# Airy equation and solution

1. Apr 25, 2010

### LBloom

1. The problem statement, all variables and given/known data

So the Airy equation states that y''-xy'=0. My problem is to proof that the improper integral
1/$$\pi$$ $$\int$$cos(1/3*t^3+x*t) from 0 to $$\infty$$ satisfies this equation.

I've tried differentiating under the integral sign, but all I've gotten is the integrand to be
y''+ty = -1/pi $$\int$$ cos(1/3*t^3+x*t)(t^2+x) from zero to infinity. Naturally I do u substitution, but my final answer comes out to be -sin(infinity)/Pi

What should I be doing?

2. Relevant equations

3. The attempt at a solution

2. Apr 25, 2010

### vela

Staff Emeritus
After the substitution, you have

$$y''-xy=\int_0^\infty \cos u\,du$$

That integral doesn't converge because the integrand doesn't vanish as $u\rightarrow\infty$. The trick is to introduce an integrating factor $e^{-\lambda u}$ and then take the limit as $\lambda\rightarrow 0^+$.

3. Apr 25, 2010

### LBloom

I'm sorry, but i don't quite understand what I do from there. If the integrand is cos u e^-k*u
it'll converge, but do i find k and why you take the limit as k go to 0 from the right?

this problem is supposed to be over the heads of students in our class b/c its extra credit and I'm a freshman in a sophomore class, so I don't know if I'm missing anything.

4. Apr 25, 2010

### vela

Staff Emeritus
No, you don't find k; you're taking a limit as k goes to 0. The idea is that e-ku will go to 1 as k goes to 0, so the integrand is again just cos u. You have to approach from the right so that -ku<0, otherwise e-ku would go to positive infinity when you plug in the upper limit of the integral.

5. Apr 25, 2010

### LBloom

ahhh, my answer works out to zero now. thank you.

6. Apr 30, 2010

### LBloom

The only thing I dont get is, if I'm trying to do a proof of y''+xy=0 for a given equation, why am i allowed to put an integrating factor into the equation?? Wouldn't that show I was unable to prove what I wanted to prove?

7. Apr 30, 2010

### vela

Staff Emeritus
I should have said "convergence factor," not "integrating factor."

Honestly, I don't know the mathematical justification for when such factors are allowed and what their requirements are. It's just a technique that I occasionally ran across in physics, but the mathematical theory behind it was never explained.

8. Apr 30, 2010

### LBloom

I'll make sure to read up on convergence factors once my tests are over. Thanks for the help!