Results 1 to 3 of 3

Math Help - Y''/y = (-λ^2 - 1)

  1. #1
    Junior Member
    Joined
    Mar 2009
    Posts
    31

    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:

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

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

    Used separation of variables:

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

    And I got:

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

    And I don't know how to solve for Y

    Here's my equation:

    Y'' = (-\lambda^2 - 1)Y
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Member
    Joined
    Jan 2009
    Posts
    145
    Notice that  \lambda is just an eigenvalue, so this is really just a second order linear ODE. So, treating  (- \lambda^2 -1) as a single coefficient, we get that

     Y(y) = C*e^{y  \sqrt{-\lambda^2 -1}} + D*e^{-y\sqrt{-\lambda^2 -1}}
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Mar 2009
    Posts
    31
    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:

    Y''(y) = ( -\lambda^2 - 1)Y(y)
    Y''(y) = -( \lambda^2 + 1)Y(y)
    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
    Follow Math Help Forum on Facebook and Google+

Search Tags


/mathhelpforum @mathhelpforum