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Two Questions

  1. Apr 1, 2009 #1
    What is the general solution of

    [itex]x^2y''+xy'-y=0[/itex]
    i tried a series solution [itex]y=\sum_{n=0}^{\infty} a_n x^n[/itex]
    and whitteld it down to

    [itex]\sum_{n=0}^{\infty} (n^2-1) a_n x^n=0[/itex] but this isn't getting me anywhere






    secondly how do i show that x=0 is a regular point of
    [itex](1-x^2)y''-2xy'+2y=0[/itex]

    the definition of a regular point is that :
    x0 is a regular point of y''+a1(x)y'+a0(x)y=0 if the coefficients a1 and a0 can be expressed as convergent power series in (x-x0) for this value of x0.

    so if i rearrange our equation into the desired form i get that

    [itex]a_1(x)=\frac{-2x}{1-x^2},a_2(x)=\frac{2}{1-x^2}[/itex]
    for x0=0, a1=0 and a2=2.

    what do i say to round of my argument. something like:

    "clearly 0 and 2 can be expressed as a convergent power series in x" or
    "clearly a1 and a0 arent singular for this value of x0"
    how do i round this off???
     
  2. jcsd
  3. Apr 1, 2009 #2

    lanedance

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    hi latentcorpse - does you series solution imply all [tex]a_n[/tex] are zero except for a certain n?
     
  4. Apr 2, 2009 #3

    HallsofIvy

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    A series solution is not the best way to do this. Are you required to use series?

    If so, look closely at lanedance's suggestion. If not, try y= xm for some number m.

    For your second question, remember that a geometric series, [itex]\sum r^n[/itex] has sum 1/(1- r).
     
  5. Apr 2, 2009 #4
    ok if i was just using series solution, i see now that [itex]a_1[/itex] will be non zero - is there any way of finding its value or is the solution just going to be [itex]y=a_1x[/itex]?

    i amen't required to use series solutions but it was a past paper exam question and i couldnt think of any other way to do it - why did you decide to use that substitution?



    i dont see how the geometric sum fits into this last bit either.
     
  6. Apr 2, 2009 #5

    HallsofIvy

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    You said you had [itex]\sum_{n=0}^{\infty} (n^2-1) a_n x^n=0[/itex].

    For Every n, either [itex]a_n= 0[/itex] or [itex]n^2- 1= 0[/itex].

    The reason I suggested xm is that (xm)'= mxm-1 and (xm)"= m(m-1)xm-1. so that x(xm)'= mxm and x2(xm)"= m(m-1)xm so every term has the same power of x.

    Finally, since the sum of the geometric series [itex]\sum ar^n[/itex] is [itex]a/(1- r)[/itex], [itex]2/(1-x^2)[/itex] can be written as [itex]\sum 2(x^2)^n= \sum 2x^{2n}[/itex]. [itex]2x/(1- x^2)[/itex] is just that multiplied by x: [itex]\sum 2x^{2n+1}[/itex].
     
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