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

problem number 51

Added June 3, 2019.

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

Solve for \(u(x,y)\)

\[ u u_x-u u_y=u^2 + (x+y)^2 \]

With \(u(x,0)=1\).

Mathematica ✓

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

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

Maple ✓

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

\[u \left (x , y\right ) = \sqrt {{\mathrm e}^{-2 y} x^{2}+2 \,{\mathrm e}^{-2 y} x y +{\mathrm e}^{-2 y} y^{2}-x^{2}-2 x y -y^{2}+{\mathrm e}^{-2 y}}\]

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