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Suppose $f(x,y)$ is a rational function. Typically the solution curves of the differential equation $$ \frac{dy}{dx}=f(x,y) $$ are not algebraic (for example, if $f(x,y)=y$, the solutions are the non-algebraic curves $y=ce^x$). There are certain $f$, however, for which all solutions are algebraic (for example, if $f$ is a rational function in just $x$, having $0$ residue at every pole).

Given $f$, is there a way to check whether all the solutions to $\frac{dy}{dx}=f(x,y)$ are algebraic?

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  • $\begingroup$ I don't think so, since if y' is a rational function, y could be rational, algebraic, logarithmic, trigonometric, inverse trigonometric or many other things $\endgroup$ – Seth Nov 5 '18 at 0:44
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    $\begingroup$ Just a tongue-in-cheek answer: a good way to check would be solving the equation. $\endgroup$ – Yuriy S Nov 5 '18 at 4:47
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    $\begingroup$ This problem has a long and storied history going back to the 1800s in Berlin: projecteuclid.org/download/pdf_1/euclid.bams/1183551413 . Although this brief history doesn't contain detailed statements of the results, it might help you with where to look. $\endgroup$ – KReiser Nov 6 '18 at 3:42
  • $\begingroup$ I'd attack such a problem by using Picard-Vessiot theory (see the first chapter of Singer and van der Put) to determine if the equation can be solved with integrals, then apply the Risch algorithm to solve the integrals. $\endgroup$ – Brent Baccala Feb 12 at 22:56

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