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A simple differential equation problem.

  1. Nov 24, 2006 #1
    is there a solution to this euqation: (i got to in a problem in physics so the notation is appropiate):
    [tex](m_1+m_2)\frac{dv}{dt}=m_1g-m_2gsin(a)-f_s-\beta v-V\rho g[/tex]
    is there a solution or perhaps i got it wrong in equations?
    thanks in advance.
     
  2. jcsd
  3. Nov 24, 2006 #2

    arildno

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    Well, the notation might be appropriate, but your definition of the symbols is not! :grumpy:
    Assuming that everything is constant apart from v,
    just rewrite your diff. eq as:
    [tex]\frac{dv}{dt}+Av=B[/tex]
    where A and B are constants.
    You may use an integrating factor to solve this problem.
    This will also work for non-constant A and B (depending solely on t, and not on v!!), but you might not get an explicit formula in terms of elementary functions then.
     
    Last edited: Nov 24, 2006
  4. Nov 24, 2006 #3
    what's wrong with my interpratation of the symbols?
     
  5. Nov 24, 2006 #4

    arildno

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    It is lacking. How should I know what you mean by your symbols when you don't define them?
     
  6. Nov 24, 2006 #5
    well, the m's are masses, the angle a doesnt change with time (perhaps here lies your critic), V stands for volume and rho stand for water density.
    that's it.
     
  7. Nov 24, 2006 #6
    perhaps, v=Acos(b)+Bsin(b) is a possible solution here?
     
  8. Nov 24, 2006 #7

    arildno

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    Hmm..and I guess that g is the acceleration due to gravity and beta is the coefficient of air resistance. But you didn't sayu that.
    Furthermore f_s is some sort of applied force I don't know what is.
     
  9. Nov 24, 2006 #8
    what integrating factor to use here?
    i mean: dv/v+Adt=Bdt/v dv/v+(A-B/v)dt=0
    dv/v(A+B/v)+dt=0 will that suffice here?
     
  10. Nov 24, 2006 #9
    this is a pretty simple integrating factor

    if you mmove the resistance term (Bv) to the left hand side and divide by (m1+m2) than you have an quation of the form


    dv/dt +Pv = Q

    where Q is some function of t, and so is P

    in a problem like this the integrating factor is e^(integral of P dt)

    multiplying by that and simplifying will get you an equation of the form


    d(ve^(integral P dt))/dt =Qe^(integral P dt)

    integrate, and then you'll have a simple algebra problem

    also notice that the solution only has 1 parameter, all first order differential equation have only one parameter in the solution.
     
  11. Nov 30, 2006 #10
    Why use an integrating factor? The ODE's clearly separable...
     
  12. Dec 1, 2006 #11

    arildno

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    Why use separation of variables?
    You can clearly use an integrating factor..:smile:

    It's just a matter of preference what method you use.
     
  13. Dec 1, 2006 #12
    hmm it doesn't appear to be seperable.

    if you note its of the form dv/dt +Bv=C

    as somebody else mentioned, that is not a seperable equation
     
  14. Dec 1, 2006 #13

    arildno

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    So, you disagree that the diff.eq may be rewritten as:
    [tex]\frac{1}{C-Bv}\frac{dv}{dt}=1[/tex] ??
     
  15. Dec 1, 2006 #14
    oh my mistake, didn't see that manipulation
     
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