{cases}x^2X''(x)-\dfrac{4s^2-1}{4}X(x)=0\\y^2Y''(y)-\dfrac{4s^2-1}{4}Y(y)=0\end{cases}$
$\begin{cases}X(x)=\begin{cases}c_1(s)x^{\frac{1}{2}+s}+c_2(s)x^{\frac{1}{2}-s}&\text{when}~s\neq0\\c_1\sqrt{x}\ln x+c_2\sqrt{x}&\text{when}~s=0\end{cases}\\Y(y)=\begin{cases}c_3(s)y^{\frac{1}{2}+s}+c_4(s)y^{\frac{1}{2}-s}&\text{when}~s\neq0\\c_3\sqrt{y}\ln y+c_4\sqrt{y}&\text{when}~s=0\end{cases}\end{cases}$
$\therefore u(x,y)=\int_sC_1(s)(xy)^{\frac{1}{2}+s}~ds+\int_sC_2(s)(xy)^\frac{1}{2}\left(\dfrac{x}{y}\right)^s~ds+\int_sC_3(s)(xy)^\frac{1}{2}\left(\dfrac{y}{x}\right)^s~ds+\int_sC_4(s)(xy)^{\frac{1}{2}-s}~ds$
or $\sum\limits_sC_1(s)(xy)^{\frac{1}{2}+s}+\sum\limits_sC_2(s)(xy)^\frac{1}{2}\left(\dfrac{x}{y}\right)^s+\sum\limits_sC_3(s)(xy)^\frac{1}{2}\left(\dfrac{y}{x}\right)^s+\sum\limits_sC_4(s)(xy)^{\frac{1}{2}-s}$
i.e. $u(x,y)=F(xy)+\sqrt{xy}~G\left(\dfrac{x}{y}\right)$
A: *

*After change $\quad \xi=\log(x),\quad \eta=\log(y)\quad$ we get equation with constant coeficients
$$u_{\xi\xi}-u_\xi-u_{\eta\eta}+u_\eta=0$$

*$${{D}_{\xi}^{2}}-{D_{\xi}}-{{D}_{\eta}^{2}}+{D_{\eta}}=\left( {D_{\xi}}-{D_{\eta}}\right) \, \left( {D_{\xi}}+{D_{\eta}}-1\right)$$

*solution of $\;u_\xi-u_\eta=0\;$ is $\;u_1=f(\xi+\eta)$

*solution of $\;u_\xi+u_\eta-u=0\;$ is $\;u_2=e^\eta g(\xi-\eta)$

*$$u=u_1+u_2=f(\xi+\eta)+e^\eta g(\xi-\eta)\\
=f\left(\log(x)+\log(y)\right)+e^{\log(y)}g(\log(x)-\log(y))\\=
f(\log(xy))+y\,g\left(\log(\frac{x}{y})\right)\\
=F(xy)+y\,G(\frac{x}{y})
$$

