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.
Best Answer
$x=0$ 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 $$y_1(x) = c_0 (x-1)^2.$$