Added May 21, 2019.
This is same problem as above, but without I.C. given.
Solve for \(u(x,y)\) \begin {align*} y u_x - x u_y = e^u \end {align*}
No IC are given.
Mathematica ✓
ClearAll["Global`*"]; pde = y*D[u[x, y], x] - x*D[u[x, y], y] == Exp[u[x, y]]; sol = AbsoluteTiming[TimeConstrained[DSolve[pde, u[x, y], {x, y}], 60*10]];
\[\left \{\left \{u(x,y)\to -\log \left (\tan ^{-1}\left (\frac {x}{\sqrt {y^2}}\right )-c_1\left (\frac {1}{2} \left (x^2+y^2\right )\right )\right )\right \}\right \}\]
Maple ✓
restart; pde :=y*diff(u(x,y),x)-x*diff(u(x,y),y)=exp(u(x,y)); cpu_time := timelimit(60*10,CodeTools[Usage](assign('sol',pdsolve(pde,u(x,y))),output='realtime'));
\[u \left ( x,y \right ) =\ln \left ( - \left ( \arctan \left ( {\frac {x}{y}} \right ) +{\it \_F1} \left ( {x}^{2}+{y}^{2} \right ) \right ) ^{-1} \right ) \]
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 \(2\) 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*}
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