How to solve these type of differential equations?

[vamsikilaru]

find a solution of (a^2-x^2)y''-2xy'+n(n+1)y=0, a not 0 by reduction to the legendre equation?

 

[HallsofIvy]

Since Legendre's differential equation starts "(1- x^2)d^2y/dx^2" an obvious first step would be to define a new variable, t= x/a. Then dy/dx= (dy/dt)(dt/dx)= 1/a dy/dt, d^2y/dx^2= 1/a^2 d^2y/dt^2, and a^2- x^2= a^2(1- (x/a)^2)= a^2(1- t^2) so that (a^2- x^2)d^2y/dx^2= (1- t^2)d^2y/dt^2.

2xy'= 2(at)((1/a)dy/dt)= 2t dy/dt.

 

[vamsikilaru]

if dx/dt=1/a then d^2 t/dx^2=0 how will it will be 1/a^2??

 

[HallsofIvy]

I have no clue what you are talking about. There is no "d^2t/dx^2" in what I wrote.

I did say that d^2y/dx^2= (1/a^2) d^2y/dt^2. That is true because
d^2y/dx^2= d/dx(dy/dx)= d/dx((1/a)dy/dt)= (1/a) d/dx( dy/dt)= (1/a)((1/a)d/dt)(dy/dt)= (1/a^2) d^2/dt^2