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]];
 

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

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'));
 

$$u(x,y)={\mathrm{e}}^{\frac{x\ln \left(x\right)-y}{x}}x-1+x{\mathrm{e}}^{\frac{-x\ln \left(x\right)+y}{x}}$$

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