Lazy mathematician meets linear 2nd order ODE (2)

Part 2: you're not lucky

As explained here:

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

Today, we suppose you're still 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)

Most textbooks provide a simple transformation achieving this reduction, but doing so they require a(x) to have a continuous derivative. As we don't want to go beyond continuity, we'll have to find a different way. On the blackboard below is a simple and explicit way to switch between (2) and (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 x₀ in the interval under consideration, and for any continuous function f we denote by If the antiderivative of f vanishing in x₀. To save parentheses, we concatenate antiderivatives and products from right to left. Thus, for instance, I²bI²b 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-I²b+I²bI²b-I²bI²bI²b+... 

Full proof on this blackboard:

With φ obtained, you made yourself lucky. Proceed as earlier or read on.

Full solution of (3)

While we're at it, why not solve (3) completely? The same proof shows that

ψ =-x₀+x-I²bx+I²bxI²bx-I²bxI²bxI²bx+... 

(we have written x instead of the function mapping x to x) is also a solution of (3). From

φ(x₀)=1,  φ'(x₀)=0, ψ(x₀)=0, ψ'(x₀)=1

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

c₁φ+c₂ψ

with 2 arbitrary constants. Examples with x₀=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!!)

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