This is using V12 on windows.
I solved this PDE by hand and wanted to ask Mathematica to verify my solution. But Mathematica solution does not look like mine. And when I tried to verify Mathematica own solution I get False.
Is it possible Mathematica solution is not correct? Could some expert help shed some light on what is the issue here? The PDE is
Solve for $u(x,y)$ in $$ y u_x - x u_y = e^u $$
Here is Mathematica code
ClearAll[x,y,u]
pde=y*D[u[x,y],x]-x*D[u[x,y],y]==Exp[u[x,y]]
sol=DSolve[pde,u,{x,y}]
Assuming[y>0,Simplify[pde/.sol]]
Here is the solution obtained by hand (derivation below), which Mathemaica verifies ok.
mySol = u -> Function[{x, y}, -Log[C[1][(x^2 + y^2)] - ArcTan[x/y]]]
Simplify[pde /. mySol]
Note on Mathematica solution: Using $\frac{1}{2}$ or not does not matter. Since this is absorbed by the constant of integration.
The question is: Why Mathematica does not verify its own solution? Is the solution given by Mathematica not correct? The difference between my solution and Mathematica solution seems to be a sign difference only! i.e. if I swap the terms inside Mathematica own solution, it becomes same as my solution and now Mathematica verifies it. So it looks like a sign problem somewhere?
Appendix
FYI, here is my hand solution
Solve $$ yu_{x}-xu_{y}=e^{u} $$ Since no initial conditions are given, I find using Lagrange-charpit method is better here than using characteristic equations with parameter $s$. The Lagrange-charpit equations for the above PDE are
$$ \frac{dx}{y}=\frac{-dy}{x}=\frac{du}{e^{u}} $$
The first two equations gives
\begin{align} xdx & =-ydy\nonumber\\ \frac{x^{2}}{2} & =-\frac{y^{2}}{2}+C_{1}\nonumber\\ C_{1} & =\frac{1}{2}\left( x^{2}+y^{2}\right) \nonumber\\ C_{1} & =\left( x^{2}+y^{2}\right) \tag{1} \end{align}
Where the $\frac{1}{2}$ is absorbed by the constant. We now need to decide to either solve $\frac{-dy}{x}=\frac{du}{e^{u}}$ together or $\frac{dx}{y}=\frac {du}{e^{u}}$. It does not matter which pair to pick. Using the second pair gives
$$ \frac{dx}{y}=\frac{du}{e^{u}} $$
But from (1), $y=\sqrt{C_{1}-x^{2}}$ (taking only the positive root) and the above equation now becomes
$$ \frac{dx}{\sqrt{C_{1}-x^{2}}}=\frac{du}{e^{u}} $$
Integrating gives
\begin{align*} \arctan\left( \frac{x}{\sqrt{C_{1}-x^{2}}}\right) & =-e^{-u}+C_{2}\\ \arctan\left( \frac{x}{y}\right) & =-e^{-u}+C_{2}\\ C_{2} & =\arctan\left( \frac{x}{y}\right) +e^{-u} \end{align*}
In this method, the constants $C_{1},C_{2}$ are always related by $C_{2}=F\left( C_{1}\right) $ where $F$ is an arbitrary function. Hence we obtain
\begin{align*} \arctan\left( \frac{x}{y}\right) +e^{-u} & =F\left( x^{2}+y^{2}\right) \\ e^{-u} & =F\left( x^{2}+y^{2}\right) -\arctan\left( \frac{x}{y}\right) \end{align*}
For positive $u$ the above simplifies to
\begin{align*} -u & =\ln\left( F\left( x^{2}+y^{2}\right) -\arctan\left( \frac{x} {y}\right) \right) \\ u\left( x,y\right) & =-\ln\left( F\left( x^{2}+y^{2}\right) -\arctan\left( \frac{x}{y}\right) \right) \end{align*}
