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Solving 1st order differential equation

  1. Apr 12, 2012 #1
    1. The problem statement, all variables and given/known data

    A ball is thrown upwards, its acceleration is given below.

    Determine the time t in terms of v, b, g and v_0 (v_0 is initial speed of the ball) -

    dv/dt = -(g/b^2)(v^2 + b^2)

    in terms of v, b, g and v_0 (v_0 is initial speed of the ball)

    2. Relevant equations



    3. The attempt at a solution

    Solving a differential equation shouldnt be so hard, but I dont know where to get t from and also what to do with all the variables.

    g = a constant (acc. due to gravity)
    b = (mg)(0.2D^2)
    v = velocity
     
  2. jcsd
  3. Apr 12, 2012 #2
    isolate dt and integrate over t i believe.
     
  4. Apr 12, 2012 #3
    I thought that in solving this I should get an expression like v(t) = some expression containing only constants and t.

    Can you point me to any worked examples of similar questions? I also punched this into wolphram alpha and got a reasonable looking answer, I just have no idea where it came from
     
  5. Apr 12, 2012 #4
    Aah, sorry it looks a bit like your problem is nonlinear cause of the v^2 term. Maybe you can use som calculus of variations or nummerical methods. the differential operator representing the equation is
    [itex] \hat{N}v = \dfrac{dv}{dt} +\dfrac{g}{b^2}(v^2+b^2)\\
    \text{equation:}\, \hat{N}(v) = 0[/itex]
    The operator is nonlinear cause:
    [itex] \hat{N}(c_1f + c_2g) \neq c_1\hat{N}f+c_2\hat{N}g[/itex]
     
  6. Apr 12, 2012 #5
    Thanks for the reply, but I dont understand it, I think it is above the level I am doing. Can you make it simpler please?

    I think I need to get v over to the LHS before integrating, is this correct?
     
    Last edited: Apr 12, 2012
  7. Apr 12, 2012 #6
    Hmm, yeah im just not that familiar with nonlinear equations, but i think it makes sense to get the v over. But a general solution is:
    [itex] -b\tan(\dfrac{g(t+c_1)}{b})[/itex]
     
  8. Apr 12, 2012 #7
    This is very similar to the standard solution given by wolphram alpha, but there is a difference.

    v(t) = b.tan((b^2c_1-gt)/b))

    Can you please explain what the difference is? I know velocity is going to be > 0 I know there is normally a choice of standard integrals depending on your range of values. Can you please post a link to where you took this from?
     
  9. Apr 12, 2012 #8
    I took this form maple 15 with dsolve command.
    The method for solving is to collect the t dependence on one side and the dv/dt on the other, as it is and integrating both sides.
    Your eq is in quadrature format so see this and search the net maybe: http://www.maplesoft.com/support/help/Maple/view.aspx?path=odeadvisor/quadrature
    And the difference is maybe caused by a typo somewhere? I just typed in the eq from this forum
     
  10. Apr 12, 2012 #9
    Another problem i can see is that you asked for the time t, that is determined by integrating over t with dt isolated
    And that is my intrepetation of your question.
     
    Last edited: Apr 12, 2012
  11. Apr 12, 2012 #10

    HallsofIvy

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    With first order differential equations the difference between linear and non-linear equations is not that important. What is important is that this equation is "separable". You can write it
    [tex]\frac{dv}{v^2+ b^2}= -\frac{g}{b^2}dt[/tex]

    integrating,
    [tex]\frac{1}{b^2}arctan(v/b)= -\frac{g}{b^2}t+ C[/tex]
    for some constant C. Taking t= 0, [itex]v= v_0[/itex] so
    [tex]\frac{1}{b^2}arctan(v_0/b)= C[/tex]

    Since the problem says "Determine the time", solve the equation
    [tex]\frac{1}{b^2}(arctan(v/b)- arctan(v_0/b))= -\frac{g}{b^2}t[/tex]
    for t, which is very simple- multply both sides by [itex]-b^2/g[/itex].
     
  12. Apr 12, 2012 #11
    Thanks, this post helped a lot.

    Just to be sure, when you integrated the RHS. The t came from the integration of -g/b^2 as two numbers, and not from the dt on the right? I think this is correct as if I differentiated -(g/b^2)t i would get -(g/b^2)

    I get confused with the notation dy/dx, as I am constantly told it is one symbol, it is not to be treated as dy over dx. Then I see in some cases it can be moved around, but apparantly only if doing so enables you to eliminate it later.

    The expression on the LHS is not so tough now I see it before me, the integral of dv must be 1, leaving 1/(v^2 + b^2) which is there is a standard integral for.

    This leaves me wondering only why there is no constant of integration on the LHS.

    I also dont follow this bit

    I got instead

    [tex]\frac{1}{b}(arctan(v/b)- arctan(v_0/b))= -\frac{g}{b^2}t[/tex]

    Please let me know if this is the correct way of seeing it! :)

    The last part of the question is

    By writing a = v dv/dx, solve the resulting differential equation
    and determine the height above the point of projection at time t
    in terms of v, b, g and v0.

    I do not understand the notation here. They ask for the position, so I assume this is the second derivative of the acceleration expression, or the derivative of the velocity expression. Why is it multiplied by v?
     
    Last edited: Apr 12, 2012
  13. Apr 13, 2012 #12
    Kawakaze, you're right, that should be 1/b in front of the left side of the equation (check a table of integrals to confirm). Just a small error by HoI, like we all do time to time https://www.physicsforums.com/Nexus/editor/smilie.png [Broken]
     
    Last edited by a moderator: May 5, 2017
  14. Apr 13, 2012 #13
    Thanks for confirming that, I know I am sure making use of my little mistake quota this week :)
     
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