I started watching a course on ODE, In the course lecture first video, the teacher gave some motivations, this example of Non-Linear population model is first among them, the equation is given by $$\frac{dy}{dt}=ay-by^2 (a,b>0)$$
$t_0$ is the intial time and let $y(t_0)=t_0$
and got the solution as $$\frac{1}{a}\ln\Big|\frac{y(a-by_0)}{y_0(a-by)}\Big|=t-t_0 $$ and found out the equilibrium points as $y=0$ and $y=\frac{a}{b}$
And then he gave the solution analysis as exercise, I proceeded and did as follows:
Case I: $0<y_0<\frac{a}{b}$
$$\Bigg|\frac{y(a-by_0)}{y_0(a-by)}\Bigg|=e^{a(t-t_0)}$$ from this we note that $y \neq 0$ $$\implies \frac{1}{|\frac{a}{y}-b|}=\frac{y_0 e^{a(t-t_0)}}{a-by_0}$$ $$\implies {\Big|\frac{a}{y}-b\Big|}=\frac{a-by_0}{y_0 e^{a(t-t_0)}}$$ since RHS is a continuous function and RHS$>0$ and RHS $\neq 0$, LHS$>0$ for all values of y, or LHS<0 for all values of y
Since $\frac{a}{y_0}-b >0$ $$\frac{a}{y}-b >0 ,\forall y$$ $$\implies y<\frac{a}{b}$$ and $$\frac{a}{y}=\frac{(a-by_0) (e^{-a(t-t_0)})+by_0}{y_0}$$ $$y=\frac{ay_0}{(a-by_0) (e^{-a(t-t_0)})+by_0}$$
Since RHS$>0$, $y>0$ and hence $0<y<\frac{a}{b}$
Case II: $y_0>\frac{a}{b}$
$$\Bigg|\frac{y(by_0-a)}{y_0(by-a)}\Bigg|=e^{a(t-t_0)}$$ from this we note that $y \neq 0$ $$\implies \frac{1}{|\frac{a}{y}-b|}=\frac{y_0 e^{a(t-t_0)}}{by_0-a}$$ $$\implies {\Big|\frac{a}{y}-b\Big|}=\frac{by_0-a}{y_0 e^{a(t-t_0)}}$$ since RHS is a continuous function and RHS$>0$ and RHS$\neq 0$, LHS$>0$ for all values of y, or LHS<0 for all values of y
Since $\frac{a}{y_0}-b <0$ $$\frac{a}{y}-b <0 ,\forall y$$ $$\implies y >\frac{a}{b}$$ and $$b-\frac{a}{y}=\frac{(by_0-a) (e^{-a(t-t_0)})}{y_0}$$ $$\frac{a}{y}=\frac{by_0-(by_0-a) (e^{-a(t-t_0)})}{y_0}$$ $$y=\frac{ay_0}{(a-by_0) (e^{-a(t-t_0)})+by_0}$$
in conclusion $y>\frac{a}{b}$.
Is this things done and observed in both the cases correct and neat? If not where is the mistake and how to proceed and how to make it much neater