A Second Order Differential Equation

In summary: Overall, great job on your analysis of the problem and finding a solution using Bi(x)! Keep up the good work!
  • #1
cyunus
1
0

Homework Statement


Hi, this problem is from first chapter of Mathematical Methods of Physics by Mathews and Walker. (Problem 1-36, second edition)

Consider the differential equation [itex]y'' - xy + y^3 = 0 [/itex] for large positive x.

a-) Find an oscillating solution with two arbitrary constants.

b-) Find a (non trivial) particular nonoscillating solution.

Homework Equations



[itex] x \rightarrow \infty [/itex] [itex] Ai(x) \rightarrow \frac{e^{-\frac{2}{3}x^{\frac{3}{2}}}}{2 \sqrt{\pi}x^{\frac{1}{4}}} [/itex] and [itex] Bi(x) \rightarrow \frac{e^{\frac{2}{3}x^{\frac{3}{2}}}}{\sqrt{\pi}x^{\frac{1}{4}}} [/itex]

The Attempt at a Solution



If we assume y is small then we get airy equation and [itex]Ai(x)[/itex] is a solution while [itex]Bi(x)[/itex] is not a solution since for [itex] x \rightarrow \infty { } Bi(x) \rightarrow \infty [/itex]. Also it should be noted that [itex]Ai(x)[/itex] does not have a root for large x therefore cannot be considered as an oscillated solution. Also WKB method didn't give me an oscillating solution but I can't say I know it well.

Thanks
 
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  • #2
for the post! I agree with your approach so far. For part a), you are correct that Ai(x) does not have a root for large x, so it cannot be considered an oscillating solution. However, there is another solution to the Airy equation that does oscillate: Bi(x). This is the solution that you mentioned in your post, but there is one small mistake. For large x, Bi(x) does not go to infinity, it actually oscillates between positive and negative values. So, for part a), you can use Bi(x) as your oscillating solution and add two arbitrary constants to it to get a general solution.

For part b), you are correct that Ai(x) and Bi(x) are not solutions for large x, but there is another approach you can take. You can use the fact that for large x, the term y'' becomes negligible compared to the other terms in the equation. This means you can approximate the equation as xy - y^3 = 0. Solving this equation for y will give you a non-trivial particular solution. Can you try this approach and see if you can find a solution?

As for the WKB method, it is a powerful tool for finding solutions to differential equations, but it may not be necessary for this problem. However, if you are interested in learning more about it, I recommend looking into it further.
 

Related to A Second Order Differential Equation

1. What is a second order differential equation?

A second order differential equation is a mathematical equation that involves the second derivative of a function. It is used to describe the relationship between a function and its derivatives, and is commonly used in many areas of science and engineering.

2. What is the difference between a first and second order differential equation?

The main difference between a first and second order differential equation is the highest order derivative that appears in the equation. In a first order differential equation, only the first derivative appears, while in a second order differential equation, the second derivative appears.

3. How is a second order differential equation solved?

A second order differential equation can be solved by using various techniques, such as separation of variables, substitution, or the method of undetermined coefficients. The specific method used depends on the form of the equation and the initial conditions given.

4. What are the applications of second order differential equations?

Second order differential equations are used in many areas of science and engineering, including physics, chemistry, biology, and economics. They are particularly useful in modeling systems that involve acceleration, such as motion of objects and electrical circuits.

5. What are the initial conditions in a second order differential equation?

The initial conditions in a second order differential equation refer to the values of the function and its derivative(s) at a specific point. These conditions are necessary to find a specific solution to the equation, as they provide a starting point for solving the equation.

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