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Thread: Integration by parts and IVPs

  1. #1
    Oct 2008

    Integration by parts and IVPs

    I have been trying to do these questions for ages now and have only got more and more confused...

    First off an integration by parts problem:
    Use integration by parts to prove that
    $\displaystyle \int{x^3\exp{\frac{x^2}{2}}dx} = x^2\exp{\frac{x^2}{2}} - 2\exp{\frac{x^2}{2}} + c$
    I know how to do integration by parts (or I thought I did!) but I just keep getting a longer and longer equation that doesn't look anything like the proper end result.

    Secondly, two initial value problems I am having trouble with. I guessed I was meant to use Laplace transforms but I don't know what to do with $\displaystyle y^-3$ in the first, and the second I just can't do for some reason. Also in both I don't know how to use the initial value since before we were always given y(0) = [whatever], which could be slotted into the laplace transform.

    $\displaystyle \frac{dy}{dx} = y^-3(x^2 - x^-2)$ given y(1) = 2

    $\displaystyle xy'(x) + y(x) = 1$ given y(1) = 2

    Hope you guys can help
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  2. #2
    MHF Contributor
    Nov 2008

    $\displaystyle \int{x^3\exp{\frac{x^2}{2}}dx}$ can be solved by parts using
    $\displaystyle u(x)= x^2$ and $\displaystyle dv = x\exp{\frac{x^2}{2}}dx = d\left(exp{\frac{x^2}{2}}\right)$

    $\displaystyle \frac{dy}{dx} = y^{-3}(x^2 - x^{-2})$ given y(1) = 2, can be solved without using Laplace transforms but maybe it is the goal of your exercise ...
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