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Solve Series and Find General Solution

  1. Apr 5, 2006 #1

    Zem

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    These are differential equations problems. They are now "redo's", so I have hints from the grader that I don't understand.

    First problem:
    (x^2 + 1)y'' + 6xy' + 4y = 0
    After isolating C_n+2, I have this.

    (n+2)(n+1)C_n+2 * X^n = -n(n-1)C_n * X^n - 6nC_n * X^n + 4C_n
    C_n+2 = [-n(n-1)C_n - 6nC_n + 4C_n] / (n+2)(n+1)

    Since the grader is saying my solution for C_n+2 is equal to
    [-(n+4) / (n+2)]C_n, it appears that I am right up until this point.

    I have simplified the numerator, and it still doesn't look right.
    C_n+2 = C_n[-n(n-1) - 6n + 4] / (n+2)(n+1)
    It looks like it should be
    C_n+2 = C_n[(-n)^2 - 5n + 4]
    C_n+2 = C_n[(n-4)(n-1)] / (n+2)(n+1)
    Is there a (n+1) in the numerator that cancels?

    Second problem:
    y'' + xy' + y = 0
    y'' = Sigma_n=2 n(n-1)C_nX^n-2
    y' = Sigma_n=1 nC_nX^n-1
    y = Sigma_n=0 C_nX^n
    Sigma_n=2 n(n-1)C_nX^n-2 + Sigma_n=1 nC_nX^n-1 + Sigma_n=0 C_nX^n = 0

    Sigma_n=0 (n+2)(n+1)C_n+2X^n + Sigma_n=0 nC_nX^n + Sigma_n=0C_nX^n = 0

    X^n[(n+2)(n+1)C_n+2 + nC_nX^n + C_nX^n = 0

    X^n(n+2)(n+1)C_n+2 = -X^n(n)C_nX^n - X^nC_n
    C_n+2 = -X^n(n)C_n - X^nC_n
    C_n+2 = (-nC_n - C_n) / (n+2)(n+1)
    Is this incorrect? The grader says this should be
    C_n+2 = -C_n / (n+2)
     
    Last edited: Apr 5, 2006
  2. jcsd
  3. Apr 5, 2006 #2

    dextercioby

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    You got

    [tex] C_{n+2}=\frac{-nC_{n}-C_{n}}{(n+1)(n+2)} =-\frac{C_{n}(n+1)}{(n+1)(n+2)} =-\frac{C_{n}}{n+2}} [/tex]

    which is what you were supposed to get, right...?

    Daniel.
     
  4. Apr 5, 2006 #3

    Zem

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    Yes, so I'll make my patterns...

    [tex] C_{n+2}=\frac{-C_{n}}{(n+2)}[/tex]
    [tex] n=0: C_{2} = \frac{-C_{0}}{2}[/tex]
    [tex] n=1: C_{3} = \frac{-C_{1}}{3}[/tex]
    [tex] n=2: C_{4} = \frac{(-1)C_{0}}{2*4}[/tex]
    [tex] n=3: C_{5} = \frac{(-1)C_{1}}{3*5}[/tex]
    [tex] n=4: C_{6} = \frac{(-1)C_{0}}{4*6}[/tex]

    The even ones look like [tex]C_{2n} = \frac{(-1)^nC_{0}}{2^{n}n!}[/tex]

    But the series for the odd ones is strange. Here is what is in the denominator when n=1, n=3, n=5:
    1*3*5*7
    This looks like [tex](2n+1)(2n-1)(2n-3)[/tex]
    How do I represent that in the series?
     
  5. Apr 6, 2006 #4

    dextercioby

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    To me it looks like

    [tex] (2k \pm 1) !! [/tex]

    Daniel.
     
  6. Apr 8, 2006 #5

    Zem

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    Thanks! I have found the radius of convergence and general solution of the second problem.
    Back to the first problem. Note:

    Code (Text):
    [tex](x^2 + 1)y'' + 6xy' + 4y = 0[/tex]
    [tex](\sum_{n = 2}^{\infty} \{n(n-1)C_n*X^n}) + (\sum_{n = 2}^{\infty} \{n(n-1)C_n*X^{{n-2}}) + 6(\sum_{n=1}^{\infty} \{n}{C_n}*X^n}) + 4(\sum_{n=0}^{\infty} \{C_n*X^n}) = 0[/tex]
    [tex](\sum_{n = 2}^{\infty} \{n(n-1)C_n*X^n}) + (\sum_{n = 0}^{\infty} \{(n+2)(n+1)C_{n+2}*X^n}) + 6(\sum_{n=0}^{\infty} \{n}{C_n}*X^n}) + 4(\sum_{n=0}^{\infty} \{C_n*X^n}) = 0[/tex]
    [tex](\sum_{n = 0}^{\infty} \{n(n-1)C_n*X^n}) + (\sum_{n = 0}^{\infty} \{(n+2)(n+1)C_{n+2}*X^n}) + 6(\sum_{n=0}^{\infty} \{n}{C_n}*X^n}) + 4(\sum_{n=0}^{\infty} \{C_n*X^n}) = 0[/tex]
    [tex](n+2)(n+1)C_{n+2}*X^n = -n(n-1)C_n * X^n - 6nC_n * X^n + 4C_n * X^n[/tex]

    [tex](n+2)(n+1)C_{n+2} = C_n[-n(n-1) - 6n + 4]/(n+2)(n+1)[/tex]
    [tex]C_{n+2} = C_n[(-n)^2 - 5n + 4]/(n+2)(n+1)[/tex]
    [tex]C_{n+2} = C_n[(n-4)(n-1)]/(n+2)(n+1)[/tex]

    The simplified equation for C_n+2 is
    Code (Text):
    [tex]C_{n+2} = C_n(n-4)/(n+2)[/tex]
    but I don't see how because (n+1) does not cancel (n-1). Have I made a mistake somewhere?
     
    Last edited: Apr 8, 2006
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