In principle, can every second order ODE be solved?

Question

In studying differential equations, I came across a formula to solve the equation: $$(A\cdot y)(x)+(B\cdot d_xy)(x)+C(x)=0$$ Where $\{A,B,C,y\}$ are all elements in an algebra of continuous functions, and $d_x=\frac{d}{dx}$. To solve this equation, we can represent the solution wrt $\{A,B,C\}$ and an arbitrary constant $\gamma$ in the following way: $$y_\gamma(x)=\gamma\Lambda(x)-\left(\Lambda\int\frac{C}{B\cdot\Lambda}dx\right)(x)$$ Where $\Lambda(x)=e^{-(\int\frac{A}{B}dx)(x)}$. If all the integrals exist, then $y_\gamma$ is a solution to the equation. This solution relies on algebraic manipulations that are not particularly complex, however convoluted. So, my question is: is it possible to do the same to find the solutions for the following equation?: $$(A\cdot y)(x)+(B\cdot d_xy)(x)+(C\cdot d_x^2y)(x)+D(x)=0$$ This is, is it possible to define a class of functions $S$ in terms of integrals, constants, and the functions $\{A,B,C,D\}$ such that, provided all the integrals exist, every $y\in S$ satisfies the second-order ODE?

Answer 1

Picard-Lindelöf guarantees a solution exists under fairly light assumptions. Second order ODEs are much less friendly than first order, even in the linear case. Every first order linear is solvable with integrating factors as you have seen. Integrating factors turn a first order linear ODE into a product rule that can be undone by integrating.

You can use a compounded integrating factor technique on some second order ODEs but not all. An example that can be solved in the way you are thinking is $y'' - 5y' + 6y = 0$. There are some very simply stated second and high order linear ODEs that are not solvable with elementary functions. The classic example is the Bessel differential equation given by

$$ x^2 y'' + xy' + (x^2 - \nu^2)y = 0. $$

This equation is solvable with elementary functions for half integer order $\nu$ and involves powers and sinusoids. Otherwise its solutions require special functions (shockingly called Bessel functions) which are not elementary, meaning they cannot be written in terms of trigonometric, exponentials, logarithms, rational functions, powers, etc. Every second order linear ODE

$$ ay'' + by' + cy = d $$

can be written in a "monic" form

$$ y'' + by' + cy = d $$

by dividing through by $a$ (these are not the same coefficients unless $a=1$ of course!). The Bessel ODE becomes

$$ y'' + \frac{1}{x} y' + \bigg(1-\frac{\nu^2}{x^2}\bigg)y = 0. $$

A second order linear ODE in this monic form can be further simplified to a second order linear ODE of the form

$$u'' + Cu = D$$

with the substitution

$$ y = \exp\bigg(-\frac{1}{2} \int b\bigg)u. $$

In the Bessel case, this is

$$ y = x^{-\frac{1}{2}} u, $$

giving the new ODE

$$ u'' + \bigg(\bigg(\frac{1}{4}-\nu^2\bigg) \frac{1}{x^2} +1\bigg)u = 0 $$

(if I haven't made any silly math errors).

I do not believe there is a direct way to extract a Bessel function from this equation with elementary calculus operations on the $C$ term (which is the only nonconstant team remaining). I suspect a true proof of this would require an appeal to differential algebra (or differential Galois theory) which is partially concerned with the kind of functions that can be obtained via integrating simple elementary functions.