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How did my professor get from here to here? DiffEq

  • #1
464
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Homework Statement


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))


Homework Equations


separation of variables

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?
 

Answers and Replies

  • #2
TSny
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B can be "absorbed" by the arbitrary constants C and D.
 
  • #3
464
8
B can be "absorbed" by the arbitrary constants C and D.
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
TSny
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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?
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
LCKurtz
Science Advisor
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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.
 

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