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Homework Help: Legendre Differential Equation Transformation by Substitution

  1. Sep 28, 2011 #1
    1. The problem statement, all variables and given/known data

    Show that the differential equation:

    sin(theta)y'' + cos(theta)y' + n(n+1)(sin(theta))y = 0

    can be transformed into Legendre's equation by means of the substitution x = cos(theta).

    2. Relevant equations

    Legendre's Equation:

    (1 - x^2)y'' - 2xy' + n(n+1)y = 0

    3. The attempt at a solution

    Using the Chain Rule for the first derivative:

    dy/d(theta) = (dy/dx)(dx/d(theta))

    x = cos(theta)
    dx/d(theta) = -sin(theta)

    dy/d(theta) = -sin(theta)dy/dx

    Using the Chain Rule for the second derivative:

    d2y/d(theta)2 = (d2y/dx2)(d2x/d(theta2))

    x = cos(theta)
    d2x/d(theta)2 = -cos(theta)

    d2y/d(theta)2 = d2y/dx2(-cos(theta))

    After substitution I get:

    -(sin(theta)*cos(theta))d2y/dx2 + -(sin(theta)*cos(theta))dy/dx + n(n+1)sin(theta)y = 0

    I've played around with this and tried various trigonometric identities to get it into a form that can be translated as the Legendre's Equation, but I still can't quite get it in the form of:

    (1-cos2(theta))d2y/dx2 - 2cos(theta)dy/dx + n(n+1)y = 0.
  2. jcsd
  3. Sep 28, 2011 #2
    So I went back and checked my differentiation on the second derivative. Turns out I need the product rule:


    By the Product Rule:

    uv' - vu'


    u = dy/dx
    du = d2y/dx2

    v = dx/d(theta)
    dv = d2x/d(theta)2


    d2y/d(theta)2 = [d2x/d(theta)2)*(dy/dx)] - [(dx/d(theta))*d2y/dx2]

    which equals:

    -cos(theta)dy/dx + sin(theta)d2y/dx2

    After fixing the Chain Rule of the second derivative, and substituting into the differential equation I have:

    (1-cos2(theta))d2y/dx2 - 2sin(theta)cos(theta)dy/dx + sin(theta)n(n+1)y = 0

    I guess my issue now would be figuring out how to get rid of the sin(theta) that is multiplied in the first derivative and the sin(theta) multiplied in the n(n+1)y term, to finally get it into the form of Legendre's Equation.
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