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For a polynomial equation in one variable over $\mathbb{Q}$, it is well known that the equation is solvable by radicals if and only if the equation's Galois group (which is a finite group) is solvable. The 'only if' part is important - we need it to prove that an equation of the fifth degree exists which is not solvable by radicals.

For differential equations, we have the following theorem, which I quote from [1] (theorem 2.64, page 151):

If an n-th order ordinary differential equation admits an n-parameter solvable group of symmetries, then the equation is solvable by quadratures - that is, the general solution to the equation can be expressed by integrals.

Here by an 'n-parameter solvable group' we mean a Lie group of dimension n which is solvable. By the equation "admitting" the group we mean that the group, acting on $\mathbb{R}^2=X \times Y$ (where $x$ is the independent variable in the equation and $y$ is the dependent variable), transforms solutions of the equation to other solutions of the equation.

Is the converse to the above theorem true? that is, If an n-th order ordinary differential equation is not solvable by quadratures, does it follow that the equation does not admit a solvable n-parameter group of symmetries?

If this converse is not true, is it because there are differential equations solvable by quadratures which do not admit an n-parameter group of symmetries at all? Or is it because there are differential equations solvable by quadratures which admit a nonsolvable n-parameter group of symmetries? Whatever of these cases is correct, is there a concrete example of such a differential equation?

[1] Olver, Peter J., Applications of Lie groups to differential equations., Graduate Texts in Mathematics. 107. New York: Springer-Verlag. xxviii, 513 p. (1993). ZBL0785.58003.

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  • $\begingroup$ The book by Willard Miller Lie Theory and Special Functions (1968) might well go there... $\endgroup$ Mar 7 at 16:30

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I stumbled on your question several times in the past, but I was never able to find an example. I just found one written by Olver himself. It's Exercise 4.7 in here. The equation is given by $$ y''=\left((x+x^2)e^y\right)'. $$ It can be solved by quadrature. Less obvious is that it does not have continuous symmetry. However, if you have a program finding the Lie symmetries, you can check that.

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