How to solve $ u'(t)=(u(t))^2-5u(t)+6 $ rigorously? I have the following homework question: 
Show that the Cauchy problem:
$$
u'(t)=(u(t))^2-5u(t)+6
$$
for $t\in[0,\infty[$ and with $u(0)=u_0$ has a global solution for $u_0≤3$ and a maximal, non-global solution for $u_0>3$.
In the solution, they proceed as follows:
$$
\frac{du(t)}{dt}=(u(t))^2-5u(t)+6\implies \frac{du(t)}{(u(t))^2-5u(t)+6}=dt \implies \int \frac{du(t)}{(u(t))^2-5u(t)+6}=\int dt \implies \log\left(\frac{u(t)-3}{u(t)-2}\right)=t+C
$$
And they deduce the solutions for $u$ from there.
I'm a bit surprised by this approach, because as I'm studying maths (and not physics) this seems a bit vague… Why can we divide by $(u(t))^2-5u(t)+6$ even though it could be $0$? How to obtain the constant solutions $u\equiv 2$ and $u\equiv 3$ because they don't seem to be covered by this approach?
If somebody could provide an approach without this kind of uncertainty, i would be very grateful.
 A: The constant solution $u\equiv 2$ is the unique solution to the Cauchy problem for $t_0\in \mathbb{R}$, 
$$ \begin{cases} \frac{du}{dt}=(u(t))^2-5u(t)+ 6, \\ u(t_0)=2.\end{cases}$$
Consequently (by Cauchy-Lipschitz theorem), a solution of $\frac{du}{dt}=(u(t))^2-5u(t)+ 6$ is either always equal to $2$ or never equal to $2$.
The same holds for the constant solution $u\equiv 3$. So by continuity, a solution $u$ of your equation is such that, for all $t$, $u(t)$ belongs to $]-\infty,2)$ or to $\{ 2 \}$, or to $(2,3)$, or to $\{ 3 \}$ or to $(3,+\infty[$. Now that you know this, the integration can be done easily in the region $]-\infty,2)$ (or $(2,3)$ ; $(3,+\infty[$).
Remark: this phenomenon is due to the fact that if $t\mapsto u(t)$ is a solution of your equation, then, for a fixed parameter $\tau$, $t\mapsto u(t+\tau)$ is also a solution of the equation.
A: $$u'=u^2-5u+6$$
This is a Riccati ODE. So, change of function : $\quad u(t)= -\frac{y'(t)}{y(t)}$
$u'=-\frac{y''}{y}+\frac{y'^2}{y^2} = \left(-\frac{y'(t)}{y(t)} \right)^2 -5\left(-\frac{y'(t)}{y(t)} \right)+6$
$$y'' + 5 y' +6 y = 0$$
$$y=c_1e^{-3t}+c_2e^{-2t}$$
$$u(t)=-\frac{y'}{y}=\frac{3c_1e^{-3t}+2c_2e^{-2t}}{c_1e^{-3t}+c_2e^{-2t}}$$
$$u(t)=\frac{3C+2e^{t}}{C+e^{t}}$$
