Legendre Differential Equation Transformation by Substitution

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Homework Statement

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).

Homework Equations

Legendre's Equation:

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

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.

 

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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.