Solving Airy's Equation and Applying the Sturm Comparison Theorem

[Ted123]

Homework Statement

(a) By using a suitable transformation, show that the normal form of the DE y'' - 2y' + (x+1)y = 0\;\;\;\;\;(*) is Airy's equation u'' + xu = 0.
(b) State the Sturm comparison theorem for zeros of 2 second order linear DEs in normal form.

(c) By comparing with the DE v'' + v = 0 prove that every solution y(x) of (*) has infinitely many positive zeros.

The Attempt at a Solution

I've done (a). For (b), the theorem is:
[PLAIN]http://img101.imageshack.us/img101/1237/sturmh.png

I'm not sure how to proceed with (c).

 

[HallsofIvy]

It's pretty straightforward, isn't it? The two functions to be compared are q(x)= x and r(x)= 1. It is certainly true that x> 1 for all x in, say, [2, \infty). Now, how many times do solutions of y''+ y= 0 vanish in that interval?

 

[Ted123]

It's pretty straightforward, isn't it? The two functions to be compared are q(x)= x and r(x)= 1. It is certainly true that x> 1 for all x in, say, [2, \infty). Now, how many times do solutions of y''+ y= 0 vanish in that interval?

The general solution of v'' + v = 0 is v = A\sin x + B\cos x

It has successive zeros at x = n\pi - \frac{\pi}{4} where n\in\mathbb{Z}.

Does this prove that every solution y of (*) has infinitely many positive zeros?

(By the comparison theorem, any solution of u'' + xu=0 and therefore of (*) has a solution in between those successive zeros. Since there are infinitely many positive zeros, there are infinitely many positive zeros in (*))

 

[Ted123]

I've finally got hold of the 'solution' but does using an example of a solution like it does with u=\sin x prove that the equation has infinitely many (positive) zeroes for all solutions?

[PLAIN]http://img94.imageshack.us/img94/9296/sturmd.jpg