Legendre Differential Equation Transformation by Substitution

1. Sep 28, 2011

schamp

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. Sep 28, 2011

schamp

So I went back and checked my differentiation on the second derivative. Turns out I need the product rule:

d/d(theta)[dy/d(theta)]=d/dtheta(dy/dx*dx/d(theta))

By the Product Rule:

uv' - vu'

where

u = dy/dx
du = d2y/dx2

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

Therefore:

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.