# Wronskian to determine L.D

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1. Mar 18, 2015

### BrettJimison

1. The problem statement, all variables and given/known data
Hello,

I was just looking for a quick tip:
If I have three distinct solutions to a second order linear homogeneous d.e, how would I show that the wronskian of (y1,y2,y3)(x)=0?

I know how to show the wronskian is not zero for a linearly independent set, but I'm confused since I'm given three distinct solutions to a second order.
2. Relevant equations
I know y=c1y1+c2y2+c3y3 by the principle of super position but I'm not sure how I would go about showing a wronskian evaluated at these is zero...any tips? just need a point in the right direction

3. The attempt at a solution

2. Mar 18, 2015

### LCKurtz

The Wronskian in the case of a 2nd order DE is a 2 by 2 determinant. What does the Wronskian of three solutions mean?
 Or maybe you mean the 3 by 3 Wronskian anyway? In which case you know they are linearly dependent from the DE so...

Last edited: Mar 18, 2015
3. Mar 18, 2015

### BrettJimison

I'm trying to find a 3x3 wronskian with y1 y2 y3 that is equal to zero.....not sure how to do it symbolically. It's a lot easier when I know the functions......

4. Mar 18, 2015

### LCKurtz

What theorems do you have about Wronskians and linear dependence/independence?

5. Mar 18, 2015

### BrettJimison

Well I know the set will be linearly dependet since the wronskian will equal zero. I also know that the set has to be L.D since it has 3 elements and the D.E is only 2nd order.

If
L2[ y1]=0
L2[y2]=0
l2[y3]=0

Then
y1=c1y2+c2y3
y2=c1y1+c2y3
y3=c1y1+c2y2

I then have three functions that I can plug into the wronskian....I tried this and got a soltion that doesnt seem to simplify to zero.

6. Mar 18, 2015

### BrettJimison

Also letting a=y1, b=y2, c=y3

My determinatnt simplfies to: a"(bc'-b'c)+a'(b"c-c"c)+a(b'c"-b"c')

7. Mar 18, 2015

### LCKurtz

No, you don't know that. You are trying to show the Wronskian is zero.

Yes. You do know the 3 functions are linearly dependent and that is why.

Now think about what being linearly dependent tells you about the system of equations
$c_1y_1 + c_2y_2 + c_3y_3 = 0$
$c_1y_1' + c_2y_2' + c_3y_3' = 0$
$c_1y_1'' + c_2y_2'' + c_3y_3'' = 0$
and what that implies about the Wronskian.

8. Mar 18, 2015

### BrettJimison

Well...I know If I put that into matrix form its determintant will be the wronskian. Im taking its determinant now but its not looking like its going to cancell out....

edit: I know that those equations are L.D which means they can be wriiten as linear combos of each other..

Last edited: Mar 18, 2015
9. Mar 18, 2015

### BrettJimison

I actually get exactly what I posted previously.......hmmm

The thing is: I know the set of equations is L.D and I know the wronskian will be zero. I have no idea how to show its zero though. When I take the determinant of the matrix, I dont get something that cancells

 I understand why its not zero: Im basically evaluating a wronskian of a set of solutions for L3[y] so I wont equal zero.....I need to find a way to get the three equations into a different form?

Last edited: Mar 18, 2015
10. Mar 18, 2015

### LCKurtz

You aren't expected to actually calculate the determinant. You have $n$ equations in the $n$ unknown $c_i$'s. What would you know about the $c_i$'s if the determinant of coefficients (the Wronskian) wasn't zero? What would Cramer's rule tell you?

11. Mar 18, 2015

### BrettJimison

If the determinant wasnt zero (wronskian) then that means that there is non trivial solutions to the system wich means linear dependence but I dont know how to show that since I cant compute the wronskian to show that it isnt zero.......

12. Mar 18, 2015

### BrettJimison

Im pretty sure cramers rule would just tell us we could find non trivial solutions....

13. Mar 18, 2015

### LCKurtz

No. Think about what you are saying. Try a little 2 by 2 system:
$x + y = 0$
$2x - y = 0$
Here the determinant of coefficients isn't zero. Solve it with Cramer's rule.

14. Mar 18, 2015

### BrettJimison

The SPECIFIC question states:

Suppose y1,y2,y3 are three distinct solutions of L2[y] =0 on I = (a,b)
Show that W(y1,y2,y3)(x)=0 on I.

15. Mar 18, 2015

### BrettJimison

Sorry meant to say if the determinant WAS zero....typo, this is what I have been trying to prove the whole time.

16. Mar 18, 2015

### LCKurtz

But what if the determinant isn't zero. What does that tell you? What does that say about your system of equations?

I have to run for now. Back in a couple of hours.

17. Mar 18, 2015

### BrettJimison

If the Det wasnt zero, then equations would be linearly independent and ONLY the trivial solution would work.
I can explain my reasoning in words, but Im having a hard time putting it into math...

Thanks!

18. Mar 18, 2015

### LCKurtz

There isn't much math to write, but you need to be careful about your notation. You don't talk about "equations being linearly independent" in this context. You talk about the functions $y_1,...y_n$ being independent. What you have stated above that is correct and relevant is that if the determinant is zero, you can get only trivial solutions for the $c_i$'s. But you know the solutions are linearly dependent so there is a non-trivial solution for the $c_i$'s. So, what do you conclude?

19. Mar 18, 2015

### LCKurtz

There isn't much math to write, but you need to be careful about your notation. You don't talk about "equations being linearly independent" in this context. You talk about the functions $y_1,...y_n$ being independent. What you have stated above that is correct and relevant is that if the determinant is nonzero, you can get only trivial solutions for the $c_i$'s. But you know the solutions are linearly dependent so there is a non-trivial solution for the $c_i$'s. So, what do you conclude?

Last edited: Mar 18, 2015
20. Mar 19, 2015

### BrettJimison

W(y1,y2,y3)(x)=0 Thanks for your time and help!