Added June 3, 2019.
Problem 3.14(d) nonlinear pde’s by Lokenath Debnath, 3rd edition.
Solve for \(u(x,y)\) \[ 2 x u_x+(x+1) u_y=y \] With \(u(1,y)=2 y\) with \(x>0\)
Mathematica ✓
ClearAll["Global`*"]; pde = 2*x*D[u[x, y], x] +(x+1)*D[u[x, y], y]== y; ic=u[1,y]==2*y; sol = AbsoluteTiming[TimeConstrained[DSolve[{pde,ic} ,u[x, y], {x, y}, Assumptions->x>0], 60*10]];
\[\left \{\left \{u(x,y)\to \frac {1}{8} \left (-2 (x-2 y+4) \log (x)-6 x-\log ^2(x)+16 y+6\right )\right \}\right \}\]
Maple ✓
restart; pde :=2*x*diff(u(x,y),x) + (x+1)*diff(u(x,y),y)= y; ic := u(1,y)=2*y; cpu_time := timelimit(60*10,CodeTools[Usage](assign('sol',pdsolve([pde,ic],u(x,y)) assuming x>0),output='realtime'));
\[u \left ( x,y \right ) =-{\frac { \left ( \ln \left ( x \right ) \right ) ^{2}}{8}}+{\frac { \left ( 4\,y-2\,x-8 \right ) \ln \left ( x \right ) }{8}}+2\,y-{\frac {3\,x}{4}}+{\frac {3}{4}}\]
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