# How to solve these type of differential equations?

1. Oct 7, 2012

### 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???

2. Oct 7, 2012

### HallsofIvy

Staff Emeritus
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.

3. Oct 7, 2012

### vamsikilaru

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

4. Oct 10, 2012

### HallsofIvy

Staff Emeritus
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
$$d^2y/dx^2= d/dx(dy/dx)= d/dx((1/a)dy/dt)= (1/a) d/dx( dy/dt)= (1/a)((1/a)d/dt)(dy/dt)= (1/a^2) d^2/dt^2$$