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The Constant of Integration

  1. Apr 15, 2012 #1
    While taking an AP physics practice exam, I encountered a difference in the way I solve a differential equation and the way the exam's rubric solves it.

    The equation is as follows:

    [itex]\frac{dv}{dt}[/itex] = [itex]\frac{F-KV}{m}[/itex]

    My solution:

    [itex]\int[/itex][itex]\frac{dv}{F-KV}[/itex] = [itex]\int[/itex] [itex]\frac{dt}{m}[/itex]

    u = F-KV

    [itex]\frac{du}{-K}[/itex] = dv

    [itex]\frac{-1}{K}[/itex] [itex]\int[/itex][itex]\frac{1}{u}[/itex]du = [itex]\int[/itex][itex]\frac{dt}{m}[/itex]

    Integrate that to find

    ln|F-KV|+C = -K[itex]\frac{t}{m}[/itex]

    But before I go any further, the 1993 Exam's Rubric shows that by integrating [itex]\int[/itex][itex]\frac{dv}{F-KV}[/itex] should yield ln|F-KV|-lnC

    To me, this makes no sense. The constant of integration should be ln|u| + C, not ln|u|-lnC

    Here's what I'm talking about:
    I've also attached the '93's rubric to this post. The problem I'm referring to is problem #2.

    Can any of the math/physics gurus out there help me out? :P

    Attached Files:

  2. jcsd
  3. Apr 15, 2012 #2


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    Staff Emeritus
    Science Advisor
    Gold Member

    What's the difference? :confused:

    Maybe you're confused because you're using C for two different things. Try comparing
    • ln|u| + D, and
    • ln|u|-lnC
  4. Apr 15, 2012 #3


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    Science Advisor

    Call your constant of integration K (remember K is completely arbitrary). Now define another constant C by K = -lnC. This gives the formula in the book. Since C is also completely arbitrary, it doesn't matter.
  5. Apr 15, 2012 #4
    Ahh. Got it. :P
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