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Falling skydiver differential equation

  1. Apr 12, 2010 #1
    1. The problem statement, all variables and given/known data
    The velocity v(t) of a skydiver falling to the ground is governed by

    m dv/dt = mg - kv^2

    where m is the skydiver's mass, g is the acceleration due to gravity, k > 0 is the drag
    coefficient, and v(t) >= 0.

    Solve this equation for v(t) with the initial condition v(0) = 0.


    2. Relevant equations

    I have been doing other diff equations but when its a word problem it confuses me.. So, i treat m, g and k as constants.
    I rearranged equation into this:

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

    so now i can use integrating factor..

    mu(x) = e^integ (k/m) ...but im going to stop here already, in case im already on wrong track.. The fact that v is squared changes things, no? (cant use integrating factor?) If this is ok though, i will proceed.. Thanks alot for any help/tips

    3. The attempt at a solution
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Apr 12, 2010 #2

    rock.freak667

    User Avatar
    Homework Helper

    Instead of putting it in that form, put it in a different form

    [tex]\frac{dv}{dt} = g- \frac{k}{m}v^2[/tex]

    [tex]\frac{1}{g- \frac{k}{m}v^2} \frac{dv}{dt} = 1[/tex]

    Split into partial fractions.
     
  4. Apr 12, 2010 #3
    I dont understand the reason for putting it in that form?
     
  5. Apr 12, 2010 #4

    rock.freak667

    User Avatar
    Homework Helper

    You had it in the form of

    dv/dt + (k/m)v2 = g


    this will not give you an integrating factor of e∫(k/m)dt


    had the v2, just been 'v', then it would work fine.
     
  6. Apr 12, 2010 #5
    ok thankjs, .so in the form you have it i can get an integ factor? I mean, this is the method i need to use to solve right? (integrating factor) (as opposed to separable equation)
     
  7. Apr 12, 2010 #6

    Mark44

    Staff: Mentor

    Forget integrating factors for this problem. The integrating factor business applies to problems in this form:
    y' + P(x)y = Q(x)
    What you have doesn't fit in this form.

    rock.freak667 has shown you how to separate the equation and has given you advice on how to continue.
     
  8. Apr 14, 2010 #7
    I need to confirm something before I proceed solving: in this equation, t in independent variable, v is dependent variable. k,m, and g are all constants. Is that correct?

    Also, if i compare this to the previous equations ive been solving, t is like x (independent), and v is like y (dependent). and also t will always be independent because its time in these motion differential equations right? thanks
     
  9. Apr 14, 2010 #8

    rock.freak667

    User Avatar
    Homework Helper

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