Added June 2, 2019.
From example 3.5.3, page 211 nonlinear pde’s by Lokenath Debnath, 3rd edition.
Solve for \(u(x,y)\) \begin {align*} x^2 u_x+y^2 u_y&=(x+y) u \end {align*}
Mathematica ✓
ClearAll["Global`*"]; pde = x^2*D[u[x, y], x] + y^2*D[u[x, y], y] ==(x+y)*u[x, y]; sol = AbsoluteTiming[TimeConstrained[DSolve[pde, u[x, y], {x, y}], 60*10]];
\[\left \{\left \{u(x,y)\to x y c_1\left (\frac {1}{x}-\frac {1}{y}\right )\right \}\right \}\]
Maple ✓
restart; pde :=x^2*diff(u(x,y),x)+y^2*diff(u(x,y),y)=(x+y)*u(x,y); cpu_time := timelimit(60*10,CodeTools[Usage](assign('sol',pdsolve(pde,u(x,y))),output='realtime'));
\[u \left (x , y\right ) = x y \textit {\_F1} \left (\frac {x -y}{x y}\right )\]
Hand solution
Solve \[ x^{2}u_{x}+y^{2}u_{y}=\left ( x+y\right ) u \] Using the Lagrange-charpit method\[ \frac {dx}{x^{2}}=\frac {dy}{y^{2}}=\frac {du}{\left ( x+y\right ) u}\] The first pair of equations gives \begin {equation} -\frac {1}{x}=-\frac {1}{y}+C_{1}\tag {1} \end {equation} And \(\frac {dx}{x^{2}}=\frac {du}{\left ( x+y\right ) u}\) gives\[ \frac {\left ( x+y\right ) }{x^{2}}dx=\frac {du}{u}\] But from (1) \(\frac {1}{y}=\frac {1}{x}+C_{1}\) or \(\frac {1}{y}=\frac {1+xC_{1}}{x}\) or \(y=\frac {x}{1+xC_{1}}\)Therefore the above becomes\begin {align*} \frac {\left ( x+\frac {x}{1+xC_{1}}\right ) }{x^{2}}dx & =\frac {du}{u}\\ \frac {\left ( x\left ( 1+xC_{1}\right ) +x\right ) }{x^{2}\left ( 1+xC_{1}\right ) }dx & =\frac {du}{u}\\ \frac {2x+x^{2}C_{1}}{x^{2}+x^{3}C_{1}}dx & =\frac {du}{u}\\ \frac {2+xC_{1}}{x+x^{2}C_{1}}dx & =\frac {du}{u}\\ 2\ln x-\ln \left ( 1+C_{1}x\right ) & =\ln u+C_{2}\\ \ln \frac {x^{2}}{1+C_{1}x} & =\ln u+C_{2}\\ \frac {x^{2}}{1+C_{1}x} & =C_{2}u \end {align*}
But \(C_{1}=\frac {1}{y}-\frac {1}{x}\) hence the above becomes\begin {align*} \frac {x^{2}}{1+\left ( \frac {1}{y}-\frac {1}{x}\right ) x} & =C_{2}u\\ \frac {x^{2}}{1+\left ( \frac {x-y}{yx}\right ) x} & =C_{2}u\\ \frac {yx^{2}}{y+\left ( x-y\right ) } & =C_{2}u\\ yx & =C_{2}u \end {align*}
Since \(C_{2}=G\left ( C_{1}\right ) \) then \(\frac {yx}{u}=G\left ( \frac {1}{y}-\frac {1}{x}\right ) \) or \begin {align*} u & =yxG^{-1}\left ( \frac {1}{y}-\frac {1}{x}\right ) \\ & =yxG^{-1}\left ( \frac {y-x}{yx}\right ) \end {align*}
Let \(G^{-1}=F\) \[ u=yxF\left ( \frac {y-x}{yx}\right ) \]
Where \(F\) is arbitrary function.
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