My reason is as follows - If I plug y = a into (Csin(ky)+Dcos(ky)), I can solve for C. Then if I plug in y = -a and my value for C, I can get -Dcos(ak) = Dcos(ak) (because sin is odd function and cos is even function) I'm not sure if my reasoning is correct so feedback is appreciated. I'm not...
I'm trying to solve laplaces equation to find a solution for the potential in a pipe with the given boundary conditions:
at x=b, V=V_0
at x= -b, V = -V_0
at y=a, V=0
at y=-a, V=0
(Assume this configuration is centered on the origin, pipe as dimensions -b<=x<=b...