2.1.53 \(x u_x+(x+y)u_y=u+1\) with \(u(x,0)=x^2\) Problem 3.5(i) Lokenath Debnath

problem number 53

Added June 3, 2019.

Problem 3.5(i) nonlinear pde’s by Lokenath Debnath, 3rd edition.

Solve for \(u(x,y)\) \[ x u_x+(x+y)u_y=u+1 \] With \(u(x,0)=x^2\).

Mathematica

ClearAll["Global`*"]; 
pde =  x*D[u[x, y], x] +(x+y)*D[u[x, y], y]== u[x,y]+1; 
ic  = u[x,0]==x^2; 
sol =  AbsoluteTiming[TimeConstrained[DSolve[{pde,ic} ,u[x, y], {x, y}], 60*10]];
 

\[\left \{\left \{u(x,y)\to x^2 e^{-\frac {y}{x}}+e^{\frac {y}{x}}-1\right \}\right \}\]

Maple

restart; 
pde :=x*diff(u(x,y),x)+(x+y)*diff(u(x,y),y)= u(x,y)+1; 
ic  := u(x,0)=x^2; 
cpu_time := timelimit(60*10,CodeTools[Usage](assign('sol',pdsolve([pde,ic],u(x,y))),output='realtime'));
 

\[u \left ( x,y \right ) ={{\rm e}^{{\frac {x\ln \left ( x \right ) -y}{x}}}}x-1+x{{\rm e}^{{\frac {-x\ln \left ( x \right ) +y}{x}}}}\]

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