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When do I use y=Acoshx+Bsinhx?

  1. Dec 5, 2014 #1
    When I have:
    y''-yu^2=0

    Why can't I just solve it using y=Ae^ux+Be^-ux ?
    Is there any specific reason for me to use y=Acoshx + Bsinhx?

    Also, for questions such as ut=uxx-u with BC u(0,t)=0 and u(L,t)=u1
    When I am setting up u=v+w do I have to set V as Acoshx+Bsinhx?
    I saw the solutions do that, but why can't I set V as u1x/L in this case?
     
  2. jcsd
  3. Dec 5, 2014 #2
    When I am separating variables, my understanding is that I can use either +lambda^2 or -lambda^2 depending on how I want my Xn or Tn to look like based on what boundary conditions I have. Is this line of thinking correct?

    For example:
    u=XT
    X''/X=T''/T=+/-lambda^2


    Sorry for so many questions at once.
     
    Last edited: Dec 5, 2014
  4. Dec 5, 2014 #3

    LCKurtz

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    If you have ##X'' - \mu^2 X=0## you may use either ##X = Ae^{\mu x} + Be^{-\mu x}## or ##X= A\cosh (\mu x) + B\sinh(\mu x)##. The advantage of one over the other is in solving for the constants with the boundary conditions. For example if you have ##X(0) = c## and ##X'(0)= d##, in the first case you would have$$X(0) = Ae^0 + Be^0 = A +B = c$$and in the second case you would have$$
    X(0)= A\cosh(0) + B\sinh(0) = A\cdot 1 + B\cdot 0 = A = c$$So you already know ##A## and the second boundary condition will be easy to solve for ##B##. In the exponential case you have two linked equations in two unknowns and it will be slightly more work to solve for the constants.
     
  5. Dec 5, 2014 #4
    I see, thanks!!
     
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