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Math Help - Second order differential equation

  1. #1
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    Second order differential equation

    Hi everybody, I have absolutely no idea how to tackle this problem:

    Let p(x) be a solution of the differential equation y\prime\prime + yy\prime = x^{3} with initial conditions y(-1) = 1 and y\prime(-1)=2. Find p\prime\prime(-1) and p\prime\prime\prime(-1).

    Any help would be greatly appreciated.
    Thank you!
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  2. #2
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    Quote Originally Posted by gundu24 View Post
    Hi everybody, I have absolutely no idea how to tackle this problem:

    Let p(x) be a solution of the differential equation y\prime\prime + yy\prime = x^{3} with initial conditions y(-1) = 1 and y\prime(-1)=2. Find p\prime\prime(-1) and p\prime\prime\prime(-1).

    Any help would be greatly appreciated.
    Thank you!
    Did you try anything on this?

    Since p(x) satisfies the equation, p''(x)= x^3- p(x)p'(x). You know the values of x, y, and y' at -1 so you can directly calculate the right side.

    To find p'"(-1), Differentiate both sides of the equation above and evaluate at x=-1.
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  3. #3
    MHF Contributor chisigma's Avatar
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    The problem is quite interesting because, applying the procedure described by HalsofIvy, You can iteratively compute the quantities p^{(n)} (-1) and that allows You to obtain the Taylor expansion of p(x) around x=-1 ...

    p(x)= \sum_{n=0}^{\infty} \frac{p^{(n)} (-1)}{n!}\cdot (x+1)^{n} (1)

    Kind regards

    \chi \sigma
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  4. #4
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    Yes, that's a standard, though very slow and not very practical, method of finding power series solutions to differential equations.
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