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Math Help - Solve for nth root first, or initial value first?

  1. #1
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    Solve for nth root first, or initial value first?

    I know how to handle seperable first order ODEs, nth roots, and initial value problems. But I'm a touch confused about what to do when they all appear in the same problem.

    For this:

    xy^3y' - 2x = 3

    where:

    x > 0

    y(1) = -1


    We can get the general solution of

    y = (12ln(x) + 8x + C)^{\frac{1}{4}}


    But I'm a bit confused at to whether I should be getting the four roots of y with C first, or if I should be trying to solve for C using the initial value given first, etc...

    How is this best handled?
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  2. #2
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    Re: Solve for nth root first, or initial value first?

    Solve for C first.
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  3. #3
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    Re: Solve for nth root first, or initial value first?

    With, of course, the information you are given. I have no idea what "four roots of y" you are talking about!
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    Re: Solve for nth root first, or initial value first?

    Quote Originally Posted by HallsofIvy View Post
    With, of course, the information you are given. I have no idea what "four roots of y" you are talking about!
    Sorry, working with y^4 = ..., there are four roots of y.
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  5. #5
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    Re: Solve for nth root first, or initial value first?

    Quote Originally Posted by pickslides View Post
    Solve for C first.
    I have this:

    y(1) = -1 = (C + 8)^{1/4}

    Do I solve for four values of C, do I just raise both sides by a power of 4 to get a single value of C, or something else?
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  6. #6
    MHF Contributor chisigma's Avatar
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    Re: Solve for nth root first, or initial value first?

    Quote Originally Posted by Lancet View Post
    I have this:

    y(1) = -1 = (C + 8)^{1/4}

    Do I solve for four values of C, do I just raise both sides by a power of 4 to get a single value of C, or something else?
    The problem formulation that immediately conducts to the solution is: what is the number x one of the fourth roots of which is -1?...

    Kind regards

    \chi \sigma
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