# What justification can you give for the fact that “most ODEs do not have an explicit solution”?

What justification can you give for the fact that "most ODEs do not have an explicit solution"?

• First of all, one would need to say what is meant by "explicit". Could you please say where you saw this quote or phrase, and in what context? – Yemon Choi Sep 22 '14 at 14:42
• I added the tag "soft question" as it is purposefully vague. – Ryan Budney Sep 22 '14 at 17:32
• Solving an ODE has to be harder than integrating or solving an equation for an unknown number, and these assignments are already typically hopeless as far as explicit solutions are concerned. – Christian Remling Sep 22 '14 at 17:44
• As is often the case, "explicit" is a misnomer for "elementary", where the latter more literally refers to polynomials, exponentials-and-logs, trig functions, roots, ... As in is implicit in Loic Teyssier's answer, the coefficients of a differential equation can be elementary, while the solutions are demonstrably not. – paul garrett Sep 22 '14 at 19:21

If the ODE is linear --and the notion of «explicit» refers to Liouvillian solutions (towers of iterated quadrature and exponential of meromorphic functions)-- then its differential Galois group (Picard-Vessiot theory) must be a solvable algebraic subgroup of $GL_n (\mathbb C)$. Such subgroups are rare: they define a proper algebraic subvariety. The defining equations correspond to trivial commutations relations, e.g $[G,G]=0$ in the Abelian case, encoding the tower of normal subgroups intervening in the definition of solvable group.
• In other words, it is much like the case for solutions to polynomials in one variable over $\mathbf{Q}$. Just as an aside, sometimes the notion of "explicit" permits the intervention of algebraic functions, corresponding to the finite component group of the differential Galois group being non-trivial; that finite group might also be non-solvable, but that is a separate can of worms. :) – user27920 Sep 22 '14 at 17:04
On the non-linear side, there is a theorem of Hudai-Verenov MR0147699, which says that for a generic (open dense set) equation $y'=R(x,y)$ the graph of a generic solution is dense. See also Ilyashenko, MR0247176. This implies that this equations do not have first integrals, so they are not "solvable" in the sense that we teach in the elementary courses.