Solving the logistic equation I need to solve the logistic equation 
$$\frac{dP}{dt} = P(a-b\ln P)$$  
How do I go about solving this?

$$\frac{dP}{dt}=P(a-b\ln P)$$
$$\int\frac1{P(a-b\ln P)}\ dP = \int1\ dt=t+K_1$$
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.
