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Math Help - Trivial Problem...

  1. #1
    T12
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    Trivial Problem...

    Apologies in advance, I'm rusty and having some trouble and really want to make more of an effort to keep up with my current year.

    Have a simple differential:

    dx/dt = x^3

    Initial condition x(0) = x0

    Apparently a simple integration by parts yields:

    x(t) = x0 / (1-2x0^2 t)^1/2

    I know i'm falling at something trivial and just not recognising the form. Could someone explain this expression? (how has it retained t terms?)
    Thank you for bearing with me. Once I get the ball rolling, maybe I can come up with better questions
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  2. #2
    A Plied Mathematician
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    I'm not sure I would say that you use integration by parts. The differential equation you have there yields to separation of variables thus:

    \displaystyle \frac{dx}{dt}=x^{3}, subject to x(0)=x_{0}. Thus,

    \displaystyle \frac{dx}{x^{3}}=dt, and hence

    \displaystyle \int\frac{dx}{x^{3}}=\int dt. Can you continue from here?
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  3. #3
    MHF Contributor chisigma's Avatar
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    Quote Originally Posted by Ackbeet View Post
    I'm not sure I would say that you use integration by parts. The differential equation you have there yields to separation of variables thus:

    \displaystyle \frac{dx}{dt}=x^{3}, subject to x(0)=x_{0}. Thus,

    \displaystyle \frac{dx}{x^{3}}=dt, and hence

    \displaystyle \int\frac{dx}{x^{3}}=\int dt. Can you continue from here?
    All that is correct if x_{0} \ne 0. In the case x_{0} = 0 the solution to the DE x^{'} = x^{3} is x=0...

    Kind regards

    \chi \sigma
    Last edited by chisigma; October 7th 2010 at 11:27 PM.
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  4. #4
    T12
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    Quote Originally Posted by Ackbeet View Post
    I'm not sure I would say that you use integration by parts. The differential equation you have there yields to separation of variables thus:

    \displaystyle \frac{dx}{dt}=x^{3}, subject to x(0)=x_{0}. Thus,

    \displaystyle \frac{dx}{x^{3}}=dt, and hence

    \displaystyle \int\frac{dx}{x^{3}}=\int dt. Can you continue from here?
    erm....

    t = -1/2x^2 + c
    1/x^2 = 2c -2t
    x^2 = 1/ (C-2t)
    x = 1/ sqt(C-2t)

    x(0) = xsubscript0
    which gives: C = 1/ (xsub0 ^2)

    substitute:
    x = 1/ sqt(1/xsub0^2 - 2t)

    I think the last step i got confused mostly:

    multiply by xsub0^2(rooted) on RHS top and bottom

    x = x0 / sqt(1-2x0^2 x t)
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  5. #5
    A Plied Mathematician
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    So is it all clear now?
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