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Math Help - Integrating after seperating vars

  1. #1
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    Integrating after seperating vars

    Haven't done much integration for a couple of years and am a bit rusty,
    have this differential equation and have to integrate but can't get my head around the suggested substitution any help would be great, cheers.

    After seperating vars i get;

    (1/γv^2 - g).dv/dt = 1

    then it asks me to integrate both sides given v(0) = 0 , using the substitution;

    u= tanh^-1(sqrt(γ/g).v))

    can someone please explain next step for me, would be greatly appreciated, thanks.
    (γ is just a constant as is g for gravity)
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  2. #2
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    Re: Integrating after seperating vars

    Is it \displaystyle \left(\frac{1}{\gamma v^2 - g}\right)\,\frac{dv}{dt} = 1 or \displaystyle \left(\frac{1}{\gamma v^2} - g\right)\,\frac{dv}{dt} = 1?
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  3. #3
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    Re: Integrating after seperating vars

    sorrys its the first of your 2 options with -g on the denominator, i had trouble with the latex commands.
    cheers
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    Re: Integrating after seperating vars

    Quote Originally Posted by Prove It View Post
    Is it \displaystyle \left(\frac{1}{\gamma v^2 - g}\right)\,\frac{dv}{dt} = 1 or \displaystyle \left(\frac{1}{\gamma v^2} - g\right)\,\frac{dv}{dt} = 1?
    Quote Originally Posted by monster View Post
    sorrys its the first of your 2 options with -g on the denominator, i had trouble with the latex commands.
    cheers
    Having trouble with latex commands is no excuse for not using brackets correctly.

    Anyway

    \displaystyle \begin{align*} \left(\frac{1}{\gamma v^2 - g}\right)\frac{dv}{dt} &= 1 \\ \int{\left(\frac{1}{\gamma v^2 - g}\right)\frac{dv}{dt}\,dt} &= \int{1\,dt} \\ \int{\frac{1}{\gamma v^2 - g}\,dv} &= t + C_1 \\ -\frac{1}{\gamma}\int{\frac{1}{\frac{g}{\gamma} - v^2}\,dv} &= t + C_1\end{align*}

    Now make the substitution \displaystyle v = \sqrt{\frac{g}{\gamma}}\tanh{u} \implies dv = \sqrt{\frac{g}{\gamma}}\,\textrm{sech}^2\,{u}\,du and the DE becomes

    \displaystyle \begin{align*} -\frac{1}{\gamma}\int{\frac{\sqrt{\frac{g}{\gamma}}  \,\textrm{sech}^2\,{u}}{\frac{g}{\gamma} - \left(\sqrt{\frac{g}{\gamma}}\tanh{u}\right)^2}\,d  u} &= t + C_1 \\ -\frac{\sqrt{\frac{g}{\gamma}}}{\gamma}\int{\frac{ \textrm{sech}^2\,{u}}{\frac{g}{\gamma} - \frac{g}{\gamma}\tanh^2{u}}\,du}&= t + C_1 \\ -\frac{\sqrt{\frac{g}{\gamma}}}{g}\int{\frac{ \textrm{sech}^2\,{u}}{1 - \tanh^2{u}}\,du}  &= t + C_1 \\ -\frac{\sqrt{\frac{g}{\gamma}}}{g} \int{\frac{ \textrm{sech}^2\,{u}}{ \textrm{sech}^2\,{u}}\,du} &= t + C_1 \\ -\frac{\sqrt{\frac{g}{\gamma}}}{g} \int{1\,du} &= t + C_1\end{align*}

    Go from here. Make sure to convert your answer back to a function of \displaystyle v and write it with a single constant.
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  5. #5
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    Re: Integrating after seperating vars

    Alternatively, integrate \int \frac{1}{\frac{g}{\gamma}- v^2} dv using "partial fractions":
    \frac{1}{\frac{g}{\gamma}- v^2} = \frac{A}{\sqrt{\frac{g}{\gamma}}- v}+ \frac{B}{\sqrt{\frac{g}{\gamma}}+ v}.

    That will give you a solution in terms of logarithms rather than inverse hyperbolic functions.
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