Verify equations (10.3) and (10.4).

(10.3) : (1-x²) u"-2 (m+1) xu'+[l(l+1) - m(m+1)] u=0

(10.4) : (1-x²) (u')" -2 [(m+1)+1] x(u')'+ [l(l+1) - (m+1)(m+2)]u'=0

An equation similar to Legendre’s equation is-

(1-x²) y"-2xy'+[l(l+1)-m²/1-x²] y=0 (∀ m²≤ l²)

The solution to this equation is of the form

P_l^m (x)=(1-x²)^m/2 d^m/dx^m (P_l(x))

These solutions are called Associated Legendre Functions.

Legendre equation is,

(1-x²) y"-2xy'+[l(l+1)-m²/1-x²] y=0 …… (1)

Let, y=(1-x²) ^m/2u.

y'=(1-x²)^m/2u-mx=(1-x²)^m/2-1u

y"=(1-x²)^m/2u'-2mx (1-x²)^m/2-1u' -m(1-x²)^m/2-1u+2mx² (m/2-1) (1-x²)^m/2-2u

Substituting the above values in equation (1):

[(1-x²)^m/2+1 u"-2mx (1-x²)^m/2 u'-m (1-x²)^m/2u+2mx² (m/2-1) (1-x²)^m/2-1 u -2x(1-x²)^m/2u'+2mx² (1-x²)^m/2+1u+ l(l+1)-m²/1-x²] (1-x²)^m/2u=0

Taking (1-x²)^m/2 common from the above equation-

(1-x²)u"-2mxu'-mu+[2mx²(m/2-1) u/(1-x²)]-2xu'+2mx²u/(1-x²)+[l(l+!)-m²/1-x²]u=0

(1-x²)u"- 2(m+1) xu'-mu+m²x²u/(1-x²) + [l(l+1)-m²/1-x²] u=0

(1-x²)u"- 2(m+1) xu'+u {l (l+1)- (m²x²-m²-m+mx²)/1-x²}=0

(1-x²)u"- 2(m+1) xu'+u {l (l+1)-m(m+1)}=0

Hence, the equation is verified.

(1-x²)u"- 2(m+1) xu'+u {l (l+1)-m(m+1)}=0

Now differentiate above derived equation:

(1-x²) u"-2xu'-2(m+1) {xu"+u'} + u' {l(l+1)-m(m+1)}=0

(1-x²) (u')"-2(m+2) x {u'+u' {l (l+1)-(m+1) (m+2)}} =0

Hence, the second equation is also verified.