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Determine the singularities of the differential equation

$(x-1)^3x^2y''-2 (x-2)xy'-3y=0$

My Attempt:

$y''-\frac {2 (x-2)}{(x-1)^3x}y'-\frac {3}{(x-1)^3x^2}y=0$

at $x_0 =0$, we have

$(x-x_0)p=(x-x_0)\frac{-2 (x-2)}{(x-1)^3x}$,

$(x-x_0)^2q=(x-x_0)^2\frac {-3}{(x-1)^3x^2}$

$xp=\frac {-2 (x-2)}{(x-1)^3}$,

$x^2q=\frac {-3}{(x-1)^3}$

What does this tell me? I followed what I learned in class, but I am having trouble understanding if that is a singular point.

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The singularities are just the values of $x$ that will make the denominator zero, because you can't divide by zero (and doing so creates a singularity). Since you are dividing through by the coefficient of $y''$, whatever value for $x$ makes the coefficient $(x-1)^3x^2=0$ will cause a singularity. If either $(x-1)^3=0$ or $x^2=0$, there will be a singularity so solving for $x$ in both of these equations gives you $x=1$ and $x=0$.

You'll then have to use both of these singularity points as your $x_0$. This will give a total of four limits that you need to solve for in order to make sure the differential equation has a power series solution.

For the first singularity ($x_0=0$) the two limits you need to check are: $$1.) \lim_{x\to x_0}\left[(x-x_0)\frac{B(x)}{A(x)}\right] = \lim_{x\to0}\left[(x-0)\frac{-2(x-2)x}{(x-1)^3x^2}\right]=\lim_{x\to0}\left[\frac{-2(x-2)}{(x-1)^3}\right]=-4$$ $$2.) \lim_{x\to x_0}\left[(x-x_0)^2\frac{C(x)}{A(x)}\right] = \lim_{x\to0}\left[(x-0)^2\frac{-3}{(x-1)^3x^2}\right]=\lim_{x\to0}\left[\frac{-3}{(x-1)^3}\right]=3$$

The solutions are regular because they are finite, as opposed to irregular if the limits were $\pm \infty.$

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  • $\begingroup$ Thank you, very much. This helped. $\endgroup$ – Kruschev Nov 3 '15 at 17:22
  • $\begingroup$ You're welcome. Make sure you take the other two limits as x goes to the other singularity (x=1). Forgot to make that clear in the examples, but I'll edit it in. $\endgroup$ – devjoco Nov 3 '15 at 19:03
  • $\begingroup$ Also, how can you tell if they are good or bad singularities? $\endgroup$ – Kruschev Nov 3 '15 at 19:06
  • $\begingroup$ By good or bad I'm assuming you mean regular or irregular. A point is a regular singularity if the functions P(x)=B(x)/A(x) and Q(x)=C(x)/A(x) are analytic at that point (the limit doesn't approach plus or minus infinity). Otherwise its an irregular singularity. $\endgroup$ – devjoco Nov 3 '15 at 19:20
  • $\begingroup$ Ah, I see. I think that's what the teacher meant. Thank you. $\endgroup$ – Kruschev Nov 3 '15 at 19:48

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