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Thread: Trying to solve dv/dt=g-kx

  1. #1
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    Trying to solve dv/dt=g-kx

    The solution x(t) is needed for $$\dot v=g-kx$$

    These apply: $$v(0) = x(0) = 0$$ and $$v = \dot x$$

    I went with $$\frac {dv} {dt}=v\frac {dv} {dx}$$ and got

    $$\frac {dv} {dt}=g-kx$$
    $$v\frac {dv} {dx}=g-kx$$
    ...
    $$v^2 = 2gx - kx^2 + C$$

    At v(0)=0 the constant C=0

    Now is $$v^2=(\frac {dx} {dt})^2$$ the only way to proceed or is there something simpler?

    Thanks
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  2. #2
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    Re: Trying to solve dv/dt=g-kx

    so we've got

    $\ddot{x} + k x = g$

    let's solve the homogeneous equation first

    $\ddot{x}+k x=0$

    the characteristic equation is

    $s^2 + k= 0$

    and this has solutions

    $s = \pm i\sqrt{k}$

    this in turn leads to time domain solutions of

    $x_h(t) = c_1 \cos( \sqrt{k} t) + c_2 \sin( \sqrt{k} t)$

    plugging this into the original equation and solving we get

    $x(t) = x_h(t) + \dfrac g k = c_1 \cos( \sqrt{k} t) + c_2 \sin( \sqrt{k} t) + \dfrac g k$
    Thanks from topsquark, sakonpure6 and flyingdog
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  3. #3
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    Re: Trying to solve dv/dt=g-kx

    Quote Originally Posted by romsek View Post
    ...
    has solutions

    $$s = \pm i\sqrt{k}$$

    this in turn leads to time domain solutions of

    $$x_h(t) = c_1 \cos( \sqrt{k} t) + c_2 \sin( \sqrt{k} t)$$
    Thanks for answering, this is good stuff

    I have not used this kind of method before and trying to search info at the moment. How does time domain solutions work?
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    Thanks from topsquark and flyingdog
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