Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: More Help yet again with Solving differential equations

  1. Mar 22, 2005 #1

    $id

    User Avatar

    Hi i posted last time about solving an equation regarding air resistance being proportional to the speed

    This time i need it to be proportional to the speed squared

    I have managed to get an equation but havent managed to solve it yet

    dv/dt -(k/m)v^2 = g


    Any ideas on how to solve this. I have tried all the methods i know, I think i need to substitute in. I will need more than just a hint here guys

    Thanks a lot

    sid
     
  2. jcsd
  3. Mar 22, 2005 #2
    The equation is seperable (you can seperate v and t). Any first order seperable equation can be reduced to a problem of indefinite integration:

    [tex] \int \frac{dv}{g - \frac{k}{m} v^2} = t + Const. [/tex]
     
  4. Mar 22, 2005 #3
    [tex] v^\prime = g + \frac{k}{m}v^2 \Longrightarrow \frac{v^\prime}{\frac{k}{m}v^2 + g} = 1[/tex]

    [tex]\Longrightarrow \int \frac{dv}{\frac{k}{m}v^2 + g} = \frac{m}{k} \int \frac{dv}{v^2 + \frac{gm}{k}} = \sqrt{\frac{m}{kg}} \arctan \left(v \sqrt{\frac{k}{mg} \right) = t + C[/tex]

    [tex] \Longrightarrow \arctan \left(v \sqrt{\frac{k}{mg}} \right) = \sqrt{\frac{kg}{m}}(t+C) \Longrightarrow v \sqrt{\frac{k}{mg}} = \tan \left( \sqrt{\frac{kg}{m}}(t+C)\right)[/tex]

    [tex] \Longrightarrow v = \sqrt{\frac{mg}{k}} \tan \left( \sqrt{\frac{kg}{m}}(t+C)\right)[/tex]
     
  5. Mar 22, 2005 #4

    dextercioby

    User Avatar
    Science Advisor
    Homework Helper


    A bit weird not to include among "all the methods" the separation of variables,which is the simplest possible. :rolleyes:

    Anyway,i hope you understood the solution Data gave you.

    Daniel.
     
  6. Mar 22, 2005 #5

    $id

    User Avatar

    Hmm i can follow Data logic despite never have seen anything like that before,

    I guess now the main problem is that equation gives me the velocity. for the model i am making the data readings are distances at a given time. Hence I will most probably to integrate that beast.

    All i know is that integral of tanx = - ln(cos x) + c or - ln (sec^2 x) or something like that

    sid
     
  7. Mar 22, 2005 #6

    dextercioby

    User Avatar
    Science Advisor
    Homework Helper

    [tex]\int \tan x \ dx=-\ln|\cos x|+C [/tex] it can be proven using the definition of tangent & the substitution [itex]\cos x=u [/itex].

    Daniel.
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook