# I Difference between Wronskian and other methods

1. Dec 6, 2016

### Mr Davis 97

I am confused about determining when two or more functions are linearly independent. My textbook notes that the Wronskian can do this, but then it also mentions the definition of linear Independence, that the linear combination $c_1 f_1 + c_2 f_2 + ... + c_n f_n = 0$ only has the trivial solution, and how we could evaluate this at n - 1 points, and see whether the corresponding system of equations only has the trivial solution. So if they both do the same thing, why do both exist? What makes the Wronskian different than just using the definition of linear independence?

2. Dec 6, 2016

### Stephen Tashi

The non-vanishing of the Wronskian of two function implies they are linearly independent, but the non-vanishing is not equivalent to the definition of linear independence. If it were then a non-non-vanishing of the Wronskian would imply the non-linear-independence of the two functions. However, the fact the Wronskian of two functions is identically zero on a interval does not imply the two functions are linearly dependent on that interval.

3. Jan 9, 2017

### MidgetDwarf

If we use the concept of linear independence (the definition), it can make it extremely tasking and may require a bit of imagination. The Wronskian allows us a formulaic approach to determine if two or more functions are linearly independent.

4. Jan 16, 2017

### LCKurtz

5. Jan 30, 2017

### rppearso

I have a very similar question so I thought I would just piggy back on to this thread to avoid creating another thread. My question is why is the equation in the red box = 0

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6. Jan 30, 2017

### LCKurtz

It isn't given to be $0$. It is set that way to prevent getting second derivatives of the unknown $u$'s in the problem. You are just looking for something that works and if that condition helps you succeed, which it does, it was a good idea.

7. Jan 31, 2017

### rppearso

thats going to take some time to digest

8. Jan 31, 2017

### LCKurtz

Remember, you have introduced two unknown $u$ functions which you have inserted to try to find a solution. They can be anything; you are trying to find some that help you solve the problem. So you can have two constraints. One constraint is that your expression must be a solution of the NH equation. You have an additional constraint you can require, and keeping second derivatives out of the equations is a good idea because it makes the method solvable.

9. Jan 31, 2017

### rppearso

thank you for your help, I am going to add parts of this to my explaination in my write up so I can go back to it later.