- #1
diffusion
- 73
- 0
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.
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.
