The Legendre function for n = 0 is y 1 ( x ) = P 0 ( x ) = 1 and y 2 ( x ) = 1 2 ln 1 + x 1 − x . Result used: Equation (6) is, y 1 ( x ) = 1 − n ( n + 1 ) 2 ! x 2 + ( n − 2 ) n ( n + 1 ) ( n + 3 ) 4 ! x 4 − ⋯ Equation (7) is, y 2 ( x ) = x − ( n − 1 ) ( n + 2 ) 3 ! x 3 + ( n − 3 ) ( n − 1 ) ( n + 2 ) ( n + 4 ) 5 ! x 5 − ⋯ Equation ( 1 1 ′ ) is, P 0 ( x ) = 1 , P 1 ( x ) = x , P 2 ( x ) = 1 2 ( 3 x 2 − 1 ) , P 3 ( x ) = 1 2 ( 5 x 3 − 3 x ) P 4 ( x ) = 1 8 ( 35 x 4 − 30 x 2 + 3 ) , P 5 ( x ) = 1 8 ( 63 x 5 − 70 x 3 + 15 x ) Calculation: First substitute n = 0 in equation (7) to obtain the Legendre function y 2 ( x ) for n = 0 as shown below. y 2 ( x ) = x − ( n − 1 ) ( n + 2 ) 3 ! x 3 + ( n − 3 ) ( n − 1 ) ( n + 2 ) ( n + 4 ) 5 ! x 5 − ⋯ = x − ( 0 − 1 ) ( 0 + 2 ) 3 ! x 3 + ( 0 − 3 ) ( 0 − 1 ) ( 0 + 2 ) ( 0 + 4 ) 5 ! x 5 − ⋯ = x − ( − 1 ) ( 2 ) 3 ! x 3 + ( − 3 ) ( − 1 ) ( 2 ) ( 4 ) 5 ! x 5 − ⋯ = x + 1 3 x 3 + 1 5 x 5 + ⋯ Obtain the expansion of 1 2 ln ( 1 + x ) ( 1 − x ) as follows. 1 2 ln ( 1 + x ) ( 1 − x ) = 1 2 ( ln ( 1 + x ) − ln ( 1 − x ) ) = 1 2 ( ( x − x 2 2 + x 3 3 − x 4 4 ⋯ ) − ( − x − x 2 2 − x 3 3 − x 4 4 ⋯ ) ) = 1 2 ( 2 x + 2 x 3 3 + 2 x 5 5 + ⋯ ) = x + x 3 3 + x 5 5 + ⋯ Therefore, the Legendre function y 2 ( x ) for n = 0 becomes, y 2 ( x ) = ( x + 1 3 x 3 + 1 5 x 5 + ⋯ ) = ( 1 2 ln ( 1 + x ) ( 1 − x ) ) = 1 2 ln 1 + x 1 − x Hence the Legendre function y 2 ( x ) for n = 0 is y 2 ( x ) = 1 2 ln 1 + x 1 − x . Now substitute n = 0 in equation (6) to obtain the Legendre function y 1 ( x ) for n = 0 as shown below. y 1 ( x ) = { 1 − n ( n + 1 ) 2 ! x 2 + ( n − 2 ) n ( n + 1 ) ( n + 3 ) 4 ! x 4 + ⋯ } = { 1 − ( 0 ) ( 0 + 1 ) 2 ! x 2 + ( 0 ) ( 0 − 2 ) ( 0 + 1 ) ( 0 + 3 ) 4 ! x 4 + ⋯ } = 1 − ( 0 ) x 2 + 0 + ⋯ = 1 Hence the Legendre function y 1 ( x ) for n = 0 is y 1 ( x ) = 1 . Consider the Legendre differential equation, ( 1 − x 2 ) y ″ − 2 x y ′ + n ( n + 1 ) y = 0 . Set n = 0 and z = y ′ in ( 1 − x 2 ) y ″ − 2 x y ′ + n ( n + 1 ) y = 0 and solve by separation of variables as shown below. ( 1 − x 2 ) z ′ − 2 x z + ( 0 ) ( 0 + 1 ) y = 0 ( 1 − x 2 ) z ′ − 2 x z = 0 z ′ = 2 x z ( 1 − x 2 ) d z d x = 2 x z ( 1 − x 2 ) On further simplifications, d z z = 2 x ( 1 − x 2 ) d x ∫ d z z = ∫ 2 x ( 1 − x 2 ) d x ln z = − ln ( 1 − x 2 ) + ln c z = c 1 − x 2 Now substitute back z = y ′ and solve for y as follows. y ′ = c 1 − x 2 d y d x = c 1 − x 2 d y = c 1 − x 2 d x ∫ d y = ∫ c 1 − x 2 d x It is further simplified as shown below. y = c ( 1 2 ln ( 1 + x 1 − x ) ) + d Therefore, the obtained solution is verified.

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