The rate of growth of the population x of a country is proportional to the product of x and (M-x), where M is a constant. Assume that x can be regarded as a continuous function of the time t with a continuous derivative at all times and show that if x=M/2 when t=50 then(adsbygoogle = window.adsbygoogle || []).push({});

[tex] x=\frac{M}{1+\exp^{-Mk(t-50)}} \\[/tex].

My attempt:

[tex]\frac{dx}{dt} \propto x(M-x) [/tex] Therefore

[tex] \int\frac{1}{x(M-x)}dx = k\int dt [/tex] Solving the first integral, using partial fractions, we get:

[tex] \int \frac{1}{x(M-x)} dx = \frac{-1}{M} \int \frac{1}{x} dx + \frac{1}{M} \int \frac{1}{m-x} dx \\ [/tex]

Solving the 2nd integral on the l.h.s., let u=M-x => -du=dx, therefore [tex]\int \frac{1}{x(M-x)} dx = \frac{-1}{M} \ln{x} + \frac{-1}{M} \int \frac{1}{u}du \\ [/tex] =-1/M(ln(x) + ln(m-x)) = -1/M*ln(x(M-x)) = [tex] k\int dt[/tex].

Therefore ln(x(M-x)) = -M*k*t - M*c. => [tex] x(M-x) = \exp{-M(kt+c)} [/tex].

As you can see x is not separated, but I cannot find the mistake/s that gave rise to this. Thanks for helping.

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# Homework Help: Separating an ODE

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