Solving U_x+U_y=1 with BC U(y,y/2)=y

[xdrgnh]

Homework Statement

U_x+U_y=1 with boundary condition U(y,y/2)=y

Homework Equations

The Attempt at a Solution

Well first I solved the homogeneous solution and got ax-ay where a is just a constant. Then for the non homogeneous solution I got ax+(1-a)y. After adding them both together and pluggin in the boundary condition I got U=x+(0)y? Is that right?

 

[LCKurtz]

Homework Statement

U_x+U_y=1 with boundary condition U(y,y/2)=y

Homework Equations

The Attempt at a Solution

Well first I solved the homogeneous solution and got ax-ay where a is just a constant. Then for the non homogeneous solution I got ax+(1-a)y. After adding them both together and pluggin in the boundary condition I got U=x+(0)y? Is that right?

It's easy enough to check it yourself. Does your solution $U(x,y)=x$ satisfy $U_x+U_y = 1$ and $U(y,\frac y 2) = y$?

 

[xdrgnh]

It does. But I'm suspicious whenever a solution to a PDE comes out that nice.

 

[LCKurtz]

Hey, if you have the solution you have the solution. There's nothing left worry about. Good going.