find a solution of (a^2-x^2)y''-2xy'+n(n+1)y=0, a not 0 by reduction to the legendre equation???
Since Legendre's differential equation starts "[itex](1- x^2)d^2y/dx^2[/itex]" 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, [itex]d^2y/dx^2= 1/a^2 d^2y/dt^2[/itex], and [itex]a^2- x^2= a^2(1- (x/a)^2)= a^2(1- t^2)[/itex] so that [itex](a^2- x^2)d^2y/dx^2= (1- t^2)d^2y/dt^2[/itex].
2xy'= 2(at)((1/a)dy/dt)= 2t dy/dt.
if dx/dt=1/a then d^2 t/dx^2=0 how will it will be 1/a^2??
I have no clue what you are talking about. There is no "[itex]d^2t/dx^2[/itex]" in what I wrote.
I did say that [itex]d^2y/dx^2= (1/a^2) d^2y/dt^2[/itex]. That is true because
[tex]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[/tex]
Separate names with a comma.