I have been going insane over the following differential equation over the past few days.
$y''\sin^2(x) + y'\tan(x) + y\cos^2(x) = 0 $
The assignment is:
$a)$ Show that $x=0$ is a regular singular point of the differential equation.
$b)$ Find the indicial roots.
$c)$ Show that $\sin(\ln(\sin x))$ and $\cos(\ln(\sin x))$ are solutions of the differential equation for some $0<x<R$ and $R>0$
I have tried many times to show $a)$:
$p(x)=\tan x/\sin^2x$ and $q(x)=\cos^2x/\sin^2x$ are both not analytic at $x=0$.
$$xp(x)=\frac{x\tan x}{\sin^2x} = \frac{x}{\sin x \cos x}$$
Trying to compute this series directly will always yield a sine in the denumerator, making this fail at x=0.
So I tried some tricks, the one I was most convinced of was determining the series of $\sin x\cos x$ and then dividing through by $x$: $$\sin x\cos x=\frac{1}{2}\sum_0^k(-1)^k\frac{(2x)^{2k+1}}{(2k+1)!}$$
So then $$\frac{x}{\sin x \cos x}= \frac{x}{\frac{1}{2}\sum_0^k(-1)^k\frac{(2x)^{2k+1}}{(2k+1)!}}=\frac{2}{\sum_0^k(-1)^k\frac{2^{2k+1}(x)^{2k}}{(2k+1)!}}$$
Setting $x=0$ here yields $\frac22=1$. So I thought this could work and that $xp(x)$ could be considered analytic at $0$ like this.
A similar argument I tried for $x^2q(x)$ was:
$$\frac{x^2\cos^2x}{\sin^2x}=\frac{x^2}{\tan^2x}$$
And then taking the series of $\tan x = x+\frac{x^3}{3}+\frac{2x^5}{15}+...$ So that $x^2q(x)$ becomes: $$\frac{x^2}{( x+\frac{x^3}{3}+\frac{2x^5}{15}+...)( x+\frac{x^3}{3}+\frac{2x^5}{15}+...)} = \frac{1}{1 + 2\frac{x^2}3+\frac{17x^4}{45}+...}$$
The series check out as far as I can tell, but I'm not convinced I can really say that $xp(x)$ and $x^2q(x)$ are analytic at the origin because of this. Any insight as to how this power series expansion conclusively shows that they are analytic is much appreciated.
If the expansions are correct, it follows that $p_0=q_0=1$, which will give indicial roots $+/-i$. I have never seen complex indicial roots and they aren't covered by the course I'm taking. It feels like these are wrong and this is the main reason I started doubting my result for a).
Showing the solutions in c) are true is simple (derive and plug in the equation). However I can't find the number R they want me to find. I thought they meant the interval of convergence of the series found in a), as this is directly related to the solution. I attempted to compute those:
I used the denumerator only, because I didn't have a clue how to calculate a radius of convergence of the reciprocal of a power series.
The expansion for $xp(x)$ gives:
$$a_k= (-1)^k\frac{2^{2k+1}x^{2k}}{(2k+1)!}$$
and
$$a_{k+1}=(-1)^{k+1}\frac{2^{2k+3}x^{2k+2}}{(2k+3)!}$$
$$\lim_{k->\infty}\frac{a_{k+1}}{a_{k}}=\lim_{k->\infty}\frac{(-1)^{k+1}\frac{2^{2k+3}x^{2k+2}}{(2k+3)!}}{(-1)^k\frac{2^{2k+1}x^{2k}}{(2k+1)!}}=\lim_{k->\infty}\frac{-4}{(2k+3)(2k+2)}x^2=0$$
So the whole thing doesn't even converge to begin with, which means my solution obtained in a) must be wrong.
I don't know what or where exactly I am going into the wrong and any advice is welcome.
The fact you are looking for is that $$ \lim_{x\to 0}\frac{\sin x}{x}=\frac{d}{dx}\sin(x)|_{x=0}=\cos(0)=1. $$ Then all the limits in the first part can be computed with minimal effort.
More generally, as you have already computed but not put in the general context, $$ \frac{\sin x}{x}=1-\frac{x^2}6+\frac{x^4}{120}\mp\dots $$ and with $$ \cos x = 1-\frac{x^2}2+\frac{x^4}{24}\mp\dots $$ all rational expressions can be written as quotients of power series with non-zero constant terms in the denominator, so that the quotients are indeed analytical themselves.
Let $\rho>0$ be small enough so that $\sum_{k=1}^\infty |a_k|ρ^k\le 1$. Then $1+\sum_{k>0}a_kz^k$ has no roots with $|z|<\rho$. The coefficients of the reciprocal series $$ 1+b_1z+b_2z^2+\dots=(1+a_1z+a+2z^2+\dots)^{-1} $$ can be computed via the Cauchy product formula as $$ b_0=1,\; b_1=-a_1,\;b_n=-\sum_{k=1}^na_kb_{n-k} $$ and thus bounded as $$ |b_n|ρ^n\le \sum_{k=1}^n |a_k|ρ^{k}\,|b_{n-k}|ρ^{n-k}\le \sum_{k=1}^n |a_k|ρ^{k}\max_{j=0,...,n-1}|b_j|ρ^j\le \max_{j=0,...,n-1}|b_j|ρ^j $$ so that by induction $|b_n|ρ^n\le|b_0|=1$. This gives $ρ$ as a lower bound for the radius of convergence of $1+\sum_{k>0} b_kz^k$.