Linear System of ODEs: Solving for n=1 or n=3

In summary, the conversation discusses a problem involving a system of ODEs and the attempt to solve it. The solutions to the equations are \alpha = r^{\frac{-(n-1)}{2} and \alpha '= r^{-(n-1)}, with the ultimate goal of showing that n=1 or n=3. However, the equations are not well-behaved when n=1 and there is an additional solution to the second equation that is not a solution to the first. The solutions match when n=3, but careful algebra is needed.
  • #1
McCoy13
74
0
Homework Statement
I'm trying to solve the following system of ODEs.

[tex]\alpha = \alpha (r)[/tex]

[tex]\alpha ' + \frac{n-1}{2r} \alpha =0[/tex]

[tex]\alpha '' + \frac{n-1}{r} \alpha ' = 0[/tex]


The attempt at a solution

The solution to the first one is

[tex]\alpha = r^{\frac{-(n-1)}{2}[/tex]

The solution to the second one is

[tex]\alpha '= r^{-(n-1)}[/tex]

Ultimately the goal is to show that n=1 or n=3 (it's a problem dealing with wave attenuation and distortion, but I'm just having problems with this step). I really can't reconcile these answers, even using arbitrary scalar factors against my solutions. When I tried substituting one equation into the other all that happened was I ended up with a factor of sqrt(2) that wasn't consistent with either equation individually.
 
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  • #2
The solutions to these equations do match when n=3, just be a bit more careful with the algebra. The equations themselves are not well-behaved when n=1, so you should be careful when trying to match the solutions for arbitrary n to the limit n->1. It happens to work for the first equation, but there is an additional solution of the second that is not a solution to the first.
 
  • #3
Gah, I see. I think I was forgetting that r was to a negative power when I was thinking about it in the case n = 3 and in the case n = 1 I wasn't looking at the equations, just the solutions, so I missed the fact that they sort of become singular. Thanks for elucidating.
 

Related to Linear System of ODEs: Solving for n=1 or n=3

1. What is a linear system of ODEs?

A linear system of ODEs is a group of differential equations that can be written in the form of a matrix equation, where the variables are dependent on time and their derivatives. The equations are considered linear because they can be expressed as a linear combination of the dependent variables and their derivatives.

2. How is a linear system of ODEs solved?

There are several methods for solving a linear system of ODEs, including substitution, elimination, and using matrix operations. The most commonly used method is the matrix method, where the system is written in matrix form and then solved using techniques such as Gaussian elimination or Cramer's rule.

3. What are the applications of linear systems of ODEs?

Linear systems of ODEs are used to model a wide range of physical systems, including electrical circuits, chemical reactions, and mechanical systems. They are also commonly used in economics, biology, and other fields to study dynamic systems and predict behavior.

4. What are the differences between a linear and a nonlinear system of ODEs?

The main difference between linear and nonlinear systems of ODEs is that the equations in a linear system are linearly related to the dependent variables and their derivatives, while the equations in a nonlinear system have nonlinear relationships. This means that the methods used to solve linear systems may not be applicable to nonlinear systems.

5. What are the advantages of using a linear system of ODEs?

One advantage of using linear systems of ODEs is that they have analytical solutions, meaning that they can be solved using mathematical techniques. This makes them useful for modeling and predicting the behavior of complex systems. Additionally, linear systems are easier to understand and manipulate than nonlinear systems, making them useful for teaching and learning purposes.

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