For the general logistic growth model that can be applied to biology and economics
$\frac{dP}{dt} = gP(1-\frac{P}{K})$
I know to separate the variables and divide by $P(1-\frac{P}{K})$ resulting in $ln(p) - ln(1-\frac{p}{k}) = gt + A$. Then just rearranging to find p.
That I understand fine, however when another term is added, uP or anything involving P, I do not know how to separate the variables.
For example how do I separate $\frac{dP}{dt} = gP(1-\frac{P}{K}) + uP$
I cannot seem to find a way to get it to result in all Ps on one side without horrendous mess and trying to integrate
$\frac{1}{(gP(1-\frac{P}{K}) +uP)}$
Both DE are Bernoulli's differential equations: $$\frac{dP}{dt} = gP(1-\frac{P}{K}) + uP$$ $$P'-P(g+u) = -g\frac{P^2}{K}$$ By integrating factor method we get: $$(Pe^{-(g+u)t})' = -g\frac{P^2}{K}e^{-(g+u)t}$$ $$\int \dfrac {d(Pe^{-(g+u)t})}{P^2e^{-2(g+u)t}} = - \frac{g}{K}\int e^{(g+u)t}dt$$ It's easy to integrate now. $$ \dfrac 1{Pe^{-(g+u)t}} = \frac{ge^{(g+u)t}}{K(g+u)} +C$$ $$ \dfrac 1{P} = \frac{g}{K(g+u)} +Ce^{-(g+u)t}$$
With fractions decomposition method: $$\frac{dP}{dt} = gP(1-\frac{P}{K}) + uP$$ $$\frac{dP}{dt} = gP(1+\dfrac ug-\frac{P}{K}) $$ $$\frac{dP}{dt} = gP(B-\frac{P}{K}) $$ $$\frac{dP}{dt} = \dfrac g KP(KB-P) $$ $$\frac{dP}{dt} = \dfrac g KP(A-P) $$ $$\int \frac{dP}{P(A-P)} = \int \dfrac g K dt$$ Where $A=BK=K(1+\dfrac ug)$. $$\dfrac 1 A \left(\int \frac{dP}P-\int \dfrac {dP}{(P-A)}\right) = \int \dfrac g Kdt$$ Now integrate.