Second order ODE problem with a series solution 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?
 A: 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 $(-\sqrt{R},\sqrt{R})$
A: 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.
