Differential equation problem. Integrating the logistic equation.

Question

I would like to know how to integrate or rather solve this:

$$ \frac{dP}{dt} = kP(L-P). $$

I have the solution, but I would like to know how to arrive at it. I have been told it involves separation of variables and partial fractions.

Answer 1

Hint/Nudge: $$\frac{dP}{dt} = kP(L-P)$$$$\iff \frac{1}{P(L-P)}dP=kdt$$ This is the seperation of variables bit. Everything that depends on $P$ is on the LHS and everything else on the RHS.

Now, do partial fractions with $\frac{1}{P(L-P)}$ on LHS and integrate.

Edit: You are almost right. You should have $\ln P-\ln (L-P)=L(kt+c)$, from here you just need to rearrange for P. (Note that you don't need constants of integration on both sides as they can just amalgamate into one (by addition/subtraction)).

From here you exponentiate both sides to get: $$\exp(\ln P-\ln (L-P))=\exp\ln(\frac{P}{L-P})=\frac{P}{L-P}=\exp (L(kt+c))$$ Which gives, $$P=(L-P)\exp (L(kt+c))$$ and then, $$P+P\exp(L(kt+c))=P(1+\exp(L(kt+c)))=L\exp(L(kt+c))$$ finally, $$P=\frac{L\exp(L(kt+c))}{1+\exp(L(kt+c))}=L\frac{1}{1+\exp(-L(kt+c))}.$$ Tidy this up by defining $A=\exp(-Lc)$ which gives your general solution as: $$P=L\frac{1}{1+Ae^{-Lkt}}.$$

Answer 2

you will get $$-{\frac {\ln \left( L-P \right) }{L}}+{\frac {\ln \left( P \right) }{L}} =kt+C$$