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:

[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:
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

Attached Files

RBCII_93.PDF (130.7 KB, 1 views)

Hurkyl

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