1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: More Help yet again with Solving differential equations
  1. Need help yet again (Replies: 1)

Loading...