2.1.62 \((y+u)u_x+y u_y=x-y\) with \(u(x,1)=1+x\). Problem 3.11 Lokenath Debnath

problem number 62

Added June 3, 2019.

Problem 3.11 nonlinear pde’s by Lokenath Debnath, 3rd edition.

Solve for \(u(x,y)\) \[ (y+u)u_x+y u_y=x-y \] With \(u(x,1)=1+x\).

Mathematica ✗

ClearAll["Global`*"]; 
pde =  (y+u[x,y])*D[u[x, y], x] +y*D[u[x, y], y]== x-y; 
 ic=u[x,1]==1+x; 
sol =  AbsoluteTiming[TimeConstrained[DSolve[{pde,ic} ,u[x, y], {x, y}], 60*10]];
 

Failed

Maple ✗

restart; 
pde :=(y+u(x,y))*diff(u(x,y),x) + y*diff(u(x,y),y)= x-y; 
ic  := u(x,1)=1+x; 
cpu_time := timelimit(60*10,CodeTools[Usage](assign('sol',pdsolve([pde,ic],u(x,y))),output='realtime'));
 

sol=()

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