# Differential Eqns. (checking an answer + some questions)

1. Dec 14, 2004

Any help is highly appreciated.

Thank you

Problem

$$y^{\prime \prime} = xy$$

My Solution

If

$$y = \sum _{n=0} ^{\infty} c_n x^n \Longrightarrow xy = \sum _{n=0} ^{\infty} c_n x^{n+1} = \sum _{n=1} ^{\infty} c_{n-1} x^n$$

and

$$y^{\prime \prime} = \sum _{n=2} ^{\infty} \left( n-1 \right) n c_n x^{n-2} = \sum _{n=0} ^{\infty} \left( n+1 \right) \left( n+2 \right) c_{n+2} x^{n}$$

Then

$$\sum _{n=0} ^{\infty} \left( n+1 \right) \left( n+2 \right) c_{n+2} x^{n} = \sum _{n=1} ^{\infty} c_{n-1} x^n$$

Hence, we may find the recursion relation

$$c_{n+2} = \frac{c_{n-1}}{\left( n+1 \right) \left( n+2 \right)} \qquad n=1,2,3,\ldots$$

and so we have:

$$c_0$$

$$c_1$$

$$c_2$$

$$c_3 = \frac{c_0}{2\cdot 3}$$

$$c_4 = \frac{c_1}{3\cdot 4}$$

$$c_5 = \frac{c_2}{4\cdot 5}$$

$$c_6 = \frac{c_3}{5\cdot 6} = \frac{c_0}{2\cdot 3\cdot 5\cdot 6}$$

$$c_7 = \frac{c_4}{6\cdot 7} = \frac{c_1}{3\cdot 4\cdot 6\cdot 7}$$

$$c_8 = \frac{c_5}{7\cdot 8} = \frac{c_2}{4\cdot 5\cdot 7\cdot 8}$$

$$\vdots$$

Well, since this is a second-degree differential equation, we must have $$c_2 = 0$$. It also follows that

$$c_{3n} = \frac{c_0}{2\cdot 3\cdot 5\cdot 6\cdot \cdots \cdot \left( 3n -1 \right) \left( 3n \right)} \qquad n=1,2,3,\ldots$$

and

$$c_{3n+1} = \frac{c_1}{3\cdot 4\cdot 6\cdot 7\cdot \cdots \cdot \left( 3n \right) \left( 3n +1 \right)} \qquad n=1,2,3,\ldots$$

Therefore

$$y = c_0 + c_1 x + c_0 \sum_{n=1} ^{\infty} \left[ \frac{x^{3n}}{2\cdot 3\cdot 5\cdot 6\cdot \cdots \cdot \left( 3n -1 \right) \left( 3n \right)} \right] + c_1 \sum_{n=1} ^{\infty} \left[ \frac{x^{3n+1}}{3\cdot 4\cdot 6\cdot 7\cdot \cdots \cdot \left( 3n \right) \left( 3n +1 \right)} \right]$$

Questions

1. Have I got it right?

2. If so, can I write the denominators in a more elegant way? I mean:

$$2\cdot 3\cdot 5\cdot 6\cdot \cdots \cdot \left( 3n -1 \right) \left( 3n \right) = \left[ 3\cdot 6\cdot 9\cdot \cdots \cdot \left( 3n \right) \right] \left[ 2\cdot 5\cdot 8\cdot \cdots \cdot \left( 3n-1 \right) \right] = n!3^n \times ( ? )$$

and

$$3\cdot 4\cdot 6\cdot 7\cdot \cdots \cdot \left( 3n \right) \left( 3n +1 \right) = \left[ 3\cdot 6\cdot 9\cdot \cdots \cdot \left( 3n \right) \right] \left[ 4\cdot 7\cdot 10\cdot \cdots \cdot \left( 3n+1 \right) \right] = n!3^n \times ( ? )$$

where $$( ? )$$ represents the other factor which I haven't been able to figure out so far.

2. Dec 14, 2004

### Tide

You essentially have it! With a little reworking you can put it into the standard form of the Airy functions (your equation is Airy's DE which describes waves travelling in a medium with a linear variation in density or water waves with linear variation of the depth, e.g.).

It's a bit much to LaTeX so you can just look here for details: http://functions.wolfram.com/BesselAiryStruveFunctions/AiryAi/06/01/01/ [Broken]

Last edited by a moderator: May 1, 2017
3. Dec 14, 2004

Thanks for your input. I've just checked out that website. The formulas there are a bit scary, but then I looked up a problem in my calc book about the Airy function. It looks simpler:

$$A(x) = 1 + \frac{x^3}{2\cdot 3} + \frac{x^6}{2\cdot 3\cdot 5\cdot 6} + \frac{x^9}{2\cdot 3\cdot 5\cdot 6\cdot 8\cdot 9} +\cdots$$

Anyway, it seems to me that I get to $$A(x)$$ from

$$y = c_0 + c_1 x + c_0 \sum_{n=1} ^{\infty} \left[ \frac{x^{3n}}{2\cdot 3\cdot 5\cdot 6\cdot \cdots \cdot \left( 3n -1 \right) \left( 3n \right)} \right] + c_1 \sum_{n=1} ^{\infty} \left[ \frac{x^{3n+1}}{3\cdot 4\cdot 6\cdot 7\cdot \cdots \cdot \left( 3n \right) \left( 3n +1 \right)} \right]$$

if I let $$c_0 = 1$$ and $$c_1 = 0$$.

Regards

4. Dec 14, 2004

### Tide

That works! By the way, the functions are usually indicated as Ai(x) and Bi(x) [not A(x) and B(x)] with Ai(x) going to 0 as x -> infinity while Bi(x) diverges in that limit.

5. Dec 14, 2004

Let me see how that works...

We get $$A_i (x)$$ if $$c_0 = 1$$ and $$c_1 = 0$$. Conversely, $$B_i (x)$$ is obtained if $$c_0 = 0$$ and $$c_1 = 1$$.

6. Dec 14, 2004

### Tide

Actually, no. The Ai and Bi are combinations of the two infinite series you wrote. One of those grows and the other one decays as x -> +infinity. I misspoke (miswrote) eariler when I compared your A(x) with Ai(x) - they are not the same.