Question

Using Wronskian determinant:

Check the following for linear dependency:

Answer 1

1. e^-x, e^x , e^2x

Answer 2

Apparently , they are not dependent , but how to prove that with Wronskain ?

Answer 3

did you write down the wronskian ? what does it look like?

Answer 4

W has to be square
you need 3 rows
f(x)
f'(x)
f''(x)

where f(x) is the row of initial functions
then evaluate it at e.g. x=0

Answer 5

I don't know how to write it ...

All I know is for something like


x^2, x^3

g(x) = x^2

g'(x) = 2x

f(x) = x^3

f'(x) = 3x^2

Answer 6

This one is a determinant because it's a square matrix.

Answer 7

But in the given case ? I have matrix 2 x 3

Answer 8

in this case
1. e^-x, e^x , e^2x
the 2nd row is the first derivative
the 3rd row is the 2nd derivative

Answer 9

yes, now evaluate at x=0 (easy value to use)
1 1 1
-1 1 2
1 1 4

Answer 10

1 * ( 4 - 2) - 1 * ( -4 - 2) + 1 * ( -1 - 1 )

2 -1 * -6 -2

2 +6 - 2 = 6

Answer 11

not zero, so they are not dependent.

Answer 12

But phi ? can I use any value ?

Answer 13

to be dependent , W must be 0 for all values
if you find just one counter-example, they are not dependent.

Answer 14

*det(W) not just W

Answer 15

How can I check for all values ?!

Answer 16

You don't. You find one counter example, thus showing the functions are not dependent, and therefore they are independent.

Answer 17

Thanks Phi.