Got Homework?
Connect with other students for help. It's a free community.
Here's the question you clicked on:
 0 viewing
consider the differential equation: x^3(x^21)^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
 10 months ago
 10 months ago
consider the differential equation: x^3(x^21)^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
 10 months ago
 10 months ago

This Question is Closed

goalie2012Best ResponseYou've already chosen the best response.0
\[x ^{3}(x^21)^2(x^2+1)y''+(x1)xy'+y=0\]
 10 months ago

goalie2012Best ResponseYou've already chosen the best response.0
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?
 10 months ago

goalie2012Best ResponseYou've already chosen the best response.0
if you know this one, please answer, but I'm going to write the other one since I'm having more trouble with it.
 10 months ago

oldrin.batakuBest ResponseYou've already chosen the best response.1
First, rewrite it in standard form:$$y''+\frac{x(x1)}{x^3(x^21)^2(x^2+1)}y+\frac1{x^3(x^21)^2(x^2+1)}y=0$$
 10 months ago

oldrin.batakuBest ResponseYou've already chosen the best response.1
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.
 10 months ago

oldrin.batakuBest ResponseYou've already chosen the best response.1
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.
 10 months ago

oldrin.batakuBest ResponseYou've already chosen the best response.1
for the above I meant the coefficients of \(y',y\) were not singular:$$y''+\underbrace{\frac{x(x1)}{x^3(x^21)^2(x^2+1)}}_{p(x)}y+\underbrace{\frac1{x^3(x^21)^2(x^2+1)}}_{q(x)}y=0$$Note that we can simplify \(p(x)\):$$p(x)=\frac{x1}{x^2(x^21)^2(x^2+1)}$$which unfortunately is still not analytic at \(x=0\). Thus it is a singular point. Observe that \(xp(x)\) is also not analytic:$$xp(x)=\frac{x1}{x(x^21)^2(x^2+1)}$$thus \(x=0\) is an irregular singular point.
 10 months ago

oldrin.batakuBest ResponseYou've already chosen the best response.1
For \(x=1\), consider simplifying \(p(x)\) a little more:$$p(x)=\frac{x1}{x^2(x^21)(x^21)(x^2+1)}=\frac1{x^2(x+1)(x^21)(x^2+1)}$$yet unfortunately it is still not analytic at \(x=1\)
 10 months ago

oldrin.batakuBest ResponseYou've already chosen the best response.1
Similarly consider \((x1)p(x)\):$$(x1)p(x)=\frac{x1}{x^2(x+1)(x^21)(x^2+1)}=\frac1{x^2(x+1)^2(x^2+1)}$$which is analytic at \(x=1\)! Now, check \((x1)^2q(x)\):$$(x1)^2q(x)=\frac{(x1)^2}{x^3(x^21)^2(x^2+1)}=\frac1{x^3(x+1)^2(x^2+1)^2}$$which is also analytic at \(x=1\) hence \(x=1\) is a regular singular point
 10 months ago

oldrin.batakuBest ResponseYou've already chosen the best response.1
To determine whether there are complex singular points, observe the roots of our denominator of \(p(x)\):$$x^2(x+1)(x^21)(x^2+1)=0$$we observe we have singularities at \(x^2+1=0\) i.e. \(x=\pm i\). To determine whether they're regular, consider whether \((xi)p(x),(x+i)p(x)\) still have the singular point:$$(xi)p(x)=\frac{xi}{x^2(x+1)^2(x^2+1)}=\frac1{x^2(x+1)^2(x+i)}\\(x+i)p(x)=\frac{x+i}{x^2(x+1)^2(x^2+1)}=\frac1{x^2(x+1)^2(xi)}$$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.
 10 months ago
See more questions >>>
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.