Differential Equation

Althought this differential equation is a particular case of Bernouilli differential equation I'm going to use the variable separated method what is not absolutely formal but it works. $\dfrac{dy}{dx} = 8y - 2y^2,\quad y(2016) = 1$ $\dfrac{dy}{8y - 2y^2} = dx \Rightarrow$ $\int \frac{1}{8y -2y^2} dy = \int dx \Rightarrow$ $\int \frac{\frac{1}{4}}{2y} dy + \int \frac{\frac{1}{8}}{4 -y} dy = x + C \space \text{ (C constant)} \Rightarrow$ $\frac{1}{8} \cdot (\ln |y| - \ln |4 - y|) = x + C \Rightarrow$ $\large {e^{\frac{1}{8} \cdot \ln \frac{|y|}{|4 - y|}}} = Ke^x \text{ (K constant)} \Rightarrow$ I'm going to consider $A = \frac{y}{4 - y} > 0$ , if it was $A < 0$ we can reason in the same way and we would get the same result. it depends on y(2016) = 1 $\left(\frac {y}{4 - y}\right)^{\frac{1}{8}} = Ke^x \quad (y(2016) = 1 \Rightarrow K \neq 0) \Rightarrow$ Powering to 8 and rearranging $y \cdot (1 + ke^{8x}) = 4ke^{8x} \quad \text{(k constant)} \quad k \neq 0 \Rightarrow$ $\displaystyle y = \frac{4ke^{8x}}{1 + ke^{8x}} \Rightarrow \lim_{x \to \infty} y = 4.$ If we consider this equation like a logistic differential equation about the population growth, it could be interpreted saying that the population in "the end of the world" will be 4 times the currently population

Unrigorous method is that as we approach carrying capacity population is constant so set dy/dx=0 and we're done