# Integrating with the substitution $u = N^{-v}$

Integrate

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

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?

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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.

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