$y' = a(y) \cdot y$ yith $a$ defined in an inequation I was wondering if there was knowledge to the following differential equation:
$$y' = a(y) \cdot y$$
with $a$ is defined as being $a_0$ when $y < b$ and $a_1$ elsewhere with $a_0 , a_1, b$ reals.
The family of functions is obviously $Span(\exp{(a_{0/1} t)})$ on each individual interval but how do you link them.
With an initial condition, the problem is just a "glueing" solution problem. However, without, I didn't managed to get more about it.
What is the family of functions satisfying the differential equations
 A: You would want the solution to be continuous at $y = b$.  If we impose the single initial condition $y(0) = y_0$, then the solution for $y<b$ is given by
$$y_1(t) = y_0e^{a_0t}.$$
Treating the solution for $y \geq b$  ($y_2$) independently, we find that
$$y_2(t) = y_{02}e^{a_1t},$$
where $y_{02}$ is the initial condition for $y_2$.  For continuity, we impose
$$y_1(t_b) = y_2(t_b),$$
where $y = b$ at $t = t_b$.  This leads to the relationship
$$y_{02} = y_{0}e^{a_0/a_1}.$$
Thus, the continuous solution is
$$y(t) = y_0e^{a_0t},y<b,$$
$$y(t) = y_0e^{a_0/a_1}e^{a_1t},y \geq b.$$
Of course, as you pointed out there is an infinite family of solutions, and in this set we can certainly include those solutions that are discontinuous at $t_b$.
One other thing to keep in mind is that you need to be careful with the signs of $a_0$ and $a_1$.  For the above continuous solution to work, you would need the two constants to have opposite signs, so that the solution decreases from $b$ in one direction, and increases from $b$ in the other.
