# Thread: Intergration of Differential Equations

1. ## Intergration of Differential Equations

Hey all, need help.
How do you find k once you have intergrated 1/NdN=kdt?
I've got to:
N=Ae^kt
And i'm told:
120=Ae^20k

Thanks
Trev

2. Originally Posted by TrevK
Hey all, need help.
How do you find k once you have intergrated 1/NdN=kdt?
I've got to:
N=Ae^kt
And i'm told:
120=Ae^20k

Thanks
Trev
if A has a value, then

$k = \frac{ ln \frac{120}{A}}{20} = \frac{ ln 120 - ln A}{20}$

3. Hello, Trev!

Did they give us an initial condition?

$\frac{dN}{N} \:=\:k\,dt$ . . . and we are given: . $N(20) \:=\:120$

We have: . $\frac{dN}{N} \;=\;k\,dt$

. . Integrate: . $\ln N \;=\;kt + c$

. . Then: . $N \;=\; e^{kt+c} \;=\;e^{kt}\cdot e^c\;=\; Ce^{kt}$

Assume that, when $t = 0,\;N \:=\:N_o$

. . Then we have: . $N_o \:=\: Ce^0\quad\Rightarrow\quad C \:=\:N_o$

. . The function (so far) is: . $N \;=\;N_oe^{kt}$

We are told that: . $N(20) \:=\:120$
. . Then we have: . $120 \;=\;N_oe^{20k}\quad\Rightarrow\quad e^{20k} \;=\;\frac{120}{N_o}$
. . Take logs: . $\ln\left(e^{20k}\right) \;=\;\ln\left(\frac{120}{N_o}\right) \quad\Rightarrow\quad 20k(\ln e) \;=\; \ln\left(\frac{120}{N_o}\right)$
. . Hence: . $k \;=\;\frac{1}{20}\ln\left(\frac{120}{N_o}\right)$

Therefore: . $N(t) \;=\;N_oe^{\left[\frac{1}{20}\ln(\frac{120}{N_o})\right]t}$

Edit: Darn, just realized that this identical to kalagota's solution.
.