16 October 2015

Lazy mathematician meets linear 2nd order ODE (1)

Mathematical laziness

Mathematicians are lazy people, they say so themselves. On the Internet, the string "lazy mathematician" yields some 1600 immediate hits, "mathematics is for lazy people" (coined in 2000 by Peter Hilton) some 300.

On Tanya Khovanova's Math Blog, among other aspects of The Virtue of Laziness, the following is worth citing here.
Mathematicians are driven by laziness. Once ancient mathematicians first solved a quadratic equation, they didn’t want to do it again. So they invented a formula that solves all quadratic equations once and for all.  
 Exactly! The Babylonians found how to solve


by writing it as

which yields first x+b/2a, then 

In any math book, the ancient reasoning is repeated once, and after that the boxed formula is used time and again. Students all over the world know the formula by heart.

Linear 1st order ODE 

Some millennia later, we enter the realm of the ODE (ordinary differential equation), where not an unknown number x but an unknown function y(x) is to be determined. Among the simplest ones is the linear ODE of first order

where the functions a(x) and R(x) are known and y(x) is to be found. Here again, it suffices to do the reasoning once, to box the formula and apply it every time you need it. (If you're extremely lazy, you could also simply verify the formula, without any reasoning at all.) When I was a student, we had to know this formula (which contains an arbitrary constant c) by heart, just as the one above.

Linear 2nd order ODE, Part 1: you're lucky

The second order version is


with a(x), b(x) and r(x) known and y(x) unknown. By definition, you are lucky if you happen to know a solution (not identically zero), say φ, of the equation

with zero as right-hand member. (If you are unlucky, see here.) Any calculus textbook will give you the steps to follow from here:
  1. determine a function u(x) such that ψ(x)=u(x)φ(x) is not simply a constant multiple of φ(x) yet satisfies the same equation as φ(x). (Because u(x) is determined from a first order ODE, this step is called reduction of the order.)
  2. determine functions v(x) and w(x) such that v(x)φ(x)+w(x)ψ(x) is a solution of the original ODE. (Because v(x) and w(x) replace arbitrary constants—which would yield nothing new—, this step is called variation of the parameters.)
Strangely enough, I could not find any textbook giving the formula by which you could avoid repeating these steps for every new 2nd order ODE. Even the first step is usually given as a strategy, while a simple formula does exist. It is to be found, for instance, in Goursat's 1917 classic Course in mathematical analysis, part Differential Equations:

But the biggest secret, well kept until now, is the general solution which emerges in the end (ta-ta!!!!):

with two arbitrary constants c1 and c2.
My colleagues used to frown upon my lazy attitude. For reasons I never understood, they preferred their students to eagerly repeat a strategy whose outcome I knew in advance. Like using the Babylonian trick instead of the formula, yes.

 Lest I forget, here is the extremely lazy proof-by-verification.