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Solving differential equations with definite integrals?

  1. Dec 28, 2007 #1
    I'm taking a calculus-based physics course, and we were solving a simple differential equation for a model of drag by separating variables: (where A is some arbitrary constant)

    [tex]m \frac {dv} {dt} = -A v^2[/tex]
    [tex]- \frac {m} {A} \frac {dv} {v^2} = dt[/tex]

    My teacher then integrates both sides, but unlike in my calculus class, he uses definite integrals:

    [tex]- \frac {m} {A} \int_{v_i}^{v_f} v^{-2} dv = \int_{t_i}^{t_f} dt[/tex]

    Initial time will be zero for simplicity, so using the FTC:

    [tex]\frac {m} {A} (\frac {1} {v_f} - \frac {1} {v_i}) = t_f[/tex]

    I understand how to solve it as is done from my calculus class, using indefinite integrals and solving for the constant of integration [itex]C = \frac {m} {A v_i}[/itex], which gives an equivalent result.

    So what's the merit of using one method as opposed to another? It seems to me like using definite integrals is quicker.

    So if I do physics problems this way, why should/shouldn't I do the initial-value problems I get in calculus using definite integrals like this? Like when it asks to solve [itex]dy/dx = 3y, y(2) = 5[/itex] or such, what's wrong with doing [itex]\frac {1} {3} \int_{5}^{y_f} y^{-1} dy = \int_{2}^{x_f} dx[/itex]? Both methods give the same result, and again, it seems quicker to do this the definite integral way rather than solving for C.

    I'm guessing that it might be harder to understand what's going on as things get more complex, or something? Or I guess using subscripts on variables to really indicate evaluating the function for the independent variable is problematic?

    (Whew, sorry for the length, but this has been bugging me for a while.)
    Last edited: Dec 28, 2007
  2. jcsd
  3. Dec 28, 2007 #2


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    It really doesn't matter. What matters is that you are given a differential equation with an initial condition. From the differential equation you can find the general solution, and then somehow you have to impose the initial condition. You can do this after writing down the general solution (by first doing an indefinite integral and then solving for the unknown integration constant) or you solve the equation with the correct starting conditions right away. Of course they (should) give the same results in the end. It works the same way for higher order equations, e.g.
    [tex]\frac{d^2x}{dt^2} = a[/tex]
    (Which I won't work out because I'm not really comfortable working with the infinitesimals as you did above.)
  4. Dec 30, 2007 #3
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