Added June 3, 2019.
Problem 3.16(a) nonlinear pde’s by Lokenath Debnath, 3rd edition.
Solve for \(u(x,y)\) \[ u_x+u u_y=1 \] With \(u(0,y)=a y\).
Mathematica ✓
ClearAll["Global`*"]; pde = D[u[x, y], x] +u[x,y]*D[u[x, y], y]== 1; ic=u[0,y]==a*y; sol = AbsoluteTiming[TimeConstrained[DSolve[{pde,ic} ,u[x, y], {x, y}], 60*10]];
\[\left \{\left \{u(x,y)\to \frac {a x^2+2 a y+2 x}{2 a x+2}\right \}\right \}\]
Maple ✓
restart; pde := diff(u(x,y),x) + u(x,y)*diff(u(x,y),y)=1; ic := u(0,y)=a*y; cpu_time := timelimit(60*10,CodeTools[Usage](assign('sol',pdsolve([pde,ic],u(x,y)) ),output='realtime'));
\[u \left ( x,y \right ) ={\frac { \left ( {x}^{2}+2\,y \right ) a+2\,x}{2\,ax+2}}\]
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