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Math Help - Particular Inegral

  1. #1
    Member billym's Avatar
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    Particular Inegral

    m\frac {d^2x}{dt^2} + 5kx = -mgsin\alpha + 11kl_0<br />

    I am trying to find the general solution.

    Setting
    \omega^2=\frac{k}{m}, I found x_c to be:

    x(t) = A~cos \left ( \sqrt 5 \omega t + \phi \right )<br />

    I have no idea how to find x_p...

    What trial solution would you suggest using?
    Last edited by billym; April 5th 2008 at 01:44 PM.
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  2. #2
    Moo
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    Hello,

    You can consider that x is a constant. Thus its second derivate will be null and you'll have an equation with xp

    I hope for you it's true that \omega^2=\frac{5k}{m}, because this is something i can't do ^^
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  3. #3
    Member billym's Avatar
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    I am totally new at this... I still have no idea.

    I divided the original equation through by m to get:

    \frac {d^2x}{dt^2} + 5\omega^2 x = -gsin\alpha + 11\omega^2 l_0

    am i getting any closer?
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  4. #4
    Moo
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    I really don't know for xc... I don't remember the rules for general solution ! If you're not sure, derivate again to see if you get xc so that xc is solution of dxc/dt+5wxc=0

    However, for xp :

    xp is a particular solution. Let X be a constant satisfying the equation.

    The second derivate of X is 0.

    We have 5kX = -mgsin\alpha + 11kl_0

    -> X=x_p=\frac{-mgsin\alpha + 11kl_0}{k}
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  5. #5
    Member billym's Avatar
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    I don't get it. Where did the 5 go?

    I did this:

    <br />
m\frac {d^2x}{dt^2} + 5kx = -mgsin\alpha + 11kl_0<br />

    <br />
\frac {d^2x}{dt^2} + 5\omega^2 x = -gsin\alpha + 11\omega^2 l_0<br />

    <br />
\frac{d^2x}{dt^2} + 5\omega^2 x = \omega^2 (11l_0-\frac{mg}{k}sin\alpha)<br />

    <br />
x_p=\frac{1}{5}(11l_0-\frac{mg}{k}sin\alpha)<br />

    How's that? Anyone? Anyone?
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  6. #6
    Moo
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    Am sorry, i forgot the 5...
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  7. #7
    Member billym's Avatar
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    So we pretty much have the same answer. Thanks for the help!
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  8. #8
    Flow Master
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    Quote Originally Posted by billym View Post
    I don't get it. Where did the 5 go?

    I did this:

    <br />
m\frac {d^2x}{dt^2} + 5kx = -mgsin\alpha + 11kl_0<br />

    <br />
\frac {d^2x}{dt^2} + 5\omega^2 x = -gsin\alpha + 11\omega^2 l_0<br />

    <br />
\frac{d^2x}{dt^2} + 5\omega^2 x = \omega^2 (11l_0-\frac{mg}{k}sin\alpha)<br />

    <br />
x_p=\frac{1}{5}(11l_0-\frac{mg}{k}sin\alpha)<br />

    How's that? Anyone? Anyone?
    Relax guy. Be vague.

    Moo dropped a 5. It was a simple error, easily spotted. As you did. If you understand what Moo was doing to get the particular solution, then all's well. So go back, put the 5 in where it's meant to go. There's the answer.

    Life's full of small mistakes, most easily spotted and corrected for. The world keeps turning. Don't have cow, man.
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  9. #9
    Moo
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    Don't have cow, man.
    Moo
    Thanks for explaining it more precisely than i did ^^
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  10. #10
    Member billym's Avatar
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    I really wasn't having a cow man! I seriously wasn't sure if she dropped the 5 on purpose or not. This site feels like I'm doing my homework with God, so I just assume everyone else is right.
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  11. #11
    Moo
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    Quote Originally Posted by billym View Post
    I really wasn't having a cow man! I seriously wasn't sure if she dropped the 5 on purpose or not. This site feels like I'm doing my homework with God, so I just assume everyone else is right.
    Not at all, we're all equal in front of maths

    errare humanum est

    Btw, what does "have a cow" means ? ^^'
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  12. #12
    Member billym's Avatar
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    He was likening my concern over your missing 5 to the ordeal of giving birth to a cow.
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