solve $f'(t) = f(t) (a-bf(t))$ 
Solve $f'(t) = f(t) (a-bf(t)), f(t_0) = y_0$, where $a,b, t_0\in\mathbb{R}$ and $f$ is a real-valued function taking real-values.

Apparently the solution is
$$f(t) = \dfrac{a y_0 e^{a(t-t_0)}}{by_0 e^{a(t-t_0)} + (a-y_0 b)},$$
but I don't know how to derive this. I know the standard approach for solving first order linear differential equations, but that's clearly not applicable in this case. I also tried doing something like multiplying both sides by some sort of integrating factor like $e^{-at}$. Doing so yields
$$(e^{-at}f(t))' = -e^{-at}b f(t)^2.$$ Therefore $$\dfrac{(e^{-at} f(t))'}{f(t)^2} = -e^{-at}b.$$
It doesn't seem like integrating both sides is useful in this case. Perhaps some sort of substitution might be useful?
 A: It is a linear DE for $g(t)=f(t)^{-1}$. To further simplify you could set $g(t)=af(t)^{-1}-b$.
Your second approach also would work well if you then set $g(t)=e^{-at}f(t)$. Then
$$
\frac{g'(t)}{g(t)^2}=-e^{at}b.
$$
A: Hint
$$\frac{d}{dt}\left[\frac{e^{at}}{f(t)}\right]=e^{at}\left[\frac{af(t)-f'(t)}{[f(t)]^2}\right]$$
A: This is Bernoulli equation:
$$y'+p(x)y=q(x)y^n,~~~n=2,3,4,...$$
Standard method to solve Bernoulli equation: Let $w=y^{1-n}$
$$\Rightarrow~~~ \frac{dw}{dx}+(1-n)p(x)\cdot w=(1-n)q(x)$$
Now you get a first-order linear ODE respect to $w$. For your case $p(x)=-a,~~q(x)=-b,~~n=2$
Can you proceed from here?
A: It seems like the solution varies depending on the parameters. For example take $a=b+1=0$. Then we have
$$
y'=y^2\Rightarrow dy/y^2=dx\Rightarrow\int dy/y^2=\int dx\Rightarrow-1/y=x+c
$$
But, as you can see, in the case $a=b=1$ we have
$$
dy/(y(1-y))=dx\Rightarrow\int(1/y+1/(1-y))dy=\int dx\Rightarrow
$$
$$
x+c=ln|y|+\ln|1-y|\Rightarrow |y(1-y)|=e^{x+c}
$$
So, the experiment shows us that you actually should care about what type of equation you have depending on the couple $(a, b)$.
