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Thread: Calculus problem

  1. #1
    Newbie
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    Calculus problem

    1) $\displaystyle y^3 -y = 2x $
    show,
    $\displaystyle 9(x^2 - \frac{1}{27})\frac{d^2 y}{dx^2} + 9x\frac{dy}{dx}=y$

    i tried this few times but i couldn't get the answer


    2)
    $\displaystyle \int_0^\pi \frac{x}{1-sin(x)cos(\alpha)}dx= \frac{\pi(\pi -\alpha)}{sin(\alpha)}$

    with substitution and using $\displaystyle \int_0^a f(x) =\int_0^a f(a-x) $
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  2. #2
    Math Engineering Student
    Krizalid's Avatar
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    first one quite straightforward so i think you should show us some progress.

    as for the second, just put $\displaystyle x\longrightarrow\pi-x$ and your integral becomes $\displaystyle \int_{0}^{\pi }{\frac{\pi -x}{1-\sin (x)\cos (\alpha) }},$ thus it remains to compute $\displaystyle \int_{0}^{\pi }{\frac{dx}{1-\sin (x)\cos (\alpha) }}$ which shouldn't be that tough.
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  3. #3
    Master Of Puppets
    pickslides's Avatar
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    hi there

    I had a look at question 1

    Using implicit differenitation I found the following

    $\displaystyle \frac{dy}{dx} = \frac{2}{3y^2-1}$

    $\displaystyle \frac{d^2y}{dx^2} = \frac{-24}{(3y^2-1)^3}$

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

    Try to sub these values into the LHS side of the DE and simplyfy, hopefully you will get y.
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