Added June 2, 2019.
From example 3.5.1, page 210 nonlinear pde’s by Lokenath Debnath, 3rd edition.
Solve for \(u(x,y)\) \begin {align*} x u_x+y u_y&=u \end {align*}
Mathematica ✓
ClearAll["Global`*"]; pde = x*D[u[x, y], x] + y*D[u[x, y], y] ==u[x, y]; sol = AbsoluteTiming[TimeConstrained[DSolve[pde, u[x, y], {x, y}], 60*10]];
\[\left \{\left \{u(x,y)\to x c_1\left (\frac {y}{x}\right )\right \}\right \}\]
Maple ✓
restart; pde :=x*diff(u(x,y),x)+y*diff(u(x,y),y)=u(x,y); cpu_time := timelimit(60*10,CodeTools[Usage](assign('sol',pdsolve(pde,u(x,y))),output='realtime'));
\[u \left (x , y\right ) = x \mathit {\_F1} \left (\frac {y}{x}\right )\]
Hand solution
Solve \[ xu_{x}+yu_{y}=u \] Using the Lagrange-charpit method\[ \frac {dx}{x}=\frac {dy}{y}=\frac {du}{u}\] The first pair of equations gives \[ x=C_{1}y \] And \(\frac {dx}{x}=\frac {du}{u}\) gives\[ x=C_{2}u \] Since \(C_{2}=G\left ( C_{1}\right ) \) then \(\frac {x}{u}=G\left ( \frac {x}{y}\right ) \) or \(u=xG^{-1}\left ( \frac {x}{y}\right ) \). Let \(G^{-1}=F\). Then the solution \[ u\left ( x,y\right ) =xF\left ( \frac {x}{y}\right ) \]
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