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In G. F. Simmons' Differential Equations book (p.141), the following claim is made: “... As a matter of fact, there is no known type of second order linear differential equation- apart from those with constant coefficients, and those reducible to these by changes of the independent variable--which can bee solved in terms of elementary functions.”

About 36 years have now passed since this statement made its published appearance. Is the remark still true or was it false even before?

My motivation is the analogous theory of integrals of finite combinations of elementary functions where you know certain large and important classes of functions whose integral is expressible as an elementary function( or as a finite combination of such). As we know, there is a somewhat complete classification as to which integral of finite combination of elementary function of one variable is expressible as a finite combination of elementary functions.(References are Piskunov's book and this ). So it seemed to me that it is natural to care whether such theorems existed for differential equations.


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I don't understand the claim. Choose two elementary functions $y(t)$ and $z(t)$, functionally independent. Then the system formed by $a(t)y'(t)+b(t)y(t)+y''(t)=0$ and $a(t)z'(t)+b(t)z(t)+z''(t)=0$ admits a unique solution $(a,b)$. Then the linear 2nd-order ODE $f''+af'+bf=0$ is solvable in terms of elementary functions. What is wrong ? – Denis Serre Feb 4 '11 at 17:28
maybe it just depends on what's meant by a type of differential equation? – Anthony Quas Feb 4 '11 at 17:36
@Denis, thanks for the edit. Is your $f$ equal to $y+z$? – Unknown Feb 4 '11 at 17:49
@Anthony, the sought-after type is a 2nd order linear ODE which is not one with constant coefficients, or one reducible to such by changes of the independent variable. – Unknown Feb 4 '11 at 18:11
@Elohemahab: not just $y+z$. By the definition of $(a,b)$, Denis's construction admits any linear combination (a 2-parameter family) of the functions $y$ and $z$. – Willie Wong Feb 5 '11 at 12:40
up vote 28 down vote accepted

This claim is not valid. The big breakthrough was the paper below:

MR0839134 (88c:12011) Kovacic, Jerald J. An algorithm for solving second order linear homogeneous differential equations. J. Symbolic Comput. 2 (1986), no. 1, 3–43. 12H05 (34A30)

Later, M. Petkovsek extended the ideas to second order difference equations.

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Thank you very much. I also found upon googling the following: – Unknown Feb 4 '11 at 23:00
Thanks for Mariano, too. I did not find the paper at first and Google did not turn anything. – Unknown Feb 4 '11 at 23:40

If we allow transformations of both the independent and the dependent variable, then the statement is tautologically true. Say you have a linear second order ODE for a function y(x). Let $y_1(x)$ and $y_2(x)$ be linearly independent solutions. Define $t=y_2(x)/y_1(x)$, and $z=y/y_1$. In the transformed variables, the general solution is $z=c_1+c_2t$, so the equation is $z''(t)=0$. Of course, this is not useful for solving any equations, since we have to know the solutions to define the transformation.

The statement as quoted focusses on the independent variable, which seems odd to me. If we have any differential equation solvable in terms of elementary functions, and we set $y=z\phi(x)$, where $\phi$ is a known elementary function, we get another equation solvable in terms of elementary functions. However, this is a change of the dependent variable, not the independent variable. Also, Denis rightly points out that any two functions can be made into solutions of a second order ODE.

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I think that the questionner could be addressed to the mathematical subject named Differential Galois Theory.

A first accessible reading could be J.F. Ritt "Integration in finite terms. Liouville's theory of elementary methods", Columbia University Press, 1948.

The corresponding wikipedia page could be another starting point.

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True, and this is what the Kovacic paper is based on. – Igor Rivin Feb 4 '11 at 21:26
Thanks. The accessibility of the book mentioned is really attractive. – Unknown Feb 4 '11 at 23:04

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