Linear dependence and Wronskian

missavvy
Messages
73
Reaction score
0

Homework Statement


Okay so the question is to show that these 2 functions are linearly dependent.
ie. they are not both solutions to the same 2nd order, linear, homogeneous differential equation for non zero choices of, say M, B and V

Homework Equations


f(x) = sin(Mx)
g(x) = Bx + V

The Attempt at a Solution



So they would be dependent if their Wronskian is equal to 0.
W = Bsin(Mx) - (Bx + V)Mcos(Mx)

I'm stuck at this point.. I can't seem to find any identities to make this = 0. I've tried expanding too.. I don't think it helps much. Any help is appreciated!
 
Physics news on Phys.org
Those functions aren't linearly dependent, they are linearly independent. That makes your job easier. Just find a value of x where the Wronskian DOESN'T vanish.
 
missavvy said:

Homework Statement


Okay so the question is to show that these 2 functions are linearly dependent.
ie. they are not both solutions to the same 2nd order, linear, homogeneous differential equation for non zero choices of, say M, B and V

From your i.e., part, I think you misunderstand the question. You aren't asked to show the functions are linearly dependent, which they aren't. You are asked to show those two functions are not solutions to the same second order linear DE. That is a different question.

The point of this exercise is to use the theorem that says two solutions of such an equation will have a Wronskian that is either identically zero or never zero. These two functions can only be solutions of the same equation if they satisfy that Wronskian property. So all you have to do is show the Wronskian is neither identically zero nor never zero. Don't you just love double negatives?

[Edited for technical accuracy]
 
Last edited:
Thanks, LCKurtz. I guess I wasn't fully plumbing the depth of the i.e. and double negative here.
 
LCKurtz said:
From your i.e., part

From the context, I don't think i.e. stands for id est. What does it mean here?
 
okay that makes much more sense. It's the way the question was phrased that had me confused. Thanks.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top