2.1.60 \((u_x+u_y)^2-u^2=0\). Problem 3.10 Lokenath Debnath

problem number 60

Added June 3, 2019.

Problem 3.10 nonlinear pde’s by Lokenath Debnath, 3rd edition.

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

Mathematica ✓

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

\begin {align*} & \left \{u(x,y)\to e^{-x} c_1(y-x)\right \}\\& \left \{u(x,y)\to e^x c_1(y-x)\right \}\\ \end {align*}

Maple ✓

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

\[u \left (x , y\right ) = c_{1} {\mathrm e}^{\frac {y \textit {\_c}_{2}+x}{\textit {\_c}_{2}+1}}\]

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