# Find interval where function is defined

Say we have the separable DE $$\frac{dy}{dx} = (1-2x)y^2, y(0) = -\frac{1}{6}$$ with explicit solution $$y = \frac{1}{x^2-x-6} = \frac{1}{(x+2)(x-3)}$$ Why is it that the interval, where the solution is defined, is said to be only $$-2<x<3$$ and not $$I = \left\{-\infty<x<\infty | x \neq -2, 3\right\}$$, since clearly $y$ is defined on $x>3$ and $x<-2$ as well?

• I presume that in your problem has been given also some initial condition. What is it? – Robert Z Sep 15 '16 at 10:10
• It could possibly be defined, except that we would say it has singularities at the point $-2$ and $3$, which is to say it blows up at those points. – астон вілла олоф мэллбэрг Sep 15 '16 at 10:12
• @RobertZ Added condition to OP. – Lozansky Sep 15 '16 at 10:19
• @Lozansky So $x_0=0\in (-2,3)$. Now see my answer. – Robert Z Sep 15 '16 at 10:20

If in your problem is given also a condition like $y(x_0)=y_0$ with $x_0\not=-2,3$ then $$y(x) = \frac{1}{x^2-x-6} = \frac{1}{(x+2)(x-3)}$$ is a solution defined in $(-2,3)$ if $x_0\in (-2,3)$. However it is defined in $(3,+\infty)$ if $x_0\in (3,+\infty)$ and, similarly it is defined in $(-\infty,-2)$ if $x_0\in (-\infty,-2)$.

Usually by definition a solution of an ODE is defined in a interval (which is connected). Take a look also here: What is the interval of definition of a solution of an ODE?

• Okay, didn't know it had to be a connected set. Thank you. – Lozansky Sep 15 '16 at 10:25