Question

Given:

$6x^{2}y^{''}+7x(1-x)y^{'}-2y = 0$

About a regular singular point x = 0,

I have solved for $p_{0} = \frac{7}{6}$ and $q_{0} = -\frac{1}{3}$ ,

I have also solved for $r_{1,2} = \frac{1}{2}, -\frac{2}{3}$

Therefore $F(r)=r(r-1) + \frac{7}{6}r-\frac{1}{3}=0$

a.) I need to find the first three non-zero terms of the two linearly independent solutions using the GENERAL RECURRENCE FORMULA: Where $a_{0}=1$

$F(r+n)a_{n} + \sum_{k=0}^{n-1}a_{k}[(r+k)p_{n-k}+q_{n-k}]=0, n\geq 1.$

NOT BY PLUGGING THE SERIES SOLUTION INTO THE DIFFERENTIAL EQUATION.

Please show ALL your steps when using the formula, every value you plug in and how to do it for the first 3 terms ($a_{1}, a_{2}, a_{3}$)

Answer 1

using frobenius method series solution of the given ode is obtained about REGULAR singular point x=0
Given O.D.E 6x^2 y^'' + 7x (1 - x) y^' -2y = 0 comparing from b_0 (x) (x -0)^2 y^'' + a_1 (x) (x -0)y^' + b_2 (x) y = 0 implies b_0 (x) = 6, b_1 (x) = 7 (1 - x), b_2 (x) = -2 clearly b_0 (0) = 6 notequalto 0 and b_0, b_1, b_2 are analytic at x_0 = 0 so x_0 = 0 is a regular singular point of the given O.D.E Let the solution is given by power series y = x^r sigma^infinity_n = 0 a_n (n + r) x^n + r -1 y^'' = sigma^infinity_n = 0 a_n (n + r) (n + r -1) x^n + r -2 implies x^2 y^'' = sigma^infinity_n = 0 a_n (n + r) (n + r -1) x^n + r 7x (1 - x) y^' = sigma^infinity_n = 0 7x (1 - x) a_n (n + r) x^n + r -1 = sigma^infinity_n = 0 7 a_n (n + r)