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Differential Equations lesson today

  1. Oct 19, 2004 #1
    Given the following equation:

    [tex]V(t) = \frac{v_{0}}{1+ktv_{0}} [/tex]

    find the position x as a function of time for an object of mass m, located at x = 0 and moving with velocity Voi (where i is the unit vector i) at time t = 0 and thereafter experiening a net force -kmv^2i

    I'm guessing I need to integrate that function however seeing as we've only started integration today in calculus and htis was assigned in physics, I'm not quite sure as to how to approach the problem.

    as far as I can see
    [tex] \frac{dx}{dt} = \frac{v_{0}}{1+ktv_{0}} [/tex]

    in which case

    [tex] ({1+ktv_{o}}) dx = {v_{0}}dt [/tex]

    but I'm not sure as to how to integrate the left and the right side of the function

    any advice would be appreciated :)
  2. jcsd
  3. Oct 19, 2004 #2
    This is not correct. It should be

    [tex] dx = \frac{v_0 dt}{1+ktv_0} [/tex].

    You should gather those [tex]t[/tex] dependence term with the same side as [tex]dt[/tex] so that you could integrate with respect to [tex]t[/tex].

    Next, integrate LHS from [tex]x=0[/tex] to [tex]x[/tex] and RHS from [tex]t=0[/tex] to [tex]t[/tex]. You will obtain a function of the position [tex]x(t)[/tex].

    Best regards,
  4. Oct 20, 2004 #3
    More generally if you have something like

    [tex]\frac{f(u)}{g(t)} = \frac{dt}{du}[/tex]

    then to integrate you'll want to bring it into the form

    [tex]f(u) du = g(t)dt[/tex]

    (sometimes, instead of the above form, you'll have [tex]\frac{g(t)}{f(u)} = \frac{dt}{du}[/tex]. To integrate, you'll have to rewrite it as [tex]\frac{dt}{g(t)} = \frac{du}{f(u)}[/tex].)
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