2.1.62 \(2 x u_x+(x+1) u_y=y\) with \(u(1,y)=2 y\). Problem 3.14(d) Lokenath Debnath

problem number 62

Added June 3, 2019.

Problem 3.14(d) nonlinear pde’s by Lokenath Debnath, 3rd edition.

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

\[ 2 x u_x+(x+1) u_y=y \]

With \(u(1,y)=2 y\) with \(x>0\)

Mathematica ✓

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

\[\left \{\left \{u(x,y)\to \frac {1}{8} \left (-2 (x-2 y+4) \log (x)-6 x-\log ^2(x)+16 y+6\right )\right \}\right \}\]

Maple ✓

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

\[u \left (x , y\right ) = -\frac {\ln \left (x \right )^{2}}{8}+\frac {\left (4 y -2 x -8\right ) \ln \left (x \right )}{8}+2 y -\frac {3 x}{4}+\frac {3}{4}\]

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