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?




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.



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


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




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

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

$\therefore$ 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.

  • $\begingroup$ 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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