ordinary differential equations
The ODE: $y' = (1-2x)y^2$
Initial Value: $y(0) = -1/6$
I've solved the particular solution, which is $1/(x^2-x-6)$. I don't understand what they mean about the solution is defined, because when I graph $1/(x^2-x-6)$, it's only discontinuous at $x = -2, 3$.
What does "Determine the interval in which the solution is defined" mean?
I find it useful to plot the function after solving for the constant from the ICs.
We have:
$$y(x) = -\dfrac{8}{e^{4 x}-2}$$
A plot of $y(x)$ shows:
Of course, analytically, we care about where that denominator is $0$, so we have:
$$e^{4 x}-2 = 0 \rightarrow x = \dfrac{\ln 2}{4} \approx 0.173286795139986$$
From the plot, you can also see what happens as $x \rightarrow \pm \infty$, but it is not bad to derive that analytically either.
One last helpful thing you may have not gotten to yet, is a phase plot, which looks like:
An interval of definition of a solution is any (open) interval on which it is defined. For example: the problem $y'=y^2$, $y(0)=1$, has solution $y(t)=1/(1-t)$. This solution is defined on the interval $(-1/2,1/3)$, so we can say this is an interval of definition of solution. It's not the largest such: the largest interval of definition is $(-\infty,1)$.
Another example: $y(t)=\sqrt{t}$ satisfies the ODE $y'=1/(2y)$. Although the domain of square root is $[0,\infty)$, when it is considered as a solution of this ODE, the interval of definition is $(0,\infty)$. At $t=0$ the ODE does not make sense, so it would not be correct to say that $y$ satisfies the ODE there. And generally, ODEs are considered on open intervals only.
Also, I'd expect the book you are using to define the term interval of definition the first time it appears.
Best Answer
It makes sense to consider solutions only on intervals that contain the initial time. The "interval where the solution is defined" (I would call it the (maximal) interval of existence) is the maximal interval of all intervals $I$ which contain 0 and there exists a solution on $I$. This maximal interval turns out to be $(-2,3)$ in your case.
So why doesn't it make sense to consider solutions defined on other subsets of $\mathbb{R}$ than intervals, e.g. $(-\infty,-2)\cup (-2,3)\cup (3,\infty)$? One reason is that you wouldn't have uniqueness of solutions, for example, $$\begin{cases}1/(x^2-x-6) & x<3 \cr 0 & x>3\end{cases}$$ would be another "solution". What happens in the other components which don't contain 0 can be rather arbitrary, so one does not allow them.