To be clear, the given relation between
time and female population is an integral:<span>
</span>
<span>
</span>
<span>The problem says that r = 1.2 and S = 400, therefore
substituting:<span>
</span>
![t = \int { \frac{P+400}{P[(1.2 - 1)P - 400]} } \, dP](https://tex.z-dn.net/?f=t%20%3D%20%5Cint%20%7B%20%5Cfrac%7BP%2B400%7D%7BP%5B%281.2%20-%201%29P%20-%20400%5D%7D%0A%7D%20%5C%2C%20dP%20)
<span>
</span>= <span><span>

In order to evaluate this integral, we need to write this rational function in
a simpler way:
</span>

</span><span>
</span>where we need to evaluate A and B. In order to do
so, let's calculate the LCD:<span>
</span>

<span>
</span>the denominators cancel out and we get:<span>
</span>P + 400 = 0.2AP - 400A + BP<span>
</span> =
P(0.2A + B) - 400A<span>
</span>The two sides must be equal to each other,
bringing the system:<span>
</span>

<span>
</span>Which can be easily solved:<span>
</span>

<span>
</span>Therefore, our integral can be written as:<span>
</span>

<span>
</span>=

<span>
</span>=

<span>
</span>= - ln |P| + 6 ln |0.2P - 400| + C<span>
</span>Now, let’s evaluate C by considering that at t = 0
P = 10000:<span>
</span>0 = - ln |10000| + 6 ln |0.2(10000) - <span>400| + C
C = ln |10000</span>| - 6 ln |1600|<span>
</span>C = ln (10⁴) - 6 ln (2⁶·5²)<span>
</span>C = 4 ln (10) - 36 ln (2) - 12 ln (5) <span><span>
</span></span><span><span>
</span>Therefore, the equation relating female population
with time requested is:<span>
</span><span>t = - ln |P| + 6 ln |0.2P - 400| + 4 ln (10) - 36 ln </span>(2) - 12 ln (5)</span></span>