use-power-series-solution-method-to-solve-the-ODE-y-xy-0-

Question Number 99646 by 24224 Opiyo Kamuki last updated on 22/Jun/20

\boldsymbol u s e \boldsymbol p o w e r \boldsymbol s e r i e s \boldsymbol s o l u t i o n \boldsymbol m e t h o d \boldsymbol t o \boldsymbol s o l v e \boldsymbol t h e \boldsymbol O D E

\boldsymbol y^{″} - \boldsymbol x y = 0

Answered by MWSuSon last updated on 22/Jun/20

\underset{k = 2}{\sum^{\infty}} k (k - 1) a_{k} x^{k - 2} - x \underset{k = 0}{\sum^{\infty}} a_{k} x^{k} = 0

\underset{k = 0}{\sum^{\infty}} (k + 1) (k + 2) a_{k + 2} x^{k} - \underset{k = 0}{\sum^{\infty}} a_{k} x^{k + 1} = 0

\underset{k = 0}{\sum^{\infty}} (k + 1) (k + 2) a_{k + 2} x^{k} - \underset{k = 1}{\sum^{\infty}} a_{k - 1} x^{k} = 0

{2 a}_{2} + \underset{k = 1}{\sum^{\infty}} [(k + 2) (k + 1) a_{k + 2} - a_{k - 1}] x^{k} = 0

a_{2} = 0

recurrence relation for k ⩾ 1

a_{k + 2} = \frac{a_{k - 1}}{(k + 2) (k + 1)}

input values for k = 1, 2, 3, 4, 5, \dots

and find a_{k} interms of a_{0} and a_{1}

Answered by mathmax by abdo last updated on 22/Jun/20

y = \sum_{n = 0}^{\infty} a_{n} x^{n} \Rightarrow y^{^{'}} = \sum_{n = 1}^{\infty} {na}_{n} x^{n - 1} \Rightarrow y^{^{″}} = \sum_{n = 2}^{\infty} n (n - 1) a_{n} x^{n - 2}

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

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

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

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

a_{2} = 0 and a_{n + 2} = \frac{a_{n - 1}}{(n + 1) (n + 2)} (n ⩾ 1) \Rightarrow a_{3} = \frac{a_{0}}{2.3}

a_{4} = \frac{a_{1}}{3.4}, a_{5} = \frac{a_{2}}{4.5}, \dots . \Rightarrow y (x) = a_{0} + a_{1} x + a_{2} x^{2} + a_{3} x^{3} + a_{4} x^{4} + \dots

= a_{0} + a_{1} x + \frac{a_{0}}{6} x^{3} + \frac{a_{1}}{12} x^{4} + \frac{a_{2}}{20} x^{5} + \dots