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Separable Differential Equations Using Initial Values

  1. Feb 19, 2014 #1
    AP Physics student here, I'm working on a problem that takes into account air resistance, where something is thrown up at initial velocity v_0, and the drag force is proportional to the velocity, so, [tex]\vec{F_{drag}}=-k\vec{v}[/tex].
    Using Newtons second law and making up positive, down negative, you get, [tex]m\vec{a}=-k\vec{v}-mg[/tex]
    Which yields the differential equation, [tex]m\frac{d\vec{v}}{dt}=-k\vec{v}-mg[/tex]
    This can be separated to become, [tex]\frac{1}{k\vec{v}+mg}d\vec{v}=-\frac{1}{m}dt[/tex].
    Now, when solving this, many sources say to use an indefinite integral on both sides, which results in a constant of integration, which you can then find by inputting t=0 and v=v_0. However, I thought that instead of using an indefinite integral and going to the trouble of finding the constant you could use a definite integral like so,
    I worked through a few problems using my method, and it seemed to yield the same solutions as indefinite integration. Now, I was wondering if integrating from the known value (ex. v_0=v(t=0)) to the variable (v(t)) on one side of the differential equation and integrating from t=0 to t on the other side could possible work for all separable differential equations so that finding the constant of integration is no longer necessary. For example, if you know that y is a function of x and y(2)=6 for the differential equation, [tex]\frac{dy}{dx}=xy^2[/tex],
    could you say that, [tex]\int_{6}^{y}\frac{1}{y^2}dy=\int_{2}^{x}xdx[/tex]
    My calculus teacher wasn't sure, so if anyone could verify that this method is or is not acceptable that would be greatly appreciated.
  2. jcsd
  3. Feb 19, 2014 #2

    But, lose the arrows over the a and v, or include an arrow over the g. Then lose all three arrows, and work only with the magnitudes of the vectors. You should not have vectorial arrows in the separated form of the equation, which should only involve scalar variables.

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