I have the folowing problem :
The population dynamics of a certain fish can be modeled using the logistic equation
$ y′ = Ay − By^2 $.
Furthermore, if it is allowed to fish 20% of the total fish population present at time; write
the differential equation including the harvesting term. Solve the resulting model taking $A = B = 1 $ and
$y(0) = 2$.
I write the equation $y' = y - y^2 - \frac 15 y$ and find $y= \frac 54 + constant $ which doesn't make any sense since the population is supposed to change
I think my mistake is when I write the equation with the harvesting term but I can't figure what the equation is suppose to be.
Hint: I will show you the solution for $y'=y-y^2$ in your case it is a similar procedure but with $y'=0.8y-y^2$.
Your differential equation is given as:
$$\dfrac{dy}{dt}=y-y^2 \implies \dfrac{dy}{y(1-y)}=dt$$ $$\text{partial fractions: }\implies \int \dfrac{dy}{y}+\int \dfrac{dy}{1-y} = \int dt$$ $$\implies \ln y-\ln(1-y)=t+\ln c_1.$$ $$\implies \ln\dfrac{y}{1-y}=t+\ln c_1.$$
Now use the initial condition $y(0)=2$: $$\ln\dfrac{2}{1}=\ln c_1 \implies c_1 = 2.$$
Hence, we obtain
$$\ln\dfrac{y}{1-y}=t+\ln 2.$$
Now, exponentiate this expression and solve for y.
$$y=\dfrac{2\exp(t)}{1+2\exp(t)}=\dfrac{1+2\exp(t)-1}{1+2\exp(t)}=1-\dfrac{1}{1+2\exp(t)}$$