**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.

Exactly! The Babylonians found how to solveMathematicians 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.

ax

^{2}+bx+c=0by writing it as

(x+b/2a)

^{2}=(b^{2}-4ac)/(4a^{2})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**

*number*x but an unknown

*function*y(x) is to be determined. Among the simplest ones is the

*linear ODE of first order*

y'+a(x)y=R(x)

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

*y''+a(x)y'+b(x)y=r(x)*

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

y''+a(x)y'+b(x)y=0,

with zero as right-hand member. (If you are unlucky, see here.) Any calculus textbook will give you the steps to follow from here:

- 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*.) - 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*.)

*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 c

_{1}and c

_{2.}

_{ }

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*.