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Given the differential equation

$$x^3y'' + y\sin x =0,$$

is $x=0$ a regular singular point or irregular singular point??

$y'' +p(x)y'+ q(x)y=0$ general equation. Now definition of regular singular point says when in a $2^{nd}$ order differential equation $xp(x)$ and $x^2 q(x)$ are analytic at $x=0$, then $x=0$ is a regular singular point.

The latter yields a function $\frac{\sin x}{x}$ which should be analytic at $x=0$ if it is a regular singular point. Answer at back of book is it is a regular singular point means that the final question is to prove that $\frac{\sin x}{x}$ is analytic at $x=0$.

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do you mean by $p(x)$ and $q(x)$, two linear independent solutions of diff.equatiom?? –  Praphulla Koushik Sep 14 '13 at 11:11
    
no no y'' +p(x)y'+ q(x)y=0 –  shivam Sep 14 '13 at 11:17
    
you better be precise in first instance itself :) –  Praphulla Koushik Sep 14 '13 at 11:30

2 Answers 2

up vote 6 down vote accepted

Are you familiar with Taylor series expansion for $\sin(x)$

$\displaystyle \sin(x)=x-\frac{x^3}{3!}+\frac{x^5}{5!}....$

can you justify that this imply $\displaystyle \frac{\sin(x)}{x}=1-\frac{x^2}{3!}+\frac{x^4}{5!}....$

do you see any conclusion from this????

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you accepted the hint.. that is fine... do you need more hints or you can work it out??? –  Praphulla Koushik Sep 14 '13 at 11:29
    
You can use \sin x to latex sine. –  Artem Sep 14 '13 at 15:27
    
@Artem : Thank you for your y=useful suggestion :) –  Praphulla Koushik Sep 14 '13 at 16:37

By the I am not agree with saying that sinx/x is analytic at x = 0. Since the function f(x) = sinx/x is not defined at x = 0. To say f is analytic first of all function should be defined at x = 0. This not the case with sinx/x.

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Technically, the statement should have been "$f(x) = \begin{cases}\frac{\sin x}{x} & x \neq 0 \\ 0 & x = 0\end{cases}$ is analytic at $x = 0$." –  JimmyK4542 Aug 16 at 6:23

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