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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?

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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 $(-10/3,7/6)$

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Woah woah woah, not so fast, you need to prove this. The domain of definition does not necessarily ends where "the derivative blows up" (ie. where $f$ such that $y'(t)=f(t,y(t))$ does not exist). Take $y' = 1 + y^2, y(0)=0$; it's well known the solution is $\tan$, which does not exists at $\frac{\pi}{2}$: something you can't guess from the equation alone the way you did. – Najib Idrissi Jul 14 '11 at 16:48

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!

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