Taken from Mathematica help pages
General solution for a quasilinear first-order PDE
Solve for \(u(x,y)\) \[ 2 u_x + 5 u_y = u^2(x,y) + 1 \]
Mathematica ✓
ClearAll["Global`*"]; pde = 2*D[u[x, y], x] + 5*D[u[x, y], y] == u[x, y]^2 + 1; sol = AbsoluteTiming[TimeConstrained[Simplify[DSolve[pde, u[x, y], {x, y}]], 60*10]];
\[\left \{\left \{u(x,y)\to \tan \left (c_1\left (y-\frac {5 x}{2}\right )+\frac {x}{2}\right )\right \}\right \}\]
Maple ✓
restart; interface(showassumed=0); pde := 2* diff(u(x, y), x) + 5*diff(u(x, y), y) = u(x, y)^2 + 1; cpu_time := timelimit(60*10,CodeTools[Usage](assign('sol',pdsolve(pde, u(x, y))),output='realtime'));
\[u \left ( x,y \right ) =\tan \left ( x/2+1/2\,{\it \_F1} \left ( -5/2\,x+y \right ) \right ) \]
Hand solution
Solve for \(u\left ( x,y\right ) \) in \(2u_{x}+5u_{y}=u^{2}+1.\) Using the Lagrange-charpit method, the characteristic equations are\[ \frac {dx}{2}=\frac {dy}{5}=\frac {du}{u^{2}+1}\] From the first pair of equation we obtain\begin {align*} 5dx & =2dy\\ 5x & =2y+C_{1}\\ C_{1} & =5x-2y \end {align*}
Now we can pick the pair \(\frac {dy}{5}=\frac {du}{u^{2}+1}\) or \(\frac {dx}{2}=\frac {du}{u^{2}+1}\) to solve for \(u\). It does not matter which. Using\[ \frac {dx}{2}=\frac {du}{u^{2}+1}\] Integrating gives\begin {align*} \frac {1}{2}x & =\arctan \left ( u\right ) +C_{2}\\ C_{2} & =\frac {1}{2}x-\arctan \left ( u\right ) \end {align*}
\(C_{1}\) and \(C_{2}\) are always related by \(C_{2}=F\left ( C_{1}\right ) \) where \(F\) is arbitrary function. Hence\begin {align*} \frac {1}{2}x-\arctan \left ( u\right ) & =F\left ( 5x-2y\right ) \\ \arctan \left ( u\right ) & =\frac {1}{2}x-F\left ( 5x-2y\right ) \\ u\left ( x,y\right ) & =\tan \left ( \frac {1}{2}x-F\left ( 5x-2y\right ) \right ) \end {align*}
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