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Math Help - a crazy Integral...

  1. #1
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    Question a crazy Integral...

    Hi!

    1. I have no idea how to solve this integral:





    2. I've got two solutions from different sources to this DE, what is the difference and why they chose to write the solutions like that (how do I know when to write them as a product or as a sum)?

    the equations is:

    y''''+y=0

    one solutions is:


    and the other:





    Thanks!
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  2. #2
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    Quote Originally Posted by dudinka View Post
    1. I have no idea how to solve this integral:

    Of course you won't have any idea if you don't know who are \mu,\,g,\,\lambda and c

    Constants?
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  3. #3
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    yes, thank you.
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  4. #4
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    Hello, dudinka!

    \int\frac{dx}{(mgx + L)^2 - c^2}

    Let: mgx + L \:=\:c\sec\theta\quad\Rightarrow\quad mg\,dx \:=\:c\sec\theta\tan\theta\,d\theta\quad\Rightarro  w\quad dx \:=\:\frac{c}{mg}\sec\theta\tan\theta\,d\theta
    . . and: . (mgx + L)^2 - c^2 \:=\:c^2\sec^2\!\theta - c^2 \:=\:c^2(\sec^2\!\theta -1) \:=\:c^2\tan^2\!\theta

    Substitute: . \int\frac{\frac{c}{mg}\sec\theta\tan\theta\,d\thet  a}{c^2\tan^2\theta} \:=\:\frac{1}{cmg}\int\frac{\sec\theta}{\tan\theta  }\,d\theta \:=\:\frac{1}{cmg}\int\csc\theta\,d\theta

    . . And we have: . \frac{1}{cmg}\ln|\csc\theta - \cot\theta| + K


    I'll let you back-substitute . . .

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  5. #5
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by dudinka View Post
    2. I've got two solutions from different sources to this DE, what is the difference and why they chose to write the solutions like that (how do I know when to write them as a product or as a sum)?

    the equations is:

    y''''+y=0

    one solutions is:


    and the other:
    I have no idea how the second formula came about. It isn't correct.

    To solve y^{\prime \prime \prime \prime} + y = 0
    you need to solve the characteristic equation:
    m^4 + 1 = 0

    This has as a solution the four 4th roots of -1:
    m = \frac{\sqrt{2}}{2} \pm i \frac{\sqrt{2}}{2}, -\frac{\sqrt{2}}{2} \pm i \frac{\sqrt{2}}{2}

    So the general solution is:
    y(x) = Ae^{\left ( \frac{\sqrt{2}}{2} + i \frac{\sqrt{2}}{2} \right ) x} + Be^{\left ( \frac{\sqrt{2}}{2} - i \frac{\sqrt{2}}{2} \right ) x} + Ce^{\left ( -\frac{\sqrt{2}}{2} + i \frac{\sqrt{2}}{2} \right ) x} + De^{\left ( -\frac{\sqrt{2}}{2} - i \frac{\sqrt{2}}{2} \right ) x}

    You may modify this form:
    y(x) = Ae^{\frac{\sqrt{2}}{2}x}e^{i \frac{\sqrt{2}}{2} x} + Be^{\frac{\sqrt{2}}{2}x}e^{-i \frac{\sqrt{2}}{2} x} + Ce^{-\frac{\sqrt{2}}{2}x}e^{i \frac{\sqrt{2}}{2} x} + De^{-\frac{\sqrt{2}}{2}x}e^{-i \frac{\sqrt{2}}{2} x}

    y(x) = e^{\frac{\sqrt{2}}{2}x} \left ( Ae^{i \frac{\sqrt{2}}{2} x} + Be^{-i \frac{\sqrt{2}}{2} x} \right ) + e^{-\frac{\sqrt{2}}{2}x} \left ( Ce^{i \frac{\sqrt{2}}{2} x} + De^{-i \frac{\sqrt{2}}{2} x} \right )

    Finally:
    y(x) = e^{\frac{\sqrt{2}}{2}x} \left ( A^{\prime}cos \left ( \frac{\sqrt{2}}{2} x \right ) + B^{\prime}sin \left ( \frac{\sqrt{2}}{2} x \right ) \right )  + e^{-\frac{\sqrt{2}}{2}x} \left ( C^{\prime}cos \left ( \frac{\sqrt{2}}{2} x \right ) + D^{\prime}sin \left ( \frac{\sqrt{2}}{2} x \right ) \right )

    These two forms represent the most commonly used forms for this kind of solution.

    -Dan
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  6. #6
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    Thumbs up

    thanks a lot!
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