# 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?

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.

• Can you please explain this line $$-\epsilon \sum_{n=1}^{\infty}na_{n-1}x^{n}=-\epsilon \sum_{n=0}^{\infty}na_{n-1}x^{n}$$ – jarhead Dec 16 '18 at 10:26
• 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. – Squanch Dec 17 '18 at 22:24

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 $$(-\sqrt{R},\sqrt{R})$$