# Differential equation for concentration of gas in the atmosphere

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1. Apr 20, 2014

### waaahboost

1. The problem statement, all variables and given/known data
Hi everyone, I'm currently studying an online course on climate science and am a bit overwhelmed by the calculus. I have studied calculus to second year of college but that was a while ago and I'm very rusty.

A few weeks ago I was a question to find the how long it would take for the concentration of a gas in the atmosphere to reduce by a given percentage. The answer was given, though steps were missing and I can't duplicate the result.

Variables
E = emission rate [Tg/yr]
C = the concentration of gas [Tg]
$τ$ = lifetime of gas in atmospher [yr]

2. Relevant equations

(eq1)$dC/dt = E - C/τ$, where E = 0 (ie. emissions are stopped)

the solution to this differential equation is given as;

(eq2)$C = C_0 exp(-t/τ)$, and

(eq3)$t = -τ log(C/C_0)$

3. The attempt at a solution

$dC/dt = - C/τ$

$∫ -τ/C dC = ∫ dt$

$-τ∫ 1/C dC = ∫ dt$

$-τ log(C) = ∫ dt$

I can't figure out how to get eq2 from eq1. If anyone can fill me in on the intermediate steps from eq1 to eq2 to eq3, I would be very grateful.

Thanks
Chris

2. Apr 20, 2014

### voko

$\int t$ is $t$ plus some constant, which we write as $- \log C_0$ for convenience. That gives you eq. 3. Transition between eq. 2 and eq. 3 is not even calculus, it involves simple algebra and the relationship between $\exp$ and $\log$.

3. Apr 20, 2014

### waaahboost

Thanks voko

The main part I'm struggling to understand is the steps between eq1 and eq2. Looks like an integration on both sides, but I can't reproduce the steps to get there.

4. Apr 20, 2014

### voko

From eq1, you go to eq3, like you (almost) did, then you go to eq2 as I said in #2.

5. Apr 20, 2014

### waaahboost

ok, thank you

6. Apr 20, 2014

### HallsofIvy

Staff Emeritus
Okay, with E= 0, this is $dC/dt= -C/\tau$ and, separating,
$dC/C= -dt/\tau$. Integrate both sides to get
$ln(C)= -t/\tau+ K$ where K is the constant of integration.

Taking the exponential of both sides, $e^{ln(C)}= C= e^{-t/\tau+ K}= e^Ke^{-t/tau}$. Let $C_0= e^K$ and you have the solution below.