Quantcast

A community for students.

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

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

  • This Question is Closed
  1. goalie2012
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  2. goalie2012
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 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?

  3. goalie2012
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 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.

  4. oldrin.bataku
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    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
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 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.

  6. oldrin.bataku
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 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.

  7. oldrin.bataku
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    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
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    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
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    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
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    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.

  11. Not the answer you are looking for?
    Search for more explanations.

    • Attachments:

Ask your own question

Sign Up
Find more explanations on OpenStudy
Privacy Policy

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
  • 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.

This is the testimonial you wrote.
You haven't written a testimonial for Owlfred.