Solving ODE with essential singularity I would like to solve the following linear ODE.
$y''(x) -\frac{2}{x} \frac{1-3x^4 +2x^3}{1+3x^4-4x^3} y'(x)+\frac{\omega^2}{(1+3x^4-4x^3)^2} y(x) = 0$
Here $x$ is a dimensionless variable which runs between $(0,1)$ and $\omega$ (frequency) is also a dimensionless quantity. This ODE has an essential singularity at $x=1$. I can't solve this even for very small $\omega$ since there is diverging $\frac{1}{(1+3x^4-4x^3)^2}$ near $x=1$. 
How to solve this ODE (at least near $x=1$)?  
 A: I propose here something simple. There are surely more sophisticated was to solve your problem. I would make a series expansion around $\omega = 0$. For convenience I name
\begin{align}
& c_1(x) = -\frac{2}{x}\frac{1-3x^4+2x^3}{1+3x^4-4x^3} \, , && c_2(x) = \frac{1}{1+3x^4-4x^3} \, .
\end{align}
Your equation becomes $ y'' + c_1 y ' + \omega^2 c_2 y = 0$.
Start by solving the equation for $\omega = 0$. With $y' = f$ your equations becomes $ f' = -c_1 f$. This is solved by variable separation,
\begin{align}
& \ln(f) + A_1 = -\int_0^x c_1(x')\text{d}x' \\
& = 2 \left(\ln(x)-\ln(x-1) - 1/2 \ln[1 + 2 x + 3 x^2]\right)\, ,\end{align}
($A_1$ is an integration constant) and provides
$$f = A_2 \frac{1}{1+2x+3x^2}\left(\frac{x}{x-1}\right)^2 \, . \qquad (*)$$
Finally I find
$$y_0 = A_3+A_2 \left(\frac{1}{3\sqrt{2}}\text{Atan}\left[\frac{4+3(x-1)}{\sqrt{2}}\right]+\ln\left[\frac{(x-1)^2}{6+8(x-1)+3(x-1)^2}\right]\right) \, .$$
Next you can compute the next order in $\omega$ by writing $y = y_0 + \omega^2 y_2$ and neglecting the term with $\omega^4$,
$$y_2'' + c_1 y_2' = - y_0 \, . \qquad (**)$$
Note that I did not include a correction proportional to $\omega$ because this is identical to $y_0$. $f = y'$ provides,
$$f' + c_1 f = - y_0 \, .$$
The homogeneous part of this equation has the same solution as before and the particular solution can be extracted with variation of constants. Add them up to get
$$f = f_0 + f_0 \int_0^x \frac{y_0(x')}{f_0(x')}\text{d}x' \, .$$
$f_0$ is given by (*) and is the derivative of $y_0$. I let you work out the details. Integrate $f$ to get $y_2$. I guess that Mathematica will be useful to compute the integrals.
When you have worked out $y_2$ you can go to the next order in $\omega$. Inserting $y = y_0 + \omega^2 y_2 + \omega^4 y_4$ into your equation and neglecting the term in $\omega^6$ provides,
$$y_4'' + c_1 y_4 = - c_2 y_2 \, .$$
This equation has the same structure as (**) and can be solved in the same way since you now know $y_2$. In principle you can keep on doing up to an arbitrary order in $\omega$ and (hoping that the series converges) reach the desires accuracy.
Note that I did not check the algebra enough. I believe that the reasoning is good, but you should redo the algebraic manipulations before you trust this solution.
