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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?

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  • $\begingroup$ I presume that in your problem has been given also some initial condition. What is it? $\endgroup$ – Robert Z Sep 15 '16 at 10:10
  • $\begingroup$ 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. $\endgroup$ – астон вілла олоф мэллбэрг Sep 15 '16 at 10:12
  • $\begingroup$ @RobertZ Added condition to OP. $\endgroup$ – Lozansky Sep 15 '16 at 10:19
  • $\begingroup$ @Lozansky So $x_0=0\in (-2,3)$. Now see my answer. $\endgroup$ – Robert Z Sep 15 '16 at 10:20
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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?

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

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