1. Limited time only! Sign up for a free 30min personal 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!

Homework Help: Differential equation problem

  1. Jan 15, 2014 #1
    I'm solving a differential equation to do with quadratic resistance and it seems to be acting very strangely - I get the opposite sign of answer than I should. If anybody could have a quick look through that would be much appreciated.

    For a particle moving downward and taking positive upwards, I have

    The terminal velocity comes out at vl=-√(mg/b) (negative as it obviously has its terminal velocity downwards).

    Substituting this in gives

    Now make the substitution z=v/vl so dv/dt=vldz/dt. Then we obtain


    Noting that 1/1-z2=0.5[(1/1+z)+(1/1-z)] and separating variables we get


    Integrating each side then gives (the initial conditions are v=0 at t=0 so the constant of integration is zero)


    Next let k=vl/2g so that








    Now, the correct answer should be




    which must be right because as t→∞, v→vl which by the definition of vl is expected.

    I've spent a lot of time trying to work out where I'm going wrong and it's driving me crazy. Thanks in advance :)
  2. jcsd
  3. Jan 15, 2014 #2


    User Avatar
    Homework Helper

    Probably better to define [itex]v_0 = (mg/b)^{1/2} > 0[/itex] so that terminal velocity is [itex]-v_0[/itex] and [itex]v[/itex] and [itex]z[/itex] increase in the same direction.

    I believe this is correct.

    I think you have confused yourself. You've defined [itex]k = v_l/2g[/itex] and you initially defined [itex]v_l[/itex] to be negative, so [itex]k[/itex] is negative. It follows that [itex]-1/k[/itex] is positive, so that [itex]e^{-t/k} \to \infty[/itex] as [itex]t \to \infty[/itex]. Thus your answer
    v=v_l \frac{e^{-t/k}-1}{e^{-t/k}+1} \to v_l < 0
    as required.

    On the other hand, if you define [itex]v_l[/itex] to be positive then [itex]-1/k[/itex] is negative, so that your answer tends to [itex]-v_l < 0[/itex] as required.
    Last edited: Jan 15, 2014
  4. Jan 15, 2014 #3
    But I defined it as vl=-(mg/b)1/2...
  5. Jan 15, 2014 #4


    User Avatar
    Homework Helper

    I only noticed that after replying, since it's extremely unnatural to do that. More normal would be to define [itex]v_l = (mg/b)^{1/2}[/itex].

    I have now edited my post accordingly.
  6. Jan 15, 2014 #5
    Ah right, thankyou very much.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted