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Homework Help: Skydiver differential equation

  1. May 24, 2010 #1
    1. The problem statement, all variables and given/known data
    the velocity of a skydiver falling to the ground is governed by:
    m dv/dt = mg - kv^2

    solve for v(t) with initial conditions v(0) = 0

    i rearranged equation into standard form:

    v' + (k/m)v^2 = g

    so i got an integrating factor, mu(x) = e^integ(k/m), but does the fact that the v is squared prohibit me from using an integrating factor?

    thanks
    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
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. May 24, 2010 #2

    Cyosis

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    Strictly a non linear differential equation doesn't prohibit you from using an integrating factor. It does however prohibit you from using the integrating factor in the way you did. You used a 'formula' for an integrating factor which was derived for linear ODEs, which does not apply in this case.

    Hint: Note that the equation is separable.
     
  4. May 24, 2010 #3
    No, you can't do that. Need y'+ay=b(x) for that. Surely you can take the expression:

    [itex]a\frac{dv}{dt}=G(v)[/itex]

    and separate variables to get [itex]a\frac{dv}{G(v)}=dt[/itex].

    Try that.
     
  5. May 24, 2010 #4
    how do you know just from looking at an equation if it is separable or not? What do you look for specifically? What is a giveaway that tells you that, nope this equation cant be separable also.
    thanks
     
  6. May 24, 2010 #5

    Cyosis

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    You look whether or not you can put all v-s on one side of the equation and all t-s on the other side.

    And yes this equation is separable.
     
  7. May 24, 2010 #6
    m dv/dt = mg - kv^2

    dv/dt + kv^2 / m = g

    dv + v^2 = mg/k dt

    its separated but it looks sort of weird to me??
    also, can u please clarify this: that k,g and m are all constants, right? thanks
     
  8. May 24, 2010 #7
    but if this is correct i will proceed, thank u.
     
  9. May 24, 2010 #8

    Cyosis

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    No that is not correct. Compare your equation to the one in post #3.
     
  10. May 24, 2010 #9
    i glossed over post 3 because i wasnt sure where the a comes from, and why the caps G.
    Is the a acceleration?
    thanks
     
  11. May 24, 2010 #10

    Cyosis

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    'a' is just a constant and with G(v) he means a function that depends on v. He might as well have called it f(v) or g(v) etc.
     
  12. May 24, 2010 #11
    Sorry you're having problems with my notation. You have got to review Separation of Variables today in some elementary DE text book. Tons in the library. You need to get the DE in the form [itex]A(x)dx+B(y)dy=0[/itex] to separate the variables. You have:

    [itex]mdv=(mg-kv^2)dt[/itex]

    now I bet you can put that into the form [itex]A(v)dv=B(t)dt[/itex] and thus separated. Wouldn't that just be [itex]\frac{m}{mg-kv^2}dv=dt[/itex]
     
  13. May 24, 2010 #12
    actually i do understand what you are doing, but i think all the constants are confusing me.

    so, just to clarify: m, g and k are all constants right?

    Also I was trying to get all the constants to go with "dt", but am i right in saying that it doesnt matter what side of the equation i put the constants?

    (i was trying to get it so that equations was v^2 dv = "something" dt and it was really difficult.

    So i now have:

    m/mg-kv^2 dv = (1)dt

    integrate both sides...

    so, because m,g and k are all constants i can take them out of the integral right?

    m/(mg-k) integ 1/(v^2) = (1)dt

    m/(mg-k) * -(1/v) = t+c

    m/(mg-k)v = t+c

    v = (t/m) (mg-k) + (c/m) (mg-k)

    is this correct so far? thanks
     
  14. May 25, 2010 #13

    Cyosis

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    You really need to brush up on basic algebraic manipulations.

    What you're claiming here is that [itex]mg-kv^2=(mg-k)v^2[/itex]. I suggest you work out the brackets so you can see it's wrong.
     
  15. May 26, 2010 #14
    i have the following equality, that i need to now integrate (correctly)

    integ: m/(mg-kv^2) dv = integ :(1)dt

    heres how i integrated LHS...

    u = mg-kv^2
    du = -2kv

    dv = m
    v = v

    integ: m/(mg-kv^2) dv = mg-kv^2(v) - integ (v)(-2kv)

    ................................= mg-kv^2(v) - (-2)((1/3)(v^3)

    simplify..

    ................................ = mgv-kv^3 +(2v^3/3)

    then add the RHS integration

    mgv-kv^3 +(2v^3/3) = t + c

    is this correct so far?
     
  16. May 26, 2010 #15

    Cyosis

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    Wrong if [itex]u=mg-kv^2[/itex] then [itex]du=-2kvdv[/itex]

    What are you doing here?

    It seems that you have tried to do something like partial integration. Notice how [itex]mg-v^2[/itex] is in the denominator on the left side and suddenly in the numerator on the right side.. I have no idea how you managed to do that, but I can tell you it is wrong.

    As a result all other steps are wrong as well.

    This is the only correct part of your post.

    I suggest you use [itex]1-v^2=(1-v)(1+v)[/itex] in combination with partial fractions.
     
    Last edited: May 26, 2010
  17. May 26, 2010 #16
    May I offer a modern approach which I believe may help Darry but is controversial. I do realize I'm new here and don't wish to cause trouble. Let me know.

    You have:

    [itex]\frac{dv}{mg-kv^2}=dt[/itex]

    and you know the right side is just t+c and you want to integrate the left side. Here comes the controversial part:

    First enter the equation in Wolfram Alpha:

    Wolfram Alpha: Integrate[m/(m g-k v^2),v]

    Now that gives you an answer in terms of Arctanh. So immediately, you go to your Calculus text and find "Integrals yielding Inverse Hyperbolic Functions" and I find the one that says:

    [itex]\int \frac{du}{a^2-u^2}=\begin{cases} 1/a \tanh^{-1} u/a+c &|u|<a \\
    1/a \coth^{-1} u/a+c & |u|>a
    \end{cases}
    [/itex]

    Now, it's not difficult to put your integral into that a^2-u^2 form which you can then integrate using the formula above. Try this approach in general when you have integration problems but always try to do it yourself first and not come to rely on a CAS to do your work for you.
     
    Last edited: May 26, 2010
  18. May 26, 2010 #17
    Last edited by a moderator: Apr 25, 2017
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