**Part 2: you're not lucky**

As explained here:

- Mathematicians are a lazy bunch.
- Facing y''+a(x)y'+b(x)y=r(x) (1)
*lucky,*if you happen to know a solution φ (not identically zero) ofy''+a(x)y'+b(x)y=0. (2)

Today, we suppose you're still

What if we managed to get rid of y' in the equation (2)? In solving third and fourth order algebraic equations, it helps a lot to eliminate the next-but-highest term. Here, getting rid of y' doesn't help

*lazy*but*not*lucky. First, let's simplify. (Lazy*and*simple-minded, euh?)**As simple as it gets**

What if we managed to get rid of y' in the equation (2)? In solving third and fourth order algebraic equations, it helps a lot to eliminate the next-but-highest term. Here, getting rid of y' doesn't help

*that*much, but is does make things a little lighter. OK, let's head for
y''+b(x)y=0. (3)

**One solution of (3)**

To obtain a solution of (3), there is the long established method of successive approximations, to be found in, e.g., E. L. Ince,

*Ordinary Differential Equations*, 1927. This strategy results in a*sequence*of functions, whose uniform convergence requires to artificially adapt the M-test, which is naturally designed for*series*.
The lazy mathematician, of course, prefers a

*formula*over a*strategy*. Our formula for φ will be a*series*, whose uniform convergence can be established most naturally by the M-test. Some notation is required. We fix x_{0}in the interval under consideration, and for any continuous function f we denote by If the antiderivative of f vanishing in x_{0}. To save parentheses, we concatenate antiderivatives and products from right to left. Thus, for instance, I^{2}bI^{2}b*stands for the function obtained by twice integrating b, multiplying the result by b, and integrating the product twice.*
And here it comes (ta-ta!!!):

φ =1-I

^{2}b+I^{2}bI^{2}b-I^{2}bI^{2}bI^{2}b+...**Full solution of (3)**

While we're at it, why not solve (3) completely?

**The same proof shows that**
ψ =-x

_{0}+x-I^{2}bx+I^{2}bxI^{2}bx-I^{2}bxI^{2}bxI^{2}bx+...
(we have written x instead of the

it is clear that both solutions are independent, and have a Wronskian determinant 1. Hence, (3) is completely solved, its solutions being

with 2 arbitrary constants. Examples with x

Nothing beats our formulas here!

If you are lazy but unlucky, you first pass from (1) to

We have kept the notation b(x) and r(x), but both functions have changed in the process of eliminating y'. Instead of b(x) we should consider B(X) as defined on the first blackboard, and change r(x) to R(X) in exactly the same way. With φ and ψ defined as above, the general solution of (8) is (ta-taa!!)

Extremely lazy proof-by-verification on the blackboard below.

*function*mapping x to x) is also a solution of (3). From
φ(x

_{0})=1, φ'(x_{0})=0, ψ(x_{0})=0, ψ'(x_{0})=1it is clear that both solutions are independent, and have a Wronskian determinant 1. Hence, (3) is completely solved, its solutions being

c

_{1}φ+c_{2}ψwith 2 arbitrary constants. Examples with x

_{0}=0 (immediately obtained):Nothing beats our formulas here!

**Full solution of the linear 2nd order ODE**

If you are lazy but unlucky, you first pass from (1) to

y''+b(x)y=r(x). (8)

We have kept the notation b(x) and r(x), but both functions have changed in the process of eliminating y'. Instead of b(x) we should consider B(X) as defined on the first blackboard, and change r(x) to R(X) in exactly the same way. With φ and ψ defined as above, the general solution of (8) is (ta-taa!!)