What is the Wronskian of the given solution vectors?

[B18]

Homework Statement

We know that X= (1 over 3)e^t+(4 over -4) te^t is a solution to X'=(2 1 over -1 0)X.

Verify that the solution vectors are linearly independent on (-∞,∞).

Homework Equations

I know that the wronskian of the solution vectors cannot be 0 if they are linearly independent.

The Attempt at a Solution

So I found X1=(1 over 3) e^t and X2= (4 over -4)te^t
when i did the wronskian i did..
l e^t 4te^t l
l 3e^t -4te^t l

I got -16te^(2t). this would be 0 at t=0. Am i missing something? i figured that it should be independent because it says to verify that it is... are these linearly dependent?

thanks to anyone for the help!

 

[Dick]

Homework Statement

We know that X= (1 over 3)e^t+(4 over -4) te^t is a solution to X'=(2 1 over -1 0)X.

Verify that the solution vectors are linearly independent on (-∞,∞).

Homework Equations

I know that the wronskian of the solution vectors cannot be 0 if they are linearly independent.

The Attempt at a Solution

So I found X1=(1 over 3) e^t and X2= (4 over -4)te^t
when i did the wronskian i did..
l e^t 4te^t l
l 3e^t -4te^t l

I got -16te^(2t). this would be 0 at t=0. Am i missing something? i figured that it should be independent because it says to verify that it is... are these linearly dependent?

thanks to anyone for the help!

If the wronskian is nonzero ANYWHERE then the functions are linearly independent. If they are dependent it will vanish everywhere.

 

[B18]

If the wronskian is nonzero ANYWHERE then the functions are linearly independent. If they are dependent it will vanish everywhere.

Okay, so if the wronskian is equal to 0 and only 0 then the solutions are linearly dependent. However if the wronskian is anything other than 0 they are linearly independent!

 

[Dick]

Okay, so if the wronskian is equal to 0 and only 0 then the solutions are linearly dependent. However if the wronskian is anything other than 0 they are linearly independent!

Yes, they are linearly independent.