Question

Can any bdy solve the last part of this one : Diff. Eq , separable variable method...

dP/dt= P-P^2

* dont forget to consider the chain rule when you integrate *

Answer 1

\[\frac{\text dP}{\text dt}= P-P^2\]
\[\frac{\text dP}{P-P^2}= \text dt\]
\[\int\frac{\text dP}{P-P^2}= \int\text dt\]

Answer 2

For the left:
\[\int \frac{dP}{P-P^2} = \int \frac{dP}{P(1-P)}=\int \frac{1}{P}+\frac{1}{1-P}dP\]