## goalie2012 2 years ago consider the differential equation: x^3(x^2-1)^2(x^2+1)y''+(x+1)xy'+y=0. determine whether x=0 is a regular singular point. determine whether x=1 is a regular singular point. are there any regular singular points that are complex numbers? justify conclusions

1. goalie2012

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

2. goalie2012

the answers I got for this one were 0, 1, and i, but I'm not sure how to justify them. Are those even right to begin with?

3. goalie2012

if you know this one, please answer, but I'm going to write the other one since I'm having more trouble with it.

4. oldrin.bataku

First, rewrite it in standard form:$$y''+\frac{x(x-1)}{x^3(x^2-1)^2(x^2+1)}y+\frac1{x^3(x^2-1)^2(x^2+1)}y=0$$

5. oldrin.bataku

To see if $$x=0$$ is ordinary, we see if the coefficients of $$y'$$ and $$y$$ are analytic at $$x=0$$; as you can see, $$y',y$$ are not analytic therefore $$x=0$$ is singular.

6. oldrin.bataku

To determine whether it is regular, we consider whether $$xp(x),x^2q(x)$$ are analytic at $$x=0$$ where $$p(x),q(x)$$ are the coefficients of $$y',y$$ respectively. Observe that neither $$xp(x),x^2q(x)$$ are analytic at $$x=0$$ (there is at least one factor of $$x$$ remaining in the denominator) thus this is an irregular singular point.

7. oldrin.bataku

for the above I meant the coefficients of $$y',y$$ were not singular:$$y''+\underbrace{\frac{x(x-1)}{x^3(x^2-1)^2(x^2+1)}}_{p(x)}y+\underbrace{\frac1{x^3(x^2-1)^2(x^2+1)}}_{q(x)}y=0$$Note that we can simplify $$p(x)$$:$$p(x)=\frac{x-1}{x^2(x^2-1)^2(x^2+1)}$$which unfortunately is still not analytic at $$x=0$$. Thus it is a singular point. Observe that $$xp(x)$$ is also not analytic:$$xp(x)=\frac{x-1}{x(x^2-1)^2(x^2+1)}$$thus $$x=0$$ is an irregular singular point.

8. oldrin.bataku

For $$x=1$$, consider simplifying $$p(x)$$ a little more:$$p(x)=\frac{x-1}{x^2(x^2-1)(x^2-1)(x^2+1)}=\frac1{x^2(x+1)(x^2-1)(x^2+1)}$$yet unfortunately it is still not analytic at $$x=1$$

9. oldrin.bataku

Similarly consider $$(x-1)p(x)$$:$$(x-1)p(x)=\frac{x-1}{x^2(x+1)(x^2-1)(x^2+1)}=\frac1{x^2(x+1)^2(x^2+1)}$$which is analytic at $$x=1$$! Now, check $$(x-1)^2q(x)$$:$$(x-1)^2q(x)=\frac{(x-1)^2}{x^3(x^2-1)^2(x^2+1)}=\frac1{x^3(x+1)^2(x^2+1)^2}$$which is also analytic at $$x=1$$ hence $$x=1$$ is a regular singular point

10. oldrin.bataku

To determine whether there are complex singular points, observe the roots of our denominator of $$p(x)$$:$$x^2(x+1)(x^2-1)(x^2+1)=0$$we observe we have singularities at $$x^2+1=0$$ i.e. $$x=\pm i$$. To determine whether they're regular, consider whether $$(x-i)p(x),(x+i)p(x)$$ still have the singular point:(x-i)p(x)=\frac{x-i}{x^2(x+1)^2(x^2+1)}=\frac1{x^2(x+1)^2(x+i)}\$$x+i)p(x)=\frac{x+i}{x^2(x+1)^2(x^2+1)}=\frac1{x^2(x+1)^2(x-i)}nope! Since \(q(x)$$'s denominator has the same root with the same multiplicity you can conclude $$(x\pm i)q(x)$$ will also be analytic at $$x=\pm i$$. Hence $$x=\pm i$$ are regular singular points.