General solution, Canonical form of PDE $$u_{xy} + yu_{yy} + \sin(x+y)=0$$
Convert the above partial differential equations into the canonical form, and then find the general solution.
The problem I am encountering is that even after making the transformations, I get a similar partial differential equation in terms of new variables.
The transformations are  -- 
$\alpha = x$ , and 
$\beta = y - e^{x}$.
If you could suggest what I could do, to successfully convert the partial differential equation into the canonical form. Sincerest gratitude.
 A: Characteristic equation would be.
$$-dxdy + y(dx)^2 = 0$$
Solving this we'll get.
$$dx[-dy + y dx] = 0$$
Note that for any $y$, the discriminant is positive. $D = 1$, hence we have a hyperbolic equation
$$\left\{\begin{matrix}
x = \xi \\ 
dy = y dx 
\end{matrix}\right.,    \left\{\begin{matrix}
x = \xi \\ 
\ln|y| = x + \eta 
\end{matrix}\right.,  \left\{\begin{matrix}
\xi = x \\ 
\eta = ln|y| - x 
\end{matrix}\right.$$
Now doing change of variables $(x,y) \mapsto (\xi,\eta)$.
$u_{x} = u_{\xi} \xi_x+u_{\eta} \eta_x = u_{\xi} - u_{\eta} $
$u_{xy} = [u_{\xi} - u_{\eta}]_y = u_{\xi\xi} \xi_y+ u_{\xi\eta}\eta_y - u_{\xi\eta}\xi_y - u_{\eta\eta}\eta_y = \frac{1}{y}(u_{\xi\eta} - u_{\eta\eta})$
$u_y = \frac{1}{y}u_{\eta}$
$u_{yy} = [u_{\eta}\frac{1}{y}]_y = \frac{1}{y}(u_{\xi\eta} \xi_y + u_{\eta\eta}\eta_y) + u_{\eta}(-\frac{1}{y^2}) = \frac{1}{y^2}(u_{\eta\eta} - u_{\eta})$
Substitute into original equation.
$$\frac{1}{y}u_{\xi\eta} - \frac{1}{y}u_{\eta\eta} + \frac{1}{y} u_{\eta\eta} - \frac{1}{y}u_{\eta} + \sin(x+y) = 0 $$
$$u_{\xi\eta} - u_{\eta} + y \sin(x+y) = 0 $$
$$u_{\xi\eta} - u_{\eta} + e^{\xi+\eta} \sin(\xi+e^{\xi+\eta}) = 0 $$
Hopefully I haven't done any computational mistake.
So we reached to canonical form of hyperbolic equations.
$u_{\xi\eta} = \Phi(\xi,\eta,u,u_{\xi},u_{\eta})$
If you're interested of more natural canonical form $u_{\xi\xi} - u_{\eta\eta} = \Phi(\xi,\eta,u,u_{\xi},u_{\eta})$, then you can do another change of variable $(\xi,\eta) \mapsto (\alpha,\beta)$
$\alpha = \frac{\xi+\eta}{2}$
$\beta = \frac{\xi-\eta}{2}$

The process is similar, I'll leave this to you :)
