What is the domain of definition for the solution of DE? We have the following Differential Equation with the initial condition
$$ \frac{dy}{dt}=\frac{1}{(y+1)(y-2)},~~ y(0)=0$$

By seperating the variables i got the general solution as follows
$$ \frac{1}{3}y^3-\frac{1}{2}y^2-2y=t+C_1 \implies 2y^3-3y^2 -12y=6t+6C_1$$

(1) What is a domain of definition?
(2) And what would be the domain of definition for the solution?
 A: I don't see the difference between "domain of definition" and "domain of definition for the solution".
The fact that the denominator is factored actually gives us a clue.  
There is a theorem that states:  If $\frac{dy}{dt} = f(t, y)$, $y(t_0) = y_0$, where both $f$ and $\frac{\partial f}{\partial y}$ are continuous in some open rectangle $(a, b) \times (c, d)$ containing the point $(t_0, y_0)$, then there exists a unique solution to the IVP for some interval of $t$-values, $(t - \epsilon, t + \epsilon) \subseteq (a, b)$.  
Now this implies that we can only expect a solution to your ODE somewhere within the infinite strip defined by $-1 < y < 2$, $t \in \mathbb{R}$.  If the solution curve leaves this strip, then all bets are off.  In practice, one finds the appropriate restrictions on $t$ based on the solution of the IVP (if one can be found).  Fortunately, your ODE was separable, and so just plugging in $y = -1$ and $y = 2$ will give the appropriate $t$-bounds.  (By the way, you can determine $C_1$, since an initial value is given.)   It helps to graph your solution... Graph $t = y^3/3 - y^2/2 - 2y$ and think "inverse functions".
Hope this helps!
A: By plugging in $y(0)=0$ you see that $C_1=0$. You still need to solve the equation for $y$, if you want to write $y$ as function of $t$, but this doesn't seem easy. From the differential equation you see that the derivative blows up at $y=-1$ and $y=2$. Plugging this into the equation you got we see that this happens at $t=7/6$ and $t=-10/3$. The domain of definition is $(-10/3,7/6)$
A: The differential equation $$y'=\frac1{(y+1)(y-2)}$$ relates the function $y$ to its derivative $y'.$ With the initial condition $y(0)=0,$ the equation implies $$y'(0)=\frac1{[y(0)+1][y(0)-2]}=-\frac12,$$ which means the domain of $y$ must be some open interval containing $0.$ The equation is equivalent to $$(y+1)(y-2)y'=(y^2-y-2)y'=1,$$ and on an open interval containing $0,$ this is equivalent to $$\left(\frac{y^3}3-\frac{y^2}2-2y\right)'=1.$$ Per the fundamental theorem of calculus, $$\frac{y(t)^3}3-\frac{y(t)^2}2-2y(t)=t,$$ since $y(0)=0.$ Let $F:\mathbb{R}\to\mathbb{R}$ be defined by $$F(x)=\frac{x^3}3-\frac{x^2}2-2x,$$ hence $$F[y(t)]=t.$$ Notice that the differential equation, together with the initial condition, constrain $y(t)\in(-1,2).$ Thus, the domain of solution must be the image of $(-1,2)$ under $F.$ To find it, notice that the stationary points of $F$ are $-1$ and $2,$ and so $$F(-1)=-\frac13-\frac12+2=\frac76$$ and $$F(2)=\frac83-2-4=-\frac{10}3,$$ hence the domain of solution is $\left(-\frac{10}3,\frac76\right),$ with $$y(t)=F^{-1}(t),$$ noting that $F^{-1}$ is well-defined when $F$ is restricted to $(-1,2).$
Finding an explcit expression for $F^{-1}$ in terms of radical expressions is possible using Cardano's method, but this is tedious, and so, I will not do that here.
