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My friend and I have conflicting answers and since his phone is off, I can't get his full solution and I don't understand his argument.

Consider this ODE

$$(x+1)y''+\frac{1}{x}y' + (x+3)y= 0$$

Basically what I did was divide out that $(x+1)$ on $y''$ and got

$$y''+\frac{1}{x(x+1)}y' + \frac{(x+3)}{(x+1)}y= 0$$

The singularities are x = -1, and 0 (both are regular)

My friend said we had to get our ODE in the form of

$$a(x-x_0)^2y'' + b(x-x_0)y' + (x-x_0)y =0$$ otherwise we cant' do anything. and he got x = 3 as an irregular singular point which i have no idea how even got this. Sorry if this is too vague, but my first source was also as vague.

what was my friend doing and who's right?

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$(x+1)y''+\dfrac{1}{x}y'+(x+3)y=0$

$y''+\dfrac{1}{x(x+1)}y'+\dfrac{x+3}{x+1}y=0$

It should be noted that the positions of finite singular points are always appear at the positions that makes one of the coefficients diverge.

$\therefore$ the positions of finite singular points in this question are $x=0$ and $x=-1$ only, absolute not like your friend said $x=3$ also is the finite singular point.

The next step is to determine whether the finite singular points are regular or irregular.

$\lim\limits_{x\to 0}\left(x\times\dfrac{1}{x(x+1)}\right)=\lim\limits_{x\to 0}\dfrac{1}{x+1}=1$

$\lim\limits_{x\to 0}\left(x^2\times\dfrac{x+3}{x+1}\right)=\lim\limits_{x\to 0}\dfrac{x^2(x+3)}{x+1}=0$

$\therefore$ the finite singular point $x=0$ is regular.

$\lim\limits_{x\to-1}\left((x+1)\times\dfrac{1}{x(x+1)}\right)=\lim\limits_{x\to-1}\dfrac{1}{x}=-1$

$\lim\limits_{x\to-1}\left((x+1)^2\times\dfrac{x+3}{x+1}\right)=\lim\limits_{x\to-1}(x+1)(x+3)=0$

$\therefore$ the finite singular point $x=-1$ is regular.

That's not the end. We should also check the singularities at infinity. Because these also act as one of the key points of distinguishing the ODE type.

Let $u=\dfrac{1}{x}$ ,

Then $\dfrac{dy}{dx}=\dfrac{dy}{du}\dfrac{du}{dx}=-\dfrac{1}{x^2}\dfrac{dy}{du}=-u^2\dfrac{dy}{du}$

$\dfrac{d^2y}{dx^2}=\dfrac{d}{dx}\left(-u^2\dfrac{dy}{du}\right)=\dfrac{d}{du}\left(-u^2\dfrac{dy}{du}\right)\dfrac{du}{dx}=\left(-u^2\dfrac{d^2y}{du^2}-2u\dfrac{dy}{du}\right)\left(-\dfrac{1}{x^2}\right)=\left(-u^2\dfrac{d^2y}{du^2}-2u\dfrac{dy}{du}\right)(-u^2)=u^4\dfrac{d^2y}{du^2}+2u^3\dfrac{dy}{du}$

$\therefore u^4\dfrac{d^2y}{du^2}+2u^3\dfrac{dy}{du}+\dfrac{1}{\dfrac{1}{u}\left(\dfrac{1}{u}+1\right)}\left(-u^2\dfrac{dy}{du}\right)+\dfrac{\dfrac{1}{u}+3}{\dfrac{1}{u}+1}y=0$

$u^4\dfrac{d^2y}{du^2}+2u^3\dfrac{dy}{du}-\dfrac{u^4}{u+1}\dfrac{dy}{du}+\dfrac{3u+1}{u+1}y=0$

$u^4\dfrac{d^2y}{du^2}+\dfrac{u^4+2u^3}{u+1}\dfrac{dy}{du}+\dfrac{3u+1}{u+1}y=0$

$\dfrac{d^2y}{du^2}+\dfrac{u+2}{u(u+1)}\dfrac{dy}{du}+\dfrac{3u+1}{u^4(u+1)}y=0$

$\lim\limits_{u\to 0}\left(u\times\dfrac{u+2}{u(u+1)}\right)=\lim\limits_{u\to 0}\dfrac{u+2}{u+1}=2$

$\lim\limits_{u\to 0}\left(u^2\times\dfrac{3u+1}{u^4(u+1)}\right)=\lim\limits_{u\to 0}\dfrac{3u+1}{u^2(u+1)}=\infty$

$\therefore$ the singularities at $x=\pm\infty$ are irregular.

So $(x+1)y''+\dfrac{1}{x}y'+(x+3)y=0$ belongs to Heun's Confluent type ODE. If you solve it in MATLAB, MATLAB will express its general solution in terms of Heun's Confluent function.

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  • $\begingroup$ I believe that your first limit should be $\lim_{x\to 0}\dfrac{1}{x+1}=1$. This does not change your conclusion, of course. Nice explanation. $\endgroup$ – bbgodfrey May 4 '15 at 12:54

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