Answer:
[tex]P(1-P) = Ae^t[/tex]
Step-by-step explanation:
given differential equation is
[tex]\frac{dP}{dt} =P-P^2\\\frac{dP}{dt} = g(P) h(t)\\g(P) = P-P^2\\h(t) = 1[/tex]
So this is separable
We separate as
[tex]\frac{dP}{P-P^2} =1dt\\\frac{dP}{P(1-P)} =1dt[/tex]
Resolve into partial fractions to solve this
[tex]dP(\frac{1}{P}+ \frac{1}{1-P} ) = 1dt\\ln P(1-P) = t+C\\P(1-P) = Ae^t[/tex]
So solution is[tex]P(1-P) = Ae^t[/tex]
