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Series Solutions | Differential Equations

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and skip the integration as the question never stated to simplify.

Series Solution

$y''+xy'+2y=0,\qquad x_0=0$

1^st) Formula

Let,

$y(x)=\sum_{n=0}^{\infty}a_n(x-x_0)^n,\qquad y'(x)=\sum_{n=0}^{\infty}a_nn(x-x_0)^{n-1},\qquad y''(x)=\sum_{n=0}^{\infty}a_nn(n-1)(x-x_0)^{n-2}$

Note(1):

$y'=\sum_{n=0}^{\infty}a_nn(x-x_0)^{n-1} = 0+\sum_{n=1}^{\infty}a_nn(x-x_0)^{n-1}$

$\sum_{n=0}^{\infty}a_nn(n-1)(x-x_0)^{n-2} = \sum_{n=1}^{\infty}a_nn(n-1)(x-x_0)^{n-2} = \sum_{n=2}^{\infty}a_nn(n-1)(x-x_0)^{n-2}$

Note(2):

$y' = \sum_{n=1}^{\infty}a_nn(x-x_0)^{n-1} = \sum_{n=0}^{\infty}a_{n+1}(n+1)(x-x_0)^n$

$y'' = \sum_{n=-1}^{\infty}a_{n+1}(n+1)((n+1)-1)(x-x_0)^{(n+1)-2} = \sum_{n=1}^{\infty}a_{n+2}(n+1)(n)(x-x_0)^{n-1}$

$= \sum_{n=0}^{\infty}a_{n+2}(n+2)(n+1)(x-x_0)^n$

Note: Geometric Series- $latex \sum_{n=0}^{\infty}ar^n=\frac{a_0}{1-r},\ |r|<1$

Thus,

$y'' = \sum_{n=0}^{\infty}a_{n+2}(n+2)(n+1)x^n, \qquad y' = \sum_{n=0}^{\infty}a_{n+1}(n+1)x^n, \qquad y = \sum_{n=0}^{\infty}a_nx^n$

Continue the Differential Equations Library

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Series Solutions | Differential Equations

Series Solution

Continue the Differential Equations Library

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