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Find series expansion of the solutions to the following DE about $x = 0$. Try to sum in closed form any infinite series that appear:

$$ y'' + (e^x - 1)y = 0 $$

My approach: OF course $x = 0$ is ordinary point, so we can find the taylor expansion of the solution assuming $$y = \sum_{n=0}^{\infty} c_nx^n $$ is a solution. And so we differentiale twice this expression and put it back into the DE to obtain a nasty equation like this: (after shifting indices)

$$ \sum_{n=0}^{\infty} c_{n+2}(n+1)(n+2)x^n + \sum_{n=0}^{\infty} \frac{1}{n!}x^n \sum_{n=0}^{\infty} c_nx^n - \sum_{n=0}^{\infty} c_nx^n = 0 $$ My question is: is this procedure fine? or is there a better way to approach this problem? and how can I find the $c_i$ in a economical manner?

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Please avoid using $$ environment in the title. – Asaf Karagila Oct 28 '12 at 0:11
I would write $$\sum_{n=0}^{\infty} \frac{1}{n!}x^n \sum_{n=0}^{\infty} c_nx^n - \sum_{n=0}^{\infty} c_nx^n$$ as $$\sum_{n=1}^{\infty} \frac{1}{n!}x^n \sum_{n=0}^{\infty} c_nx^n$$ instead. – Pedro Tamaroff Oct 28 '12 at 0:26
I think finding series expansion and trying to sum in closed form of the solutions of $y''+(e^x-1)y=0$ is just the OP's creative idea, because directly finding all the terms of the series expansion of the linear ODE with non-polynomial coefficients should be impossible. @Marvis method should be the only possible approach. – doraemonpaul Oct 28 '12 at 19:48
up vote 2 down vote accepted

Let $2e^{x/2} = t$, we then get that $$\dfrac{dy}{dx} = \dfrac{t}2\dfrac{dy}{dt}$$ $$\dfrac{d^2y}{dx^2} = \dfrac{d}{dt} \left( \dfrac{t}2\dfrac{dy}{dt}\right) \dfrac{dt}{dx} = \dfrac{t}2 \left( \dfrac{t}2 \dfrac{d^2y}{dt^2} + \dfrac12 \dfrac{dy}{dt}\right)$$ Hence, the initial ODE becomes $$\dfrac{t^2}4\dfrac{d^2y}{dt^2} + \dfrac{t}4 \dfrac{dy}{dt} + \left(\dfrac{t^2}{4} -1\right) y = 0$$ $$\underbrace{t^2 \dfrac{d^2y}{dt^2} + t \dfrac{dy}{dt} + \left(t^2 -4\right) y = 0}_{\text{Bessel's equation}}$$ \begin{align} y(x) & = c_1 J_2(2e^{x/2}) + c_2 Y_2(2e^{x/2})\\ & = c_1 J_2(2e^{x/2}) + c_2 Y_2(2e^{x/2}) \end{align}

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$$\dfrac{t^2}4\dfrac{d^2y}{dt^2} + \dfrac{t}4 \dfrac{dy}{dt} + \left(\dfrac{t^2}{4} -1\right) y = 0$$ $$\underbrace{t^2 \dfrac{d^2y}{dt^2} + t \dfrac{dy}{dt} + \left(t -4\right) y = 0}_{\text{Bessel's equation}}$$ typo??? – doraemonpaul Oct 28 '12 at 19:37
@doraemonpaul Yes. It was a typo. I have fixed it now. Thanks. – user17762 Oct 28 '12 at 19:39

This is precisely how we use the Frobenius method. Now put all the terms under the same summation, which equals zero, this means the terms must be zero. Rearrange to find a recurrence relation for ${c_{n + 2}}$ or shift $n$ to get a recurrence relation for ${c_n}$.

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I thought the Frobenius method is used when the singularity is regular? – ILoveMath Oct 28 '12 at 0:33
Series method is a special case of Frobenius, so the terminology is used interchangeably, at least in physics. – glebovg Oct 28 '12 at 0:57

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