# Homework Help: Simple question - barometric formula derivation

1. Mar 8, 2010

### diffusion

I haven't had extensive experience in solving DE's just yet; I'm curious as to where the P(0) term comes from between the 4th and 5th expressions in the following derivation:

http://en.wikipedia.org/wiki/Barometric_formula#Derivation

My attempt follows:

$\frac{dP}{P} = - \frac{M g\,dz}{kT}$

$\int\frac{1}{P}dP = - \int\frac{M g}{kT}dz$

$P = e^{-\int\frac{M g}{kT}dz}$

$P = e^{-Mgz/kT}$

Clearly I'm missing something obvious, but I don't know what exactly.

2. Mar 9, 2010

### aim1732

Limits of the integration.