1-x-d-2-y-dx-2-x-dy-dx-xy-1-1-x-x-1-has-power-series-solution-for-x-lt-1-

Question Number 119567 by abdul88 last updated on 25/Oct/20

(1 - x) \frac{d^{2} y}{{d x}^{2}} + x \frac{d y}{d x} - x y = \frac{1}{1 - x}, x \neq 1

h a s p o w e r s e r i e s s o l u t i o n f o r ∣ x ∣< 1

Answered by mathmax by abdo last updated on 26/Oct/20

e \Rightarrow {(1 - x)}^{2} y^{^{″}} + x (1 - x) y^{^{'}} - x (1 - x) y = 1 (x \neq 1) let y = \sum_{n = 0}^{\infty} a_{n} x^{n}

\Rightarrow y^{^{'}} = \sum_{n = 1}^{\infty} {na}_{n} x^{n - 1} and y^{^{″}} = \sum_{n = 2}^{\infty} n (n - 1) x^{n - 2}

e \Rightarrow (x^{2} - 2 x + 1) \sum_{n = 2}^{\infty} n (n - 1) a_{n} x^{n - 2} + (x - x^{2}) \sum_{n = 1}^{\infty} {na}_{n} x^{n - 1}

+ (x^{2} - x) \sum_{n = 0}^{\infty} a_{n} x^{n} = 1 \Rightarrow

\sum_{n = 2}^{\infty} n (n - 1) a_{n} x^{n} - 2 \sum_{n = 2}^{\infty} n (n - 1) a_{n} x^{n - 1} + \sum_{n = 2}^{\infty} n (n - 1) a_{n} x^{n - 2}

+ \sum_{n = 1}^{\infty} {na}_{n} x^{n} - \sum_{n = 1}^{\infty} {na}_{n} x^{n + 1} + \sum_{n = 0}^{\infty} a_{n} x^{n + 2} - \sum_{n = 0}^{\infty} a_{n} x^{n + 1} = 1 \Rightarrow

\Rightarrow \sum_{n = 2}^{\infty} n (n - 1) a_{n} x^{n} - 2 \sum_{n = 1}^{\infty} (n + 1) {na}_{n + 1} x^{n}

+ \sum_{n = 0}^{\infty} (n + 2) (n + 1) a_{n + 2} x^{n} + \sum_{n = 1}^{\infty} {na}_{n} x^{n} - \sum_{n = 2}^{\infty} (n - 1) a_{n - 1} x^{n}

+ \sum_{n = 2}^{\infty} a_{n - 2} x^{n} - \sum_{n = 1}^{\infty} a_{n - 1} x^{n} = 1 \Rightarrow

\sum_{n = 2}^{\infty} {n (n - 1) a_{n} - 2 n (n + 1) a_{n + 1} + (n + 1) (n + 2) a_{n + 2} + {na}_{n}

- (n - 1) a_{n - 1} + a_{n - 2} - a_{n - 1}) x^{n} - {4 a}_{2} x + {2 a}_{2} + {6 a}_{3} x

a_{1} x - a_{0} x = 1 \Rightarrow

\sum_{n = 2}^{\infty} {n^{2} a_{n} - 2 n (n + 1) a_{n + 1} + (n + 1) (n + 2) a_{n + 2} - {na}_{n - 1} + a_{n - 2}) x^{n}

+ \dots = 1 \dots . be continued \dots