Solving the logistic equation

Question

I need to solve the logistic equation

$$\frac{dP}{dt} = P(a-b\ln P)$$

How do I go about solving this?

Answer 1

$$\frac{dP}{dt}=P(a-b\ln P)$$ $$\int\frac1{P(a-b\ln P)}\ dP = \int1\ dt=t+K_1$$ The derivative of $\ln(a-b\ln P)$ with respect to $P$ is $\frac{-b}{P(a-b\ln P)}$, so: $$\int\frac1{P(a-b\ln P)}\ dP =-\frac1b\ln(a-b\ln P)+K_2=t+K_1$$ $$-\frac1b\ln(a-b\ln P)=t+K_3$$ $$\ln(a-b\ln P)=-bt+K_4$$ $$a-b\ln P=e^{-bt+K_4}=A_1e^{-bt}$$ $$\ln P=\frac{a-A_1e^{-bt}}b=\frac ab-Ae^{-bt}$$

$$P=e^{\frac ab-Ae^{-bt}}$$

where $A$ is a constant.