find one solution of x2yxyx21y0 near x0SolutionDividing both

find one solution of x²y\'\'+xy\'+(x²-1)y=0 near x=0

Solution

Dividing both sides by x² yields

                                    y\" + P(x)y\' + Q(x)y = 0,

                        with     P(x) = (1/x) and Q(x) = [(x² – 1) / x²] . Thus, x = 0 is a regular

                        singular point. Substituting

                                    y = x_n=0 a_nxⁿ                                                                                 (b)

                        and its derivatives into equation (a), we find

                        0 = a₀(² – 1)x + a₁(( + 1)² – 1)x¹⁺ + _n=0 [a_n(( + n)² – 1) + a_n–2] xⁿ⁺ .

                                    Equating all the coefficients of the powers of x to zero, we find

                                    (² – 1)a₀ = 0,                                                                                     (c)

                          (( + 1)² – 1)a₁ = 0,                                                                                      (d)

                        and, for n 2,

                                    a_n(( + n)² – 1) + a_n–2 = 0,

                        or,        a_n = [(– 1) / {( + n)² – 1}] a_n–2 ,         n 2.                                      (e)

                                    Equation (c) gives us the indicial equation

                                    ² – 1 = 0

                        with roots ₁ = – 1 and ₂ = + 1. Since ₁ – ₂ = – 2 is an integer, we

                        consider the larger root ₂ = 1. Letting = 1 in (d) and (e) gives

                                    a₁ = 0                                                                                                  (f)

                        and for n 2

                                    a_n = [(– 1) / {n(n + 2)}] a_n–2 .                                                              (g)

                                    From (f) and (g) we immediately find

                                    0 = a₁ = a₃ = a₅ ... = a_2n+1 = ...

                        and     a_2n = [(– 1) / {2n(2n + 2)}] a_2n–2 ,

                        or         a_2n = [(– 1) / {4n(n + 1)}] a_2n–2 ,                                                         (h)

                        for n 1.

                                    Applying (h) repeatedly, we find for n 1

                                    a_2n = [{(– 1)ⁿ} / {4ⁿn! (n + 1)!}] a₀ .

                        Thus   y₁(x) = a₀x _n=0 [{(– 1)ⁿ} / {4ⁿ n! (n + 1)!}] x²ⁿ

                        gives one family of solutions to (a).

                                    To find the second family of solutions to (a) we use equations (d)

                        and (e) repeatedly to obtain a_n in terms of : (d) gives a₁ = 0, hence,

                        by (e),

                                    0 = a₁ = a₃ = a₅ = … = a_2n+1 = …     .

                                    Equation (e) yields:

                                    a₂ = a₂() = [(– 1) / {( + 1) ( + 3)}] a₀ ,

                                    a₄ = a₄() = [{(– 1)²} / {( + 1) ( + 3)² ( + 5)}] a₀ ,

                                    a₆ = a₆() = [{(– 1)³} / {( + 1) ( + 3)² ( + 5)² ( + 7)}] a₀, etc.

                        Thus, since ( – ₁) = ( – (– 1)) = ( + 1), we consider

                             ( + 1)y(x, ) = ( + 1) [_n=0 a_n ()xⁿ⁺] ,

                        or ( + 1)y(x, )

                                    = [( + 1)a₀x – [1 / ( + 3)] a₀x²⁺ + [1 / {( + 3)² ( + 5)}]a₀x⁴⁺

                                                                        – [1 / {( + 3)² ( + 5)² ( + 7)}] a₀x⁶⁺ + …] .

                                    Differentiating both sides with respect to ,

                        ( / ) [( + 1)y(x, )]

                        = [ a₀x + a₀( + 1)x(log x) + [1 / {( + 3)²}]a₀x²⁺ – [(a₀x²⁺) / ( + 3)](log x)

                           – [(2 a₀x⁴⁺) / {( + 3)³( + 5)}] – [(a₀x⁴⁺) / {( + 3)²( + 5)²}]

                           + [{a₀x⁴⁺(log x)} / {( + 3)²( + 5)}] + [(2a₀x⁶⁺) / {( + 3)³( + 5)²( + 7)}]

                           + [(2a₀x⁶⁺) / {( + 3)²( + 5)³( + 7)}] + [(a₀x⁶⁺) / {( + 3)²( + 5)²( + 7)²}]

                                                                        – [{a₀x⁶⁺ (log x)} / {( + 3)² ( + 5)² ( + 7)}] ].

                                    The right hand side of the last equation was obtained with the aid of

                        Leibniz’s rule for the derivative of a product:

                        (g₁(x)g₂(x) … g_n(x))\'

                                    = g₁\'(x)g₂(x) … g_n(x) + g₁(x)g₂\'(x) … g_n(x) + … + g₁(x)g₂(x) … g_n\'(x) .

                                    We then obtain, upon substituting = – 1,

                           y₂(x) = ( / ) [( + 1)y(x, )] |_=–1

                                    = a₀[ x^–1 + 0 + (1/4) x – (1/2) x (log x) – (1 / 16) x³ – (x³ / 64)

                                                + [{x³ (log x)} / 16] + (x⁵ / 384) + (x⁵ / 768) + (x⁵ / 304)

                                                                                                            – [{x⁵ (log x)} / 384] + …],

                        or y₂(x) = a₀[x^–1 + (1/4) x – (5 / 64) x³ + (10 / 2304) x⁵ – …]

                                                                        + a₀ (– log x) [(x/2) – (x³ / 16) + (x⁵ / 384) – …] ,

                        or y₂(x) = [a₀x^–1 {1 + (1/4) x² – (5 / 64) x⁴ + (10 / 2304) x⁶ – …}

                                                                                                                        – (1/2)(log x) y₁(x)].

                                    The general solution to equation (a) is, therefore,

                                    y = b₁y₁(x) + b₂y₂(x),

                        with b₁ and b₂ arbitrary constants.

                                    Note that the expression

                                    b₁y₁(x) + b₂y₂(x)

                        would appear to contain three arbitrary constants b₁, b₂, and a₀:

                        b₁y₁ + b₂y₂ = b₁a₀(a series in x) + b₂a₀ (a sum of two series in x)

                                    But b₁a₀ and b₂a₀ both run over the real numbers. Thus b₁a₀ and

                        b₂a₀ are, therefore, only two arbitrary constants.