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## Homework Statement

Given the differential equation

[itex] (\sin x)y'' + xy' + (x - \frac{1}{2})y = 0 [/itex]

a) Determine all the regular singular points of the equation

b) Determine the indicial equation corresponding to each regular point

c) Determine the form of the two linearly independent solutions near each of the singular points

d) Give the behavior of all non-zero solutions as x approaches 0.

I'm mostly concerned about questions b) and c), as those are the ones that I'm stuck at.

## Homework Equations

Singular points of equations with the form [itex] b(x)y'' + p(x)y' + q(x)y = 0 [/itex] occur when [itex] b(x) = 0 [/itex] and [itex] \frac{(x-x_0)p(x)}{b(x)} [/itex] has a definite limit as x -> x0 and [itex]\frac{(x-x_0)^2q(x)}{b(x)} [/itex] has a definite limit as x -> x0.

Since we're looking for solutions near singular points, I am assuming we have to use the Frobenius method, as that's the only method I'm aware of for doing such a thing. So, assume [itex] y = x^r \sum_{n=0}^{\infty}a_nx^n [/itex].

## The Attempt at a Solution

So the first thing I attempted was finding the regular singular points.

Firstly, [itex] \sin x = 0 [/itex] when [itex] x = \pi n, n \in Z [/itex]. So all regular singular points must be in that set.

Looking at [itex]\frac{xp(x)}{b(x)}[/itex] we get for x = 0: [itex] \frac{x^2}{\sin(x)} = x * (\frac{\sin(x)}{x})^{-1} \\[/itex] and [itex]\frac{x^2q(x)}{b(x)} = \frac{x^3}{\sin(x)} - \frac{x^2}{2\sin(x)} [/itex]. The limit of each of these is 0 because the limit as x approaches 0 of [itex] \frac{\sin(x)}{x} = 1 [/itex]. So 0 looks like a regular singular point. However, for every other possible singular point ([itex] x = \pi, 2\pi, 3\pi ...[/itex]), I don't believe these limits exist, because instead of [itex] \frac{x}{sin(x)} = \frac{0}{0} [/itex] there would be a real number divided by 0. So I believe the only regular singular point is at x = 0.

Next I tried to find the indicial equation, but no matter how I try it, I get the indicial equation in terms of [itex] a_0 [/itex] and [itex] a_1 [/itex]. Here's my procedure:

[itex] y = x^r \sum_{n=0}^{\infty} a_nx^n = \sum_{n=0}^{\infty}a_nx^{n+r} \\ [/itex]

[itex] y' = \sum_{n=0}^{\infty} a_n(n+r)x^{n+r-1} \\[/itex]

[itex] y'' = \sum_{n=0}^{\infty} a_n(n+r)(n+r-1)x^{n+r-2} \\[/itex]

So here's the differential equation in series form: [itex]\\[/itex]

[itex] \sum_{n=0}^{\infty} a_n(n+r)(n+r-1)\sin(x)x^{n+r-2} + \sum_{n=0}^{\infty} a_n(n+r)x^{n+r} + \sum_{n=0}^{\infty} a_nx^{n+r+1} - \sum_{n=0}^{\infty} \frac{a_n}{2}x^{n+r} = 0\\[/itex]

Next, I shifted the indices so all sums were in terms of [itex] x^{n+r+1} \\[/itex]

[itex] \sum_{n=-2}^{\infty} a_{n+2}(n+r+2)(n+r+1)\sin(x)x^{n+r+1} + \sum_{n=-1}^{\infty} a_{n+1}(n+r+1)x^{n+r+1} + \sum_{n=0}^{\infty} a_nx^{n+r+1} - \sum_{n=-1}^{\infty} \frac{a_{n+1}}{2}x^{n+r+1} \\ [/itex]

Now I remove terms from each series until I can write the rest of the series starting from n = 0. These extra terms that I removed make up the indicial equation. So for the first sum, I take the first two terms because we are starting at n = -2, and from the second and last sums I remove the first term because those start at n = -1.

So, the sum of the terms I removed is: [itex] a_{2-2}(2+r-2)(-2+r+1)\sin(x)x^{r-1} + a_{2-1}(2+r-1)(1+r-1)\sin(x)x^{r} + a_{1-1}(-1+r+1)x^{-1+r+1} - \frac{a_{-1+1}}{2}x^{-1+r+1} \\[/itex]

Simplifying, and removing the factor of [itex] x^{r} \\ [/itex]

[itex]r(r-1)a_0\sin(x)x^{-1} + (r+1)(r)a_1\sin(x) + ra_0 - \frac{a_0}{2} \\[/itex]

So I'm stuck with terms with [itex] a_1 [/itex] , [itex] \sin(x) [/itex], and exponentials of x. Is this a legitimate indicial equation? If not, where did I go wrong?