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Math Help - general solution

  1. #1
    Member billym's Avatar
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    general solution

    <br />
\frac{d^2x}{dt^2} + \frac{5k}{m} \cdot x = 0<br />

    um... is this correct?

    <br />
x(t) = B~cos(\sqrt5\frac{k}{m})+C~sin(\sqrt5\frac{k}{m})<br />

    or just... wrong...
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by billym View Post
    <br />
\frac{d^2x}{dt^2} + \frac{5k}{m} \cdot x = 0<br />

    um... is this correct?

    <br />
x(t) = B~cos(\sqrt5\frac{k}{m})+C~sin(\sqrt5\frac{k}{m})<br />

    or just... wrong...
    If you mean
    x(t) = B~cos \left ( t \sqrt{\frac{5k}{m}} \right ) + C~cos \left ( t \sqrt{\frac{5k}{m}} \right )

    then yes, it is correct.

    -Dan
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