For the given initial-value problem:

$$dfrac{dy}{dt}=dfrac{1}{left(y+2right)^{2}}, quad y(0)=1$$

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:

$$y(t)=left(3t+27right)^{1/3}-2$$

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?