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I'm trying to obtain an analytical solution to the following ODE:

$$-\epsilon x y+\left(\epsilon R-x-\epsilon x^2\right)y'+\left(R-x^2\right)y''=0$$

The only method that would make sense for me is the series method where I define $$y=\sum_{n=0}^\infty a_n x^n$$

and try to obtain a recurrence relation.

If each part of the ODE is written separately we obtain:

$$-\epsilon xy = \epsilon \sum_{n=0}^{\infty}a_nx^{n+1}$$ $$\left(\epsilon R-x-\epsilon x^2\right)y'=\sum_{n=1}^{\infty}n a_n x^{n-1}-\sum_{n=1}^{\infty}na_nx^n-\epsilon \sum_{n=1}^{\infty}na_nx^{n+1}$$ $$\left(R-x^2\right)y''=R\sum_{n=2}^{\infty}n(n-1)a_nx^{n-2}-\sum_{n=2}^{\infty}n(n-1)a_nx^n$$

Each time I'm trying to work out the equation I reach a dead end in terms of the indexes. Is it possible to reduce this problem to a recurrence relation problem?

If not, is this equation solvable analytically?

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Series representation of each term: $$-\epsilon xy = -\epsilon \sum_{n=0}^{\infty}a_nx^{n+1}$$ $$\left(\epsilon R-x-\epsilon x^2\right)y'=\epsilon R\sum_{n=1}^{\infty}n a_n x^{n-1}-\sum_{n=1}^{\infty}na_nx^n-\epsilon \sum_{n=1}^{\infty}na_nx^{n+1}$$ $$\left(R-x^2\right)y''=R\sum_{n=2}^{\infty}n(n-1)a_nx^{n-2}-\sum_{n=2}^{\infty}n(n-1)a_nx^n$$

You can write these equations as follows $$-\epsilon xy = -\epsilon \sum_{n=0}^{\infty}a_nx^{n+1}$$ $$\left(\epsilon R-x-\epsilon x^2\right)y'=\epsilon R\sum_{n=0}^{\infty}(n+1)a_{n+1} x^{n}-\sum_{n=0}^{\infty}na_nx^n-\epsilon \sum_{n=0}^{\infty}na_nx^{n+1}$$ You can start 2nd and 3rd term of the above series expresions from $n=0$, since for $n=0$ these expressions equal $0$. So, it does not affect the series. $$\left(R-x^2\right)y''=R\sum_{n=0}^{\infty}(n+2)(n+1)a_{n+2}x^{n}-\sum_{n=0}^{\infty}n(n-1)a_nx^n$$ Again, you can start 2nd term of the above series from $n=0$, since for $n=0$ and $n=1$ the term gives you $0$.

Now, the sum of all series expression can be written as $$-\epsilon \sum_{n=0}^{\infty}(n+1)a_nx^{n+1}+\sum_{n=0}^{\infty}\bigg[\epsilon R(n+1)a_{n+1} -na_n+R(n+2)(n+1)a_{n+2}-n(n-1)a_n\bigg]x^n=0$$

Let's modify the 1st term of the above series. $$-\epsilon \sum_{n=0}^{\infty}(n+1)a_nx^{n+1}=-\epsilon \sum_{n=1}^{\infty}na_{n-1}x^{n}=-\epsilon \sum_{n=0}^{\infty}na_{n-1}x^{n}$$ since for $n=0$ it is $0$.

Finally all series expressions can be rewritten as $$\sum_{n=0}^{\infty}\bigg[-\epsilon na_{n-1}+\epsilon R(n+1)a_{n+1} -na_n+R(n+2)(n+1)a_{n+2}-n(n-1)a_n\bigg]x^n=0$$

From the above series, you can determine coeefficient of the series.

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  • $\begingroup$ Can you please explain this line $$-\epsilon \sum_{n=1}^{\infty}na_{n-1}x^{n}=-\epsilon \sum_{n=0}^{\infty}na_{n-1}x^{n}$$ $\endgroup$ – jarhead Dec 16 '18 at 10:26
  • $\begingroup$ Left side is $-\epsilon[(1)a_0x+(2)a_1x^2+\dots]$. Right side is $-\epsilon[(0)a_{-1}+(1)a_0x+(2)a_1x^2+\dots]$=$-\epsilon[(1)a_0x+(2)a_1x^2+\dots]$. Two series are the same. $\endgroup$ – Squanch Dec 17 '18 at 22:24
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Rewrite the equation as

$$ (R-x^2)(y'' + \epsilon y') - x (y' + \epsilon y) = 0 $$

Let $u = y' + \epsilon y$, then

$$ (R-x^2)u' - xu = 0 $$

Separate and solve to obtain

$$ u = \frac{A}{\sqrt{R-x^2}} $$

What's left is

$$ y' + \epsilon y = \frac{A}{\sqrt{R-x^2}} $$

Use the integrating factor to finish off.

$$ y = Ae^{-\epsilon x} \int_0^x \frac{Ae^{\epsilon t}}{\sqrt{R-t^2}}dt + Be^{-\epsilon x} $$

Note that the solution interval is restricted to $(-\sqrt{R},\sqrt{R})$

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