Re: differential equations help

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Homework Help Overview

The discussion revolves around a differential equations assignment, specifically focusing on expressing a linear combination of solutions in determinant form. The original poster seeks assistance with demonstrating that the differential equation for a combination of two functions can be represented as a determinant, and they also inquire about the implications of a specific case regarding the independence of those functions.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the requirement to show that the determinant equals zero and the implications of linear dependence among the functions involved. There are questions about the proof process and the interpretation of the problem statement.

Discussion Status

The discussion is active, with participants exploring different interpretations of the problem. Some provide insights into the nature of linear dependence and the requirements for independent solutions in the context of second-order differential equations. Guidance has been offered regarding the evaluation of the determinant and the relationship between the functions.

Contextual Notes

There is a mention of the original problem not explicitly stating that it involves a second-order ordinary differential equation, which may affect the participants' understanding of the requirements. Additionally, some participants express uncertainty about the proof and the necessary steps to demonstrate the relationship between the functions.

confusedM
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I'm having a couple problems with my diff eq assignment if anyone can help me. The problem is 1. (a) show that the diff eq. for y=c1u1(x) + c2u2(x), where u1(x) and u2(x) are functions which are at least twice differentiable, may be written in determinant form as
| y u1(x) u2(x) |
| y u1'(x) u2'(x) | = 0
| y u1''(x) u2''(x)|

b) what happens in the case where

w = | u1(x) u2(x) | = 0
| u1'(x) u2'(x)|
 
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confusedM said:
I'm having a couple problems with my diff eq assignment if anyone can help me. The problem is 1. (a) show that the diff eq. for y=c1u1(x) + c2u2(x), where u1(x) and u2(x) are functions which are at least twice differentiable, may be written in determinant form as
| y u1(x) u2(x) |
| y u1'(x) u2'(x) | = 0
| y u1''(x) u2''(x)|

b) what happens in the case where

w = | u1(x) u2(x) | = 0
| u1'(x) u2'(x)|

In that case, the two function, u1(x) and u2(x), are not independent. You need two independent solutions to write the general solution to a second order equation.
 
I understand that if the determinant is zero, the equations are linearly dependent, but I don't know how to go about doing the proof required to answer the problem.
 
That's part B... part A though asks for proof of the determinant of the three equations equaling zero...that's the part I'm really having issues with.
 
Well you didn't provide the DE, so I assume it's a 2nd order ODE. Is the first column of the matrix supposed to read y,y',y'' from top to bottom instead? HallsOfIvy has said you need 2 linearly independent solutions for a second order ODE, so what can you say if the determinant of that is zero?
 
yes it is supposed to read y, y', y"... the problem is as it is written in the book... it didn't say that it was a 2nd order ODE.
Clearly the original differential eq. and it's two derivatives are linearly dependent since the determinant is zero, but I don't understand what I'm exactly supposed to be showing
 
This can be verified directly. You are given y=c1u1+c2u2. Find y' and y'' and evaluate the determinant. No proof is needed because none is asked for.
 
Yes, Defennder is right. I had overlooked A and answered B.

If y= c1u1+ c2u2, then
[tex]\left|\begin{array}{ccc} y & u & v \\ y' & u' & v' \\ y" & u" & v" \end{array}\right|= \left|\begin{array}{ccc} c1u+ c2v & u & v \\ c1u1'+ c2u2' & u' & v' \\ c1u1"+ c2u2" & u" & v" \end{array}\right|[/tex]

If you know a little linear algebra, and properties of determinants, the fact that the first column is a linear combination of the second and third columns, that tells you all you need. If you don't recognise that, go ahead and do the calculations.
 

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