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>
<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>