How to find domain of definition of solution? For the given initial-value problem:
$$\dfrac{dy}{dt}=\dfrac{1}{\left(y+2\right)^{2}}, \quad y(0)=1$$
we are asked to solve it then to state the domain of definition of the solution. So first of all I separated the variables and applied the initial condition and obtained:
$$y(t)=\left(3t+27\right)^{1/3}-2$$
but now I can't figure out why does the solution exist only when $t>-9$? I am having trouble understanding also what does this have to do with $\dfrac{dy}{dt}$ not being defined at $y=-2$. Why are we letting
$$3t+27 >0 $$
what if $t=-10$? how does this make $y$ undefined?
 A: $$\dfrac{dy}{dt}=\dfrac{1}{\left(y+2\right)^{2}}$$
$$t=\int (y+2)^2dy=\frac13 (y+2)^3 +c$$
Condition : $t(1)=0=\frac13 (1+2)^3 +c \quad\implies\quad c=-9 $
$$t=\frac13 (y+2)^3-9$$
For $y$ and $t$ reals this implies $t>-9$.

If the problem is from Physic the speed $y'(t)$ is infinite at $(t=-9\:,\:y=-2)$ .  This excludes the branch $y\leq -2$.
A: It is perhaps not stated often enough outright, but in standard ODE theory, where the "O" stands for "ordinary", the domain of the ODE is supposed to be an open set that is chosen so that the right side exists, and is continuous in all variables. A solution is then a continuously differentiable function with a graph inside that set.
In this case this means that the line $y=-2$ is a boundary of the domain, and as the initial condition is $y(0)=1$, the domain is the upper part $\{(x,y):y>-2\}$. As you computed, the solution only stays inside this domain for $t>-9$. There can be no more solution in the given context.
Now it is of course quite possible to consider limits of the solution at the boundary, and solutions in the second domain $\{(x,y):y<-2\}$, pairs of solutions that can be joined together at the boundary, or result from the same algebraic expression. But these are extended solution concepts, which need to be requested explicitly.
