# Wronskian and two solutions being independent

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1. Jan 31, 2016

### bubblewrap

In the uploaded file, question 11 says that in b) the solutions y1 and y2 are linearly independent but the Wronskian equals 0. I think it said that they are independent because it's not a fixed constant times the other solution (-1 for -1<=t<=0 and 1 for 0<=t<=1) but it clearly says in the textbook that the two solutions are linearly independent if and only if Wronskian equals 0.

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2. Jan 31, 2016

### Krylov

If two (differentiable) functions are linearly dependent on an interval, then their Wronskian vanishes.

The converse is not true. So: If two functions are linearly independent on an interval, then it does not necessarily follow that their Wronskian does not vanish. The phrase
is wrong in any case, so I hope it is not really in your textbook like this.

3. Jan 31, 2016

### bubblewrap

It says so here

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4. Jan 31, 2016

### Krylov

No, it's not what is written there. It says "unequal".

In any case, if the Wronskian does not vanish, then the functions are linearly independent on the interval in question. This is just the contrapositive of what I wrote here:
The converse is not true. Maybe you know more about the functions $y_1$ and $y_2$ that solve (3) in your book to be able to draw this conclusion, but in general it is false.

5. Jan 31, 2016

### bubblewrap

I made a mistake at the end of my first post I meant 'if and only if ... unequals 0"

6. Jan 31, 2016

### bubblewrap

This is the case where it says that they are linearly independent but the Wronskian is 0

Doesn't 'if and only if' mean necessary and sufficient condition? It means that if linearly independent then the Wronskian always does not equal zero, it never is another case and there are only two cases.

Last edited: Jan 31, 2016
7. Jan 31, 2016

### bubblewrap

Also in the differential equation, do y, dy/dt, dy2/d2t and so on have to be continuous?

8. Jan 31, 2016

### Krylov

I believe the point of the exercise in post #1 is exactly to demonstrate that two differentiable functions (here: $y_1$ and $y_2$) may be linearly independent on an interval (here: $[-1,1]$) and yet their Wronskian may still vanish identically. This shows that the converse of
is not true.

In part (d) of that exercise it is then asked to show that $y_1$ and $y_2$ cannot satisfy the premise of the corollary in post #3, although I have to guess here because I don't know what equation (3) in your book is. So there is no contradiction between the exercise and the corollary. The corollary is not about arbitrary differentiable functions, but about functions that satisfy additional properties.

If P and Q are two logical statements, then "P if and only if Q" means: "P implies Q" and "Q implies P". So: Either P and Q are both true, or they are both false.

9. Feb 1, 2016

### bubblewrap

Does 'to vanish' mean being 0?
Like, does W vanishes mean W equals 0?

10. Feb 1, 2016

### Krylov

Yes. In this context "$W$ vanishes" means that $W$ is zero not just in a point but on the entire interval.