I am trying to evaluate integral of the form $dN/(N(k-N))$, where $N(t)$ is function of $t$, I know that without factor of $N$, result will be $\ln(t)$, but what can I do with another factor $N(t)$? please help me to solve it
Asked
Active
Viewed 1,087 times
1
-
1What is $k$? A function or a constant? – Davide Giraudo Jan 25 '12 at 09:24
-
constant also one more question if N(t) is a function of t,then integral of N(t) with respect to t is it equal to (t^2)/2+c? – dato datuashvili Jan 25 '12 at 09:27
-
It would help to clarify the problem better, but perhaps you are looking for a "partial fractions" expansion: $\frac{1}{N(k-N)}=\frac{a}{N}+\frac{b}{N-k}$ for constants $a,b$ ($a=-b=1/k$ I think). – bgins Jan 25 '12 at 09:29
-
So you can use partial fraction: $\frac 1{N(k-N)}=\frac 1k\frac k{N(k-N)}=\frac 1{kN}+\frac 1{k(k-N)}$. – Davide Giraudo Jan 25 '12 at 09:31
-
it is a short form of logistic equation of population growth,namely dN/dt=r*N(k-N)/K – dato datuashvili Jan 25 '12 at 09:32
-
but how to integrate it? – dato datuashvili Jan 25 '12 at 09:35
-
What do you mean when you say that you know that without the factor of $N$ the result would be $\log t$? If you are saying that $\int{dN\over k-N}=\log t+C$, you are very confused; the correct answer is $-\log(k-N)+C$. – Gerry Myerson Jan 25 '12 at 12:19
-
See also http://math.stackexchange.com/questions/78560/how-do-you-solve-the-initial-value-probelm-dp-dt-10p1-p-p0-0-1 – Hans Lundmark Jan 25 '12 at 14:23
2 Answers
2
From $${dN\over dt}={rN(k-N)\over K}$$ you get $$\int\,dt=\int{K\;dN\over rN(k-N)}$$ On the left, that's just $t$, and on the right, use Davide's comment to get $$t=\int{K\;dN\over rkN}+\int{K\;dN\over rk(k-N)}={K\over rk}\left(\int{dN\over N}+\int{dN\over k-N}\right)$$ Now, can you do those two integrals?
Gerry Myerson
- 179,216
0
Use partial fractions: $$ \frac{1}{N(k-N)} = \frac {1/k}{N} + \frac{1/k}{k-N}. $$ (And don't forget to apply the chain rule when you find the antiderivative of the second fraction.)