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Solving an ODE: The Pwer Series and Seperation of Variables

  1. Nov 30, 2008 #1

    TFM

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    1. The problem statement, all variables and given/known data

    Solve the following equation by a power series and also by separation of variables. Check that the two agree.

    2. Relevant equations

    N/A

    3. The attempt at a solution

    Power Series:

    [tex] (1+x) \frac{dy}{dx} = y [/tex]

    [tex] (1+x) \frac{1}{dx} = y \frac{1}{dy} [/tex]

    The power series is:

    [tex] (1+x) \equiv 1+x+0x^2+0x^x [/tex] ...

    Thus

    [tex] \frac{1 + x}{dx} = \frac{y}{dy} [/tex]


    Separation by Variables:

    [tex] (1+x) y' = y [/tex]

    [tex] y' = \frac{1}{((1+x)} y [/tex]

    [tex] x'=g(t)h(x) [/tex]

    [tex] H(x) = G(t) + C [/tex]

    [tex] H=\int \frac{dx}{h(x)} ; G=\int g(t)dt [/tex]

    [tex] H=\int \frac{dy}{y} \equiv \int \frac{1}{y} dy = ln y [/tex]

    [tex] G = \int \frac{1}{1+x} dx = ln(1 + x) [/tex]

    [tex] ln y = ln (1+x) + c [/tex]

    [tex] y = 1 + x + c [/tex]

    These two methods haven't agreed for this question. i think the problem lays in my Power Series.

    Anyone got any idesa?

    TFM
     
  2. jcsd
  3. Nov 30, 2008 #2

    CompuChip

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    Your second solution looks right, up to
    ln(y) = ln(1 + x) + c
    But if you exponentiate that, you won't get
    y = 1 + x + c,
    do that step again.

    As for your real question, I suppose that they mean: plug in a solution
    [tex]y(x) = \sum_{n = 0}^\infty a_n x^n[/tex]
    and determine the coefficients [itex]a_n[/itex] from the differential equation.
    Note that you can differentiate by terms, so
    [tex]\frac{dy}{dx} = \sum_{n = 0}^\infty n a_n x^{n - 1}[/tex]
    etc.
     
  4. Nov 30, 2008 #3

    TFM

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    Indeed it won't:

    ln(y) = ln(1 + x) + c

    take exponentials of both sides leaves:

    [tex] y = 1 + x + e^c [/tex]

    Okay so for the power series:

    [tex] y(x) = \sum_{n = 0}^\infty a_n x^n [/tex]

    so would the coeffieients coming from here:

    [tex] (1+x) \frac{1}{dx} = y \frac{1}{dy} [/tex]

    be 1 and 1, since you have x^0 has a coefficient of 1, and x has a coefficent of 1 also?

    TFM
     
  5. Nov 30, 2008 #4

    CompuChip

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    No,
    e^(ln(1 + x) + c) = e^ln(1+x) * e^c = ...

    And what do you mean by: "the coefficients being 1 and 1"? You have infinitely many of them! Start by plugging the sum into
    [tex]
    (1+x) \frac{dy}{dx} = y [/tex]
    - what do you get?
     
  6. Nov 30, 2008 #5

    TFM

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    Since e^c is another constant, wouldn't it not be:

    = k + kx

    ???

    So I should have:

    [tex] (1+x) \frac{dy}{dx} = \sum_{n = 0}^\infty a_n x^n [/tex]

    ???

    TFM
     
  7. Nov 30, 2008 #6

    gabbagabbahey

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    Yes, that would be fine.

    Well, that is a true statement, but not very useful to you in that form. If [tex]y(x)= \sum_{n = 0}^\infty a_n x^n[/tex]...what is [tex]\frac{dy}{dx}[/tex]?
     
  8. Dec 1, 2008 #7

    TFM

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    Would this be:

    [tex] y(x)= \sum_{n = 0}^\infty a_n x^n [/tex]

    [tex] \frac{dy}{dx} = \sum_{n = 0}^\infty a_n(n - 1) x^{n-1} [/tex]

    Does this look right???

    TFM
     
  9. Dec 1, 2008 #8

    CompuChip

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    Almost; is [itex]2 x^2[/itex] the derivative of [itex]x^3[/itex]?

    Now write down your complete differential equation.
     
  10. Dec 1, 2008 #9

    TFM

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    Okay so:

    x^3 differentiates to 3x^2

    [tex] \frac{dy}{dx} = \sum_{n = 0}^\infty a_n(n) x^{n-1} [/tex]

    so would this be:

    [tex] \frac{dy}{dx} = \sum_{n = 0}^\infty a_n(n) x^{n-1} + a_{n-1}(n+1) x^{n} + a_{n-2}(n+2) x^{n + 1} + ... [/tex]

    ???

    TFM
     
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