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Why is the interval of definition for $xy' + 4y = x^3 - x$ given by

$0 < x < \infty$ instead of $-\infty < x < 0 \; \cup \; 0 < x < \infty$?

When I solved it I got $y = \frac{1}{7}x^3 - \frac{1}{5}x + cx^{-4}$, and I thought when finding the interval of definition, you check where the function is defined in the original and its derivatives.

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Because "$-\infty < x < 0 \; \cup \; 0 < x < \infty$" is not an interval. – user53153 Feb 10 '13 at 3:20
up vote 1 down vote accepted

It goes back to what we mean by interval of existence for a solution to the DE.

Take a look at the explanations here; they should help.

The bottom line is, after the equation is put in the standard form of a first order linear equation (as demonstrated in the linked notes), there is a discontinuity in the ingredient functions at $x=0$, which subdivides the $x$ axis into two intervals: $(-\infty,0)$ and $(0,\infty)$. The interval of existence is one or the other (dictated by where you want to have the initial condition specified), but not both.

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