Differential Equations: Wronskian question.

tarmon.gaidon
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Homework Statement


Hey Everyone,

Here is a problem from my book that has my confused. I really don't understand what it wants me to do so if anyone could give me a few hints it would be greatly appreciated.

I am doing problem 34, but I included 33 since it wanted to follow the same method.

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Sorry if I seem like I am asking you to do my homework. I'm not, just looking for a place to start.

Thanks,
Rob
 
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Mark44 said:
the attachment is invalid...

Really? I am able to open it just fine. Can anyone else open this?

EDIT: I take that back. It works for me in Chrome but not Firefox. I will upload it somewhere else and fix my post.
 
Let's start by looking at 33. Do you know how to use the Wronskian to show that functions are linearly independent in general?
 
Hey office_Shredder,

I am reasonably familiar with using the Wronskian to show that functions are linearly independent. I am more used to this form:

8d606b824d0946483b111ce8935ba568.png


Where if W = 0 is true on an open interval I then the functions are linearly independent.

I don't completely understand the wronskian equation given in problem 33.
 
tarmon.gaidon said:
Hey office_Shredder,

I am reasonably familiar with using the Wronskian to show that functions are linearly independent. I am more used to this form:

8d606b824d0946483b111ce8935ba568.png


Where if W = 0 is true on an open interval I then the functions are linearly independent.

I don't completely understand the wronskian equation given in problem 33.

If you define f_i(x)=\text{exp}(r_i x), 1\leq i \leq n, what is f'_i(x)? How about f''_i(x)? What does that make the Wronskian for the n=3 case? Is there a rule for taking determinants where a column is multiplied by some factor that might help you here?
 
I understand where the matrix comes from now but I am not sure what method you are talking about for solving the determinant. Care to shed some light?
 
Use property 3 here. For example,

\begin{vmatrix} a & 2b & 3c \\ 4a & 5b & 6c \\ 7a & 8b &9c \end{vmatrix}=a\begin{vmatrix} 1 & 2b & 3c \\ 4 & 5b & 6c \\ 7 & 8b &9c \end{vmatrix}=ab\begin{vmatrix} 1 & 2 & 3c \\ 4 & 5 & 6c \\ 7 & 8 &9c \end{vmatrix}=abc\begin{vmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 &9 \end{vmatrix}
 
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