TIM B. answered • 08/31/19
The given equation [2xy'' + (1 − 2x2)y' − 4xy = 0]is put in the form y'' + P(x)y' + Q(x)y = 0 after division by 2x, which yields y'' + (1/2x − x)y' −2y = 0. Q(x) = -2 is everywhere analytic; that is, its Taylor series [∑(n = 0 to n = ∞) f(n)(x0)(x − x0)/n!] converges to f(x) = -2 in some neighborhood of any value chosen for x0.
P(x) or (1/2x − x) is not defined at x = 0, but xP(x) or (1/2 − x2) is everywhere defined and analytic. Then x0 = 0 is a regular singular point.
Note also that (1/2x)[2xy'' + (1 − 2x2)y' − 4xy = 0] can be written as y'' + (1 − 2x2)y'/2x − 4x(2x)/(2x)2 = 0 which is of the form y'' + R1(x)/P0(x) + R2(x)/(P0(x))2 = 0, with P0(x) being the coefficient function before y'' in the given equation.
The Taylor Expansion about x = 0 of R1(x) = 1 − 2x2 is given by (1 − 2(0)2)(x − 0)0/0! + -4(0)(x − 0)1/1! + -4(x − 0)2/2! + 0(x −0)3/3! + ... which gives 1 − 0 − 4x2/2 or 1 −2x2.
The Taylor Expansion about x = 0 of R2(x) = -4x(2x) or -8x2 is given by -8(0)2(x − 0)0/0! + -16(0)(x − 0)1/1! + -16(x − 0)2/2! which gives 0 − 0 − 16x2/2 or -8x2.
The Taylor Expansions for both R1(x) and R2(x) in this case do converge to their respective values of
1− 2x2 and -8x2, showing them to be analytic about x = 0. Thus, x = 0 is a regular singular point.
When x = 0 is a regular singular point of the given equation in the form of [P0(x)y'' +P1(x)y' +P2(x)y = 0], there always exists .a solution y = [xm∑(n = 0 to n = ∞)Anxn] or A0x(m + 0) + A1x(m + 1) + A2x(m + 2) + ... +
A(n - 1)x(m + n − 1) + Anx(m + n) +A(n + 1)x(m + n + 1) + ... Here A0 ≠ 0 and m and the A coefficients are determined so that y satisfies [P0(x)y'' +P1(x)y' +P2(x)y = 0].
With y expanded as shown above, also expand y' as mA0x(m − 1) + (m + 1)A1x(m) + (m + 2)A2x(m + 1) + ... +
(m + n − 1)A(n - 1)x(m + n − 2) + (m + n)Anx(m + n − 1) +(m + n + 1)A(n + 1)x(m + n) + ...
Then expand y'' as (m − 1)mA0x(m − 2) + m(m + 1)A1x(m −1) + (m + 1)(m + 2)A2x(m) + ... +
(m + n − 2)(m + n − 1)A(n - 1)x(m + n − 3) + (m + n − 1)(m + n)Anx(m + n − 2) +
(m + n)(m + n + 1)A(n + 1)x(m + n − 1) + ...
The expansions above for y, y', and y'' are placed into the given equation 2xy'' + (1 − 2x2)y' −4xy = 0. A tremendous amount of Algebraic Reorganization yields
m(2m − 1)A0x(m − 1) + (2m + 1)(m + 1)A1xm + [-2(m + 2)A0 + (2m + 3)(m + 2)A2]x(m + 1) +
-2(m + 3)A1x(m + 2) + -2(m + 4)A2x(m + 3) + [2(m + n)2 − 5(m + n) + 3]A(n − 1)x(m + n − 2) +
[2(m + n)2− m − 2(n − 1)]Anx(m + n − 1) + [-2(m + n + 1)A(n − 1) + (2(m + n)2 +3(m + n) + 1)A(n + 1)]x(m + n) +
-2(m + n + 2)Anx(m + n + 1) + -2(m + n + 3)A(n + 1) x(m + n + 2) + ... = 0.
A0 cannot equal 0; set A0 = 1 for convenience. Each term in this last expansion must be equated to 0 and solved for Ai and m. The first term m(2m − 1)A0x(m − 1) = 0 for m = 0 or m = 1/2. The second term
(2m + 1)(m + 1)A1xm = 0 only if A1 is set equal to 0. The remaining Ai terms are found by transforming the 8th term [-2(m + n + 1)A(n − 1) + (2(m + n)2 +3(m + n) + 1)A(n + 1)]x(m + n) into the Recursion Formula
A(n + 1) = 2A(n −1)/(2m + 2n +1) where n ≥ 1.
For {m, A0, A1} equal to {0, 1, 0}, the Recursion Formula gives values for A2 through A8 of 2/3, 0, 4/21, 0, 8/231, 0, & 16/3465. Placing these values into y = [xm∑(n = 0 to n = ∞)Anxn] or A0x(m + 0) + A1x(m + 1) + A2x(m + 2) + ... + A(n - 1)x(m + n − 1) + Anx(m + n) +A(n + 1)x(m + n + 1) + ... will give a result of y1 = x0[1 + 2x2/3 + 4x4/21 + 8x6/231 + 16x8/3465 + ...].
For {m, A0, A1} equal to {1/2, 1, 0}, the Recursion Formula gives values for A2 through A8 of 1/2, 0, 1/8, 0, 1/48, 0, & 1/384. Placing these values into y = [xm∑(n = 0 to n = ∞)Anxn] or A0x(m + 0) + A1x(m + 1) + A2x(m + 2) + ... + A(n - 1)x(m + n − 1) + Anx(m + n) +A(n + 1)x(m + n + 1) + ... will give a result of y2 = x1/2[1 + x2/2 + x4/8 + x6/48 + x8/384 + ...].
The complete solution is written as
y = Ay1 + By2 or Ax0[1 + 2x2/3 + 4x4/21 + 8x6/231 + 16x8/3465 + ...] + Bx1/2[1 + x2/2 + x4/8 + x6/48 + x8/384 + ...].
