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A bit of confusion regarding instructions

  1. Nov 6, 2015 #1
    Mentor note: moved the thread to the homework forum

    So I was given a few problems to take home, to complete by the professor. My question is not regarding how to solve the problems but rather, if I understand the instructions.

    We are given the following instructions.

    Find the general solutions on (0,infinity) of
    2(x^2)y''-x(x-1)y'-y=o.

    I believe, since the problem is asking me to find solutions on the interval (0,infinity), we do not need to worry about the regular singular point at x=0 (power series expansion does not exist at x=0). So my solutions will be in the form
    of power series. Method of Frobenius is not required ( solutions will have the form ΣCnX^(n+r)). Regular power series solutions is what the problem is asking me correct?

    Thanks.
     
    Last edited by a moderator: Nov 8, 2015
  2. jcsd
  3. Nov 8, 2015 #2
    bump. any help?
     
  4. Nov 8, 2015 #3

    BvU

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    No typos ? Tried to cheat using WolframAlpha ?
     
  5. Nov 8, 2015 #4

    HallsofIvy

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    Yes, since the only singular point, x= 0, is not in the domain, you do not need to use Frobenius.
     
  6. Nov 8, 2015 #5

    pasmith

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    No, the Method of Frobenius is the correct method. Restriction to [itex](0, \infty)[/itex] guarantees that [itex]x^{n+r}[/itex] will actually be real, whatever [itex]r[/itex] is.
     
  7. Nov 8, 2015 #6
    So I'm definitely overthinking this simple problem. The problem says to find general solutions on (0, infinity). (multiple problems/thats where the find general solutions came from). Since I am not including x=0 (i think), I can ignore the singular point and solve using regular power series solutions. If I do it this way, I get one solution. If i use Method of Frobenius, I get 2 solutions that are linearly independent. I'm basically teaching myself :/.

    I know that x=0, a singular point. However, I was reading my differential book, Zill: A first course in differential, p. 254/example 3.

    The DE given is:

    2xy'' + (1+x)y' + y = 0. They list no interval where we have to find solution.

    On the bottom of the page, this is what is listed.

    The series converges for x≥0; as given, the series is not defined for negative values of x because of the presence of x^1/2. For r2=0, a second solution is....

    Now this is the part that made me question myself.

    On the interval (0,infinity) the general solution is y=C1y1(x) + C2y2(x).

    For this example, the author used Method of Frobenius. If i used regular power series, 1 solution disappears.

    From this, can I make the assumption that I have to use Method of Frobenius, when I need solutions on the interval (0, infinity)?

    Thanks for everyone that replied.
     
  8. Nov 8, 2015 #7
    Also, I encountered something that was different in every other problem I did so far. This is a separate problem.

    When I have a singular point, and i am told to find the solution at this point, in this case the singular point happens at x=0. For my Indicial Equation, I get r=0, with multiplicity 2.
    This is the only r I get. I know that by the Frobenius Theorem, I am guaranteed atleast 1 solution. However, the r=0 with multiplicity 2. Does this follow the same idea of the Cauchy Euler Differential Equations, when the roots of the auxiliary equations are equal?

    Do i just multiply my first answer by a logarithm?
     
  9. Nov 8, 2015 #8

    vela

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    While x=0 may not be in the domain of the solution, you're expressing the solution as a series expanded about x=0, a regular singular point, so you need to use the method of Frobenius.
     
  10. Nov 8, 2015 #9
    Thanks Vela, I consulted the ODE book by Coddington. It was exactly what you were talking about.

    I also found something very interesting. When using the method of Frobenious, if we get an r that has multiplicity greater than 1, then our second solution will have a logarithmic term. It is in the style of solving a Cauchy-Euler ODE with repeated roots.

    Really exciting how such a simple problem, can make you learn so much.
     
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