Solving logistic differential equation! Given that the logistic differential equation is $$\frac{dx}{dt}=kx-\lambda x^2$$ The above logistic law is assume to be no emigration! $$k=\frac{1}{400}$$ $$\lambda=10^{-8}$$ $x$ is human population at year $t$! An island consist of human population such that the 100 human emigrates out of the island to the mainland yearly. Modify the differential equation such that it includes the emigration of human! It is stated that the human population at the year of $1980$ is $20,000$. My attempt is $$\frac{dx}{dt}=k(x-100)-\lambda (x-100)^2$$ But to no avail my attempt fail! The answer given at the back is $$x=\frac{(10^5)[3e^{\frac{15}{10^4}(1980-t)}-2]}{6e^{\frac{15}{10^4}(1980-t))}-1}$$

**HINT** :

Your modification should be $\displaystyle \frac{dx}{dt} = kx-\lambda x^2 - 100$.

Then, you have $\displaystyle \frac{1}{\lambda x^2 - kx + 100}=-dt$.

Can you take it from here?