solve-y-xy-0-by-using-integr-series-

Question Number 57925 by maxmathsup by imad last updated on 14/Apr/19

s o l v e y^{^{″}} - x y = 0 b y u s i n g i n t e g r s e r i e s .

Commented by maxmathsup by imad last updated on 18/Apr/19

l e t s e a r c h d e v e l l o p p a b l e a t i n t e g r s e r i e l e t y = \sum_{n = 0}^{\infty} a_{n} x^{n} \Rightarrow

y^{^{'}} (x) = \sum_{n = 1}^{\infty} {n a}_{n} x^{n - 1} a n d y^{^{″}} (x) = \sum_{n = 2}^{\infty} n (n - 1) a_{n} x^{n - 2}

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

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

\Rightarrow a_{2} = 0 a n d (p + 1) (p + 2) a_{p + 2} - a_{p - 1} = 0 \forall p ⩾ 1 \Rightarrow a_{p + 2} = \frac{a_{p - 1}}{(p + 1) (p + 2)}

\forall p ⩾ 1 a n d a_{2} = 0 \Rightarrow a_{2 n + 2} = \frac{a_{2 n - 1}}{(2 n + 1) (2 n + 2)} a n d a_{2 n + 3} = \frac{a_{2 n}}{(2 n + 2) (2 n + 3)}

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

= \sum_{n = 2}^{\infty} \frac{a_{2 n - 3}}{(2 n - 1) (2 n + 1)} x^{2 n} + a_{0} + \sum_{n = 1}^{\infty} \frac{a_{2 n - 2}}{2 n (2 n + 1)} x^{2 n + 1}

a l s o w e c a n u s e t h a t \frac{a_{2 n + 2}}{a_{2 n - 1}} = \frac{1}{(2 n + 1) (2 n + 2)} \dots b e c o n t i n u e d \dots

Answered by 121194 last updated on 14/Apr/19

y = \underset{i = 0}{\sum^{\infty}} a_{i} x^{i}

y^{'} = \underset{i = 0}{\sum^{\infty}} a_{i} {i x}^{i - 1} = \underset{i = - 1}{\sum^{\infty}} a_{i + 1} (i + 1) x^{i} = \underset{i = 0}{\sum^{\infty}} a_{i + 1} (i + 1) x^{i}

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

y^{″} - x y = 0

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

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

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

2 a_{2} = 0 \Rightarrow a_{2} = 0

(i + 2) (i + 1) a_{i + 2} - a_{i - 1} = 0; i > 0

a_{i + 2} = \frac{a_{i - 1}}{(i + 2) (i + 1)}; i > 0

a_{i} = \frac{a_{i - 3}}{i (i - 1)}; i > 2

a_{0} = c_{1}

a_{1} = c_{2}

a_{2} = 0

a_{3} = \frac{a_{0}}{6} = \frac{c_{1}}{6}

a_{4} = \frac{a_{1}}{12} = \frac{c_{2}}{12}

a_{5} = \frac{a_{2}}{20} = 0

a_{6} = \frac{a_{3}}{30} = \frac{c_{1}}{180}

\dots

y = \underset{i = 0}{\sum^{\infty}} a_{i} x^{i}

= \underset{i = 0}{\sum^{\infty}} a_{3 i} x^{3 i} + \underset{i = 0}{\sum^{\infty}} a_{3 i + 1} x^{3 i + 1} + \underset{i = 0}{\sum^{\infty}} a_{3 i + 2} x^{3 i + 2}

= c_{1} \underset{i = 0}{\sum^{\infty}} \frac{x^{3 i}}{f (i)} + c_{2} \underset{i = 0}{\sum^{\infty}} \frac{x^{3 i + 1}}{g (i)}

f (i) = {\begin{cases} 1 & i = 0 \\ \frac{f (i - 1)}{3 i (3 i - 1)} & i > 0 \end{cases}

g (i) = {\begin{cases} 1 & i = 0 \\ \frac{f (i - 1)}{3 i (3 i + 1)} & i > 0 \end{cases}

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