Solving ODE: Separate x from M-x

[John O' Meara]

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
x=\frac{M}{1+\exp^{-Mk(t-50)}} \\.
My attempt:
\frac{dx}{dt} \propto x(M-x) Therefore
\int\frac{1}{x(M-x)}dx = k\int dt Solving the first integral, using partial fractions, we get:
\int \frac{1}{x(M-x)} dx = \frac{-1}{M} \int \frac{1}{x} dx + \frac{1}{M} \int \frac{1}{m-x} dx \\
Solving the 2nd integral on the l.h.s., let u=M-x => -du=dx, therefore \int \frac{1}{x(M-x)} dx = \frac{-1}{M} \ln{x} + \frac{-1}{M} \int \frac{1}{u}du \\ =-1/M(ln(x) + ln(m-x)) = -1/M*ln(x(M-x)) = k\int dt.
Therefore ln(x(M-x)) = -M*k*t - M*c. => x(M-x) = \exp{-M(kt+c)}.
As you can see x is not separated, but I cannot find the mistake/s that gave rise to this. Thanks for helping.

 

[neutrino]

Recheck the integration by partial fractions.

 

[John O' Meara]

I now get x/(M-x) =\exp{-M(kt+c)

 

[neutrino]

What you wrote as

\int \frac{1}{x(M-x)} dx = \frac{-1}{M} \int \frac{1}{x} dx + \frac{1}{M} \int\frac{1}{m-x} dx \\

should have been...

\int \frac{1}{x(M-x)} dx = \frac{1}{M} \int \frac{1}{x} dx + \frac{1}{M} \int \frac{1}{M-x} dx \\

i.e.No minus signs.After integrating, it should read,

\frac{1}{M}\left[\ln{x} - \ln{\left(M-x\right)}\right] = k\int{dt}