Airy Function problem

**TwistedOne151** · May 27th 2008, 01:16 AM

The Airy function Ai(x) is one of the solutions to the differential equation y''-xy=0. I'm trying to figure out the (similar, corresponding) differential equation that has Ai(m*x+b) as a solution. I've tried several ways of performing the calculus and change of variables, and keep getting different answers. This one looks the most correct, but I'm not sure why I'm getting other answers:

Let $u=mx+b$ . Now, we know:
$\frac{d^2}{du^2}y=u\cdot{y}(u)$ (Airy's equation)
Now, via the chain rule, $\frac{dy}{dx}=\frac{dy}{du}\frac{du}{dx}=m\frac{dy }{du}$ ,and thus $\frac{d^2y}{dx^2}=m^2\frac{dy}{du}$
So
$\frac{d^2y}{dx^2}=m^2\frac{dy}{du}=m^2uy(u)=m^2(mx +b)y(mx+b)$
so Ai(m*x+b) would be the solution to $y''-m^2(mx+b)y=0$ by this arguement.
Does this make sense?

--Kevin C.

**mr fantastic** · May 27th 2008, 05:39 AM

Originally Posted by TwistedOne151

The Airy function Ai(x) is one of the solutions to the differential equation y''-xy=0. I'm trying to figure out the (similar, corresponding) differential equation that has Ai(m*x+b) as a solution. I've tried several ways of performing the calculus and change of variables, and keep getting different answers. This one looks the most correct, but I'm not sure why I'm getting other answers:

Let $u=mx+b$ . Now, we know:
$\frac{d^2}{du^2}y=u\cdot{y}(u)$ (Airy's equation)
Now, via the chain rule, $\frac{dy}{dx}=\frac{dy}{du}\frac{du}{dx}=m\frac{dy }{du}$ ,and thus $\frac{d^2y}{dx^2}=m^2\frac{dy}{du}$
So
$\frac{d^2y}{dx^2}=m^2\frac{dy}{du}=m^2uy(u)=m^2(mx +b)y(mx+b)$
so Ai(m*x+b) would be the solution to $y''-m^2(mx+b)y=0$ by this arguement.
Does this make sense?

--Kevin C.

I get the same answer.

I think it's easier to see when the DE is written as $\frac{1}{m^2} \frac{d^2 y}{d x^2} - (mx + b) y = 0$ . It's immediately obvious that the substitution u = mx + b gives $\frac{d^2 y}{d u^2} - u y = 0$ .....

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