# How did my professor get from here to here? DiffEq

1. Feb 28, 2016

### grandpa2390

1. The problem statement, all variables and given/known data
X(x)=(Ae^kx+Be^-kx)
Y(y)=(Csin(ky)+D(cos(ky))
V(x,y)=(Ae^kx+Be^-kx)(Csin(ky)+D(cos(ky))

2. Relevant equations
separation of variables

3. The attempt at a solution
so our boundary condition says that as x->infinity , V=0
this is only possible if A=0

so A=0 and Be^-kx= 0
so X= Be^-kx
and
V(x,y)=(e^-kx)(Csin(ky)+D(cos(ky))

why is there an e^-kx on the front. Shouldn't there also be a B?

2. Feb 28, 2016

### TSny

B can be "absorbed" by the arbitrary constants C and D.

3. Feb 28, 2016

### grandpa2390

ah! you're right. distribute the B into the the parenthesis. I knew it was something so simple otherwise he wouldn't have glossed over it. lol.
Thanks!

just one last question. so B^e^-kx we keep because the constant is not equal to zero?
because the next part we solve for Y, and D ends up being 0 but we keep Csin(ky) which equals 0 because sin(0)=0. so we keep that one but eliminate the part where D=0.

so we keep the part where the constant is not equal to zero?

4. Feb 28, 2016

### TSny

Yes. If a boundary condition forces a constant factor of a term to be zero, then you drop that term. Otherwise, keep the term.

5. Feb 28, 2016

### LCKurtz

Apparently you are solving a homogeneous PDE with boundary conditions. The identically zero function is always a solution, but you want non-trivial solutions. So you hope not to have to take all the constants be zero. So you probably have something like $Ce^{-kx}\sin(ky)$ and you get your last boundary condition to be zero by choosing particular values of $k$ so the sine term can be zero without making $C=0$. These values of $k$ determine the eigenvalues of the problem and allow nontrivial solutions.