Question

value of P1(x) is

1)1
20
3)3
4)X​

Answer 1

Answer:

1)--1 will be the awnser

Answer 2

Answer:

The value of $P_1(x)=x$.

Step-by-step explanation:

The second order differential equation given as

$(1-x^{2} )\dfrac{d^{2} y}{dx^{2} } -2x\dfrac{dy}{dx } +n(n+1)y$ for $n > 0$, $|x| < 1$

is known as Legendre’s equation.

• Solutions of Legendre’s equation are called Legendre functions of order n.
• The general solution of Legendre’s equation is given as a function of sum of two Legendre functions i.e.,

       $y=AP_n(x)+BQ_n(x)$

• where

       $P_n(x)=\dfrac{1}{2^{n} n!} \dfrac{d^{} }{dx^{n} } (x^{2} -1)^{n}$

       is the Legendre function of the first kind and

• $Q_n(x)=\dfrac{1}{2} Q_n(x)ln\dfrac{1+x}{1-x }$

        is the Legendre function of the second kind.

• If n = 0, 1, 2, 3,... the $P_n(x)$ functions are called Legendre polynomials of order n.

Given to find the value of $P_1(x)$.

So, substituting n=1 in the formula of $P_n(x)$,

$P_1(x)=\dfrac{1}{2^{1} 1!} \dfrac{d^{} }{dx^{1} } (x^{2} -1)^{1}=\dfrac{1}{2} (2x)=x$

Therefore, the value of $P_1(x)=x$.