2.1.64 \(y^2 u_x+(x y) u_y=x\) with \(u(x,1)=x^2\). Problem 3.14(f) Lokenath Debnath

problem number 64

Added June 3, 2019.

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

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

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

With \(u(x,1)=x^2\).

Mathematica ✓

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

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

Maple ✓

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

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

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