Differential Eqns. (checking an answer + some questions)

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In summary: But Ai and Bi are linear combinations of your A(x) and B(x).In summary, the conversation discusses a differential equation and its solution, which involves finding the recursion relation and using it to determine the coefficients of the infinite series. The equation is related to Airy's differential equation and its solutions, Ai(x) and Bi(x). The conversation also mentions how to obtain Ai(x) and Bi(x) from the solution discussed.
  • #1
DivGradCurl
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Any help is highly appreciated.

Thank you

Problem

[tex] y^{\prime \prime} = xy [/tex]

My Solution

If

[tex] 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 [/tex]

and

[tex] 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} [/tex]

Then

[tex] \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 [/tex]

Hence, we may find the recursion relation

[tex] c_{n+2} = \frac{c_{n-1}}{\left( n+1 \right) \left( n+2 \right)} \qquad n=1,2,3,\ldots [/tex]

and so we have:

[tex] c_0 [/tex]

[tex] c_1 [/tex]

[tex] c_2 [/tex]

[tex] c_3 = \frac{c_0}{2\cdot 3} [/tex]

[tex] c_4 = \frac{c_1}{3\cdot 4} [/tex]

[tex] c_5 = \frac{c_2}{4\cdot 5} [/tex]

[tex] c_6 = \frac{c_3}{5\cdot 6} = \frac{c_0}{2\cdot 3\cdot 5\cdot 6} [/tex]

[tex] c_7 = \frac{c_4}{6\cdot 7} = \frac{c_1}{3\cdot 4\cdot 6\cdot 7} [/tex]

[tex] c_8 = \frac{c_5}{7\cdot 8} = \frac{c_2}{4\cdot 5\cdot 7\cdot 8} [/tex]

[tex] \vdots [/tex]

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

[tex] 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 [/tex]

and

[tex] 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 [/tex]

Therefore

[tex] 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] [/tex]

Questions

1. Have I got it right? :smile:

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

[tex] 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 ( ? ) [/tex]

and

[tex] 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 ( ? ) [/tex]

where [tex] ( ? ) [/tex] represents the other factor which I haven't been able to figure out so far.
 
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  • #2
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 traveling 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]
 
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  • #3
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:

[tex] 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 [/tex]

Anyway, it seems to me that I get to [tex]A(x)[/tex] from

[tex] 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] [/tex]

if I let [tex]c_0 = 1[/tex] and [tex]c_1 = 0[/tex].

Regards
 
  • #4
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
Let me see how that works...

We get [tex]A_i (x)[/tex] if [tex]c_0 = 1[/tex] and [tex]c_1 = 0[/tex]. Conversely, [tex]B_i (x)[/tex] is obtained if [tex]c_0 = 0[/tex] and [tex]c_1 = 1[/tex].
 
  • #6
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.
 

1. What is a differential equation?

A differential equation is a mathematical equation that describes how a quantity changes over time or in relation to other variables. It involves derivatives, which represent the rate of change of the quantity being studied.

2. How do you solve a differential equation?

The methods for solving a differential equation depend on the type of equation. Some common techniques include separation of variables, using an integrating factor, and using power series. Solving a differential equation usually involves finding a general solution and then applying initial conditions to find a particular solution.

3. How do you check if an answer to a differential equation is correct?

To check if an answer to a differential equation is correct, you can substitute the solution into the original equation and see if it satisfies the equation. You can also plot the solution and compare it to the given initial conditions to ensure it meets the required criteria.

4. What are some real-world applications of differential equations?

Differential equations have many applications in physics, engineering, and other scientific fields. They are used to model various systems such as motion, heat transfer, population growth, and chemical reactions. They are also used in finance and economics to model changes in stock prices and interest rates.

5. Can you solve a differential equation without initial conditions?

No, in most cases, you need initial conditions to solve a differential equation. The initial conditions provide the starting values for the variables in the equation and help determine the specific solution. However, there are some special cases where a differential equation can be solved without initial conditions, such as when using a power series method.

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