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Find the coefficients of $ a_{n}$ and $b_{n}$ for $ 0 ≤ n ≤ 4$ for the power series expansion of two linearly independent solutions of the ODE: $y'' -(e^{x}-1)y=0$.

This is what I've tried so far:

$\sum_{n=2}^{\infty} n(n-1)c_{n}x^{n-2} - \sum_{n=0}^{\infty}\frac{1}{n!}x^{n}\sum_{n=0}^{\infty}c_{n}x^{n} + \sum_{n=0}^{\infty}c_{n}x^{n} = 0$

$\sum_{n=0}^{\infty} (n+2)(n+1)c_{n+2}x^{n} - \sum_{n=0}^{\infty}\frac{1}{n!}x^{n}\sum_{n=0}^{\infty}c_{n}x^{n} + \sum_{n=0}^{\infty}c_{n}x^{n} = 0$

Don't know how to continue from here.

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Have you tried anything? –  Daniel R Sep 20 '13 at 8:16
    
I'm not sure exactly how to start. –  Mehran Baba Sep 20 '13 at 8:48
    
Here is a start: if $y(x)=a_0+a_1x+a_2x^2+a_3x^3+a_4x^4+$some higher order terms, what is the power series expansion of $y''(x)$ up to order $x^2$? –  Did Sep 20 '13 at 8:52
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2 Answers 2

Hint: $e^x = 1 + x + \dfrac{x^2}{2!}+\dfrac{x^3}{3!}+\dfrac{x^4}{4!}+\ldots$, $y_1(x) = x + a_2 x^2 + a_3 x^3 + a_4 x^4 + \ldots$, and $y_2(x) = 1 + b_2 x^2 + b_3 x^3 + b_4 x^4 + \ldots$.

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HINT Collect like powers of $x$ from all three terms in your LHS (after figuring out how to write the product of two sums as a single sum); set each coefficient to zero; solve these equations to find the coefficients. If you get bogged down in notation & all that, write out the sums explicitly (sure, up to some order) & try to see a pattern. That's at the very core of the method so you should know it, but I somehow get the feeling that this may be where your problems begin.

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