The Constant of Integration

[fizzacist]

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:

\frac{dv}{dt} = \frac{F-KV}{m}

My solution:

\int\frac{dv}{F-KV} = \int \frac{dt}{m}

u = F-KV

\frac{du}{-K} = dv

\frac{-1}{K} \int\frac{1}{u}du = \int\frac{dt}{m}

Integrate that to find

ln|F-KV|+C = -K\frac{t}{m}

But before I go any further, the 1993 Exam's Rubric shows that by integrating \int\frac{dv}{F-KV} 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:
http://imgur.com/c83p1
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
Thanks

 

[Hurkyl]

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

What's the difference?

Maybe you're confused because you're using C for two different things. Try comparing

• ln|u| + D, and
• ln|u|-lnC

 

[mathman]

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.

 

[fizzacist]

Ahh. Got it. :P