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I'm studying for a qualifying examination and am stuck on the following question. Could anyone give me some help?

$$y''+\frac{\sin x}{x}y'+\frac{2\cos(x+x^2)-\frac{2}{(x-1)^2}+4x}{x^2}y=0$$

Find all singular points of the equation and classify them as regular/irregular. Then find the first term in a series in powers of $x-1$ for each of two linearly independent solutions as $x\rightarrow1$.

I think the singular points are 0 and 1, and they are both regular. But I am having some trouble with the series solution.

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Do you know how the ansatz for the series near a regular point looks like? – Fabian Jan 8 '13 at 21:46
By the way: 0 is not a singular point... – Fabian Jan 8 '13 at 21:47
How is 0 not a singular point? And the ansatz is $y(x)=\sum_{n=0}^{\infty}a_n x^n.$ – Alex Jan 8 '13 at 21:50
$\lim_{x \rightarrow 0} \sin{x}/x = 1$ and $\lim_{x \rightarrow 0} [2 \cos{(x +x^2)} - 2 (x-1)^{-2} + 4 x]/x^2 = -7$. Because the singularities of the coefficients of the diff eq'n are removable at $x=0$, there is no singularity of the solution at $x=0$. – Ron Gordon Jan 8 '13 at 22:39

$x=0$ is an ordinary point as the singularities of the coefficients are removable.

There is a regular singular point at $x=1$. To analyze this point, you should approximate the ODE near this point (expand the coefficients in $z=x-1$): $$y'' + O(1) y' + \left[-\frac{2}{(x-1)^2} + O(x-1)^{-1}\right] y =0 .$$

Now you know that there is at least one solution has the form ($c_0\neq0$) $$y(z)= z^{\alpha} \sum_n c_n z^n.$$ We can determine $\alpha$ by plugging this ansatz in the ODE: to lowest order $$\alpha(\alpha-1) z^{\alpha-2} c_0 - \frac{2}{z^2} c_0 z^\alpha=0 $$ which is valid when $\alpha (\alpha-1) =2$. The solutions are $$\alpha_1 =-1, \text{ and } \alpha_2 = 2.$$ So the first term for one solution is $$y_1(x) = c_0 (x-1)^2.$$

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What happened to the $x^2$ in the denominator of the "y" term? Also, can you briefly explain how you got to the "approximated differential equation? Thanks! – Alex Jan 9 '13 at 21:16
@Ales: just Taylor expansion around $x=1$... – Fabian Jan 9 '13 at 22:20

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