Solve-5x-2-y-x-1-x-y-y-0-

Question Number 217797 by Tawa11 last updated on 21/Mar/25

Solve :

{5 x}^{2} y^{″} + x (1 + x) y^{'} - y = 0

Answered by AntonCWX8 last updated on 22/Mar/25

I . F, μ (x) = \frac{1}{5 x^{2}} e^{\int \frac{x (1 + x)}{5 x^{2}} d x} = \frac{1}{5 x^{2}} e^{\int \frac{1 + x}{5 x} d x} = \frac{1}{5 x^{2}} e^{\frac{l n (x)}{5} + \frac{x}{5}}

M u l t i p l y t h r o u g h

e^{\frac{l n (x) + x}{5}} y^{″} + \frac{(1 + x)}{5 x} e^{\frac{l n (x) + x}{5}} y^{'} - \frac{1}{5 x^{2}} e^{\frac{l n (x) + x}{5}} y = 0

R e w r i t e i n S t u r m - L i o u v i l l e F o r m g i v e s

{(e^{\frac{l n (x) + x}{5}} y^{'})}^{'} - \frac{1}{5 x^{2}} e^{\frac{l n (x) + x}{5}} y = 0

S t u c k h e r e

Answered by vnm last updated on 22/Mar/25

l e t p b e a n y r e a l n u m b e r

y = p (x + \underset{n = 2}{\sum^{\infty}} \frac{{(- 1)}^{n - 1} x^{n}}{\underset{k = 2}{\prod^{n}} (5 k + 1)})

Commented by Tawa11 last updated on 22/Mar/25

Please workings sir .

Answered by vnm last updated on 22/Mar/25

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

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

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

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

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

5 x^{2} y^{″} = \underset{n = 0}{\sum^{\infty}} 5 a_{n + 2} (n + 2) (n + 1) x^{n + 2} = \underset{n = 2}{\sum^{\infty}} 5 a_{n} n (n - 1) x^{n}

a_{0} = 0

a_{1} x = a_{1} x \Rightarrow a_{1} = p, p i s a n y r e a l n u m b e r

a_{n} n + a_{n - 1} (n - 1) + 5 a_{n} n (n - 1) - a_{n} = 0

a_{n} (n - 1) + a_{n - 1} (n - 1) + 5 a_{n} n (n - 1) = 0

a_{n} + a_{n - 1} + 5 a_{n} n = 0

a_{n} = - \frac{a_{n - 1}}{5 n + 1}

a_{1} = p

a_{2} = - \frac{a_{1}}{5 \cdot 2 + 1} = - \frac{p}{11}

a_{3} = - \frac{a_{2}}{5 \cdot 3 + 1} = \frac{p}{11 \cdot 16}

a n d s o o n

Commented by Tawa11 last updated on 23/Mar/25

Thanks sir .

Facebook Tweet Pin