# The Constant of Integration

1. Apr 15, 2012

### 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

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
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2. Apr 15, 2012

### Hurkyl

Staff Emeritus
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

3. Apr 15, 2012

### 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.

4. Apr 15, 2012

### fizzacist

Ahh. Got it. :P