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Simple differential equation

  1. Feb 13, 2015 #1
    1. The problem statement, all variables and given/known data

    Given that a particle has an initial velocity v0 and then undergoes an acceleration a = - bv. , where b is a constant, obtain an expression for v = v(t) and x = x (t)



    2. Relevant equations

    Not sure

    3. The attempt at a solution

    If I integrate a = - b v

    I think I get v + vo = - b x + xo

    But then I get stuck

    The book actually gives the answer. x(t) = xo + (vo/b) (1 - e^-bt)

    But I can't figure out how they got to that. The book kind of assumes that one can solve this and I should be able to but I am stuck. Thanks
     
  2. jcsd
  3. Feb 13, 2015 #2

    Ray Vickson

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    If you knew the velocity function ##v(t)##, how would you get the acceleration ##a = a(t)##? So, if you are told that ##a = -b v## with constant ##b##, what sort of equation do you get?
     
  4. Feb 13, 2015 #3
    If you know the velocity, you can get the acceleration by differenting. dv/dt = - b v ??
     
  5. Feb 13, 2015 #4
    So v (t) = - 1/b dv/dt ?
     
  6. Feb 13, 2015 #5
    Thanks Ray -
     
  7. Feb 13, 2015 #6
    If my function for v(t) is correct, I try integrating this to get x.

    I starting with v(t) = - 1/b dv/dt

    Integrating both sides, I get

    x(t) + xo = -1/b v (t) + v0.

    But I know this is incorrect. I can't see where the e^-bt. In their answer comes from. Thanks
     
  8. Feb 13, 2015 #7

    PeroK

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    It's not incorrect, but you've expressed v in terms of x, not in terms of t. Although, you've been careless with x0 and v0.

    If you try ##v(t) = e^{-bt}## you'll find it satifies your equation.

    To get v(t) you need to separate v and t before you integrate.
     
  9. Feb 13, 2015 #8
    Thanks Perok.

    I think I've got it. At least I can show that the answer they gave is a solution.


    Starting with the answer they gave:

    x(t) = x0 + (vo/b) (1 - e^-bt)

    I differentiate this once to get:

    v(t) = vo e^-bt

    I differentiate this a second time:

    a(t) = -b vo e^-bt


    Substuting for [vo e^-bt ] gives:

    a(t) = -bv (t) as required.

    So now I can see that it would (theoretically) be possible for me (next time) to solve the problem.

    Thanks
     
  10. Feb 13, 2015 #9

    Ray Vickson

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    The DE dv/dt = -bv is among the simplest, and is one of the first you meet in a differential equations course. The solution is easy, because somebody already solved it for us in the past, and it is well documented in just about every textbook: v = const* exp(-b t). The reason is simple: (d/dt) exp(a*t) = a* exp(a*t), from Calculus 101; basically, this is just the exponential derivative (d/dx) exp(x) = exp(x), with a change of variable to x = a*t.
     
  11. Feb 13, 2015 #10
    Yes - thanks. I should have got it sooner, but it is about 7 years since I did my course in applied maths and am pretty rusty. Cheers
     
  12. Feb 13, 2015 #11
    From your differential equation,
    $$\frac{dv}{v}=-bdt$$
    Both sides of this equation are exact differentials.

    Chet
     
  13. Feb 13, 2015 #12
    Hi Chet,

    Can you explain a bit more about what you mean when you say that both sides are exact differentials?

    Thanks

    Peter
     
  14. Feb 13, 2015 #13
    What is ##\int{\frac{dv}{v}}##?

    What is ##\int{dt}##?

    Chet
     
  15. Feb 14, 2015 #14
    I am not sure. The integral of dV/V, is it
    ln V ?

    And the integral of dt is t?

    So t = lnV

    Which seem to make sense given that

    V = Vo = e^-bt (I've got to figure out latex!)


    But I still don't get the term 'exact differentials. Cheers
     
  16. Feb 14, 2015 #15

    PeroK

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    Given that you simply wrote down the integral of each term, is that not a strong hint of what an exact differential is?

    A good example is ##mv\frac{dv}{dt}##

    Can you see what that is an (exact) derivative of?

    You definitely need to revise derivatives, integration and differentials if you're trying to study DE's.
     
  17. Feb 14, 2015 #16
    No. Including the constant of integration C, the result should read:

    lnv=-bt + C

    Inverting this gives:
    $$v=e^{(C-bt)}$$

    Hope this makes some sense.

    Chet
     
  18. Feb 14, 2015 #17
    Hi Perok

    You wrote:

    mvdv/dx and then ask, what is this a derivative of.

    This is my attempt:

    I start by separating the variables:

    mvdv = dx

    Then integrating both sides, I get:

    ma + vo = t + to

    I am not sure if this is what you meant.
     
  19. Feb 14, 2015 #18
    Hi Chet

    Thanks I get that. I am still stuck though on this new term (at least it is a term that I am unfamiliar with) "exact differential". But don't worry - it's probably not so important right now. The main thing is that I am getting back into doing this maths. Cheers and thanks again.
     
  20. Feb 14, 2015 #19

    PeroK

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    First, I gave you an expression, not an equation. You've effectively set that expression = 1.

    In any case:

    ##mv\frac{dv}{dt} = \frac{d}{dt}(\frac{1}{2}mv^2)##

    So, that expression is the rate of change of Kinetic Energy. I.e. it's an exact derivative of the expression for KE. It's a good one to remember.

    As I said before, you need to revise your calculus!
     
  21. Feb 14, 2015 #20
    Thanks - I guessed that it would be something really obvious like energy but missed it.

    You are right, I do need to revise my calculus. Like I said it is 7 years since I did this stuff and now I am trying to study quantum field theory.
     
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