Question

(1-x^2)d^2y/dx^2 -2x dy/dx+n(n+1)y=0

Answer 1

$=$ $(1-x^{2})\dfrac{d^{2}y}{dx^{2}}-2x\dfrac{dy}{dx}+n(n+1)y=0$ .........................................1)

This equation is called LEGENDRE'S polynomial equation .

substituting -

$=y=a_0x^m+a_1x^{m+1}+a_2x^{m+2}+.....(a_0\neq0),$

Now , series first take of the form -

$=a_0m(m-1)x^{m-2}+a_1(m+1)mx^{m-1}+.......[a_{r+2}(m+r+2)(m+r+1)-{(m+r)(m+r+1)-n(n+1)}a_r]x^{m+r}+....=0$

Equating to zero the co-efficient of the lowest power of $x$ , i.e. of $x^{m-2},$ we get -

$=a_0m(m-1)=0, m=0,1$ $[\because$ $a_0\neq0]$

Equating to zero the coefficient of $x^{m-1}\ and \ x^{m-r}\ , we$ get $a_1(m+1)m=0.....................2)$

$a_{r+2}(m+r+2)(m+r+1)-{(m+r)(m+r+1)-n(n+1)}a_r=0...............................(3)$

When m=0, (2) is satisfied and therefore ,$a_1\neq0.$ Then (3) gives , taking r=0,1,2........in turn ,

$a_2=\dfrac{-n(n+1)}{2}a_0$ $, \ a_3=\dfrac{(n-1)(n+2)}{3!}a_1$

$a_4=\dfrac{-(n-2)(n+3)}{4.3}a_2=\dfrac{n(n-2)(n+1)(n+3)}{4!}a_0$

$a_5=\dfrac{-(n-3)(n+4)}{5.4}a_3=\dfrac{(n-1)(n-3)(n+2)(n+4)}{5!}a_1, etc$

Hence for m=0 , there are two independent solutions of (1),

$y_1=a_0({1-\dfrac{n(n+1)}{2!}}x^{2}+\dfrac{(n-2)n(n+1)(n+3)}{4!}x^{4}-............4)$

$y_2=a_1(x-\dfrac{(n-1)(n+2)}{3!}x^{3}+\dfrac{(n-3)(n-1)(n+2)(n+4)}{5!}x^{5}-.....)$ ......................5)

When m=1 , (2) shows that $a_1=0.$ Therefore (3) gives ,

$a_3=a_5=a_7=0$

$and$ $a_2=\dfrac{-(n-1)(n+2)}{3!}a_0$

$a_4=\dfrac{(n-3)(n-1)(n+2)(n+4)}{5!}a_0, etc$

Thus for m=1 , we get the solution of (5) again , Hence $y=y_1+y_2$ is a general solution of (1).

So our general solution of given differential equation is $y_1+y_2$