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Thread: Y''/y = (-λ^2 - 1)

  1. #1
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    Y''/y = (-λ^2 - 1)

    Hello! I was trying to solve a Laplace's Equation and I got stuck on this step.

    Here's the original question:

    $\displaystyle u_{xx}+u_{yy}+u=0$
    $\displaystyle 0<x<1$
    $\displaystyle 0<y<1$
    $\displaystyle u(0,y) = 0 $
    $\displaystyle u(1,y) = 0$
    $\displaystyle u(x,0) = 0$
    $\displaystyle u_y(x,1) = sin(3\pi x)$

    So I assumed $\displaystyle u(x,y) = X(x)Y(y)$

    Used separation of variables:

    $\displaystyle \frac{Y''}{Y} + 1 = \frac{X''}{X} = -\lambda^2$

    And I got:

    $\displaystyle X(x) = Acos\lambda x + Bsin\lambda x$

    And I don't know how to solve for Y

    Here's my equation:

    $\displaystyle Y'' = (-\lambda^2 - 1)Y$
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  2. #2
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    Notice that $\displaystyle \lambda$ is just an eigenvalue, so this is really just a second order linear ODE. So, treating $\displaystyle (- \lambda^2 -1) $ as a single coefficient, we get that

    $\displaystyle Y(y) = C*e^{y \sqrt{-\lambda^2 -1}} + D*e^{-y\sqrt{-\lambda^2 -1}} $
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  3. #3
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    I thought of that too but it looked really messy to me so I didn't take it as an answer.

    I think I could also do this:

    $\displaystyle Y''(y) = ( -\lambda^2 - 1)Y(y)$
    $\displaystyle Y''(y) = -( \lambda^2 + 1)Y(y)$
    $\displaystyle Y(y) = Ccos(y*\sqrt{\lambda^2+1}) + Dsin(y*\sqrt{\lambda^2+1})$

    Because I think this form works better with the initial conditions. =P
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