Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top


$$\frac{dN}{dt} = rN(1 - K^{-v}N^v)$$

where $r,v,K$ are positive constants (and so I think $N$ is a function of $t$), using the substitution $u =N^{-v}$, given that $N$ has an initial value at $N_0 < K$. Determine the behaviour of the solution for large times.

So, in accordance with the answers, I have done everything correct and got to the point where I now have

$$\frac{du}{dt} = -vr(u - K^{-v}),$$

but now I'm a little stuck. For some reason, they have integrated

$$\int_{u_0}^u \frac{du}{u - K^{-v}} = -vrt.$$

Why have they integrated between those two limits as opposed to ingetrating just $\frac{du}{u - K^{-v}}$ and then solving for $u$ to get the initial condition like normal?

share|cite|improve this question

Because integrating with the lower limit $u=u_0$ is equivalent to applying the initial condition at $t=0$. To see this, write instead

$$\int_{u_0}^u \frac{du'}{u'-K^{-v}} = -v r \int_0^t dt'$$

Note that I primed the integration variables; they are dummy variables and what we call them is of no importance to the out come of the problem. The limits, however, are important: the lower limits and upper limits of integration on each side of the equation must correspond. That is, $u(0)=u_0$ which is equivalent to $N(0)=N_0$.

You can also see this by integrating as you would expect, an indefinite integral:

$$\int \frac{du}{u-K^{-v}} = -v r t + C$$

where $C$ is an integration constant. This of course implies that

$$\log{(u-K^{-v})} = -v r t + C$$

Now, at $t=0$, $u=u_0$, which means that

$$C = \log{(u_0-K^{-v})}$$

so that

$$\log{(u-K^{-v})}-\log{(u_0-K^{-v})} = -v r t$$

which is equivalent to

$$\int_{u_0}^u \frac{du'}{u'-K^{-v}} = -v r t$$

I hope this makes sense of where the lower integration limit comes from.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.