I know that a set of vector functions [math]{\vec{v_{1}}(t), \vec{v_{2}}(t)+...+\vec{v_{n}}(t)}[/math] in a vector space [math]\mathbb{V}[/math] if [math]c_{i}= 0 [/math] for the following equation:

 

[math]c_{1}\vec{v_{1}}(t)+c_{2}\vec{v_{2}}(t)+...+c_{n}\vec{v_{n}}(t) \equiv \vec{0}[/math]

 

Where does the Wronskian come into play? Is it basically a determinant with functions and derivatives?

 

Thanks

thats only if the functions are linearly independant. by taking the determinant (wronskian) you can find out whether or not they are linearly independant.

But why does the determinant involve taking derivatives?

is just a way of filling up the matrix, just think about what conditions you would have to impose for the derivatives of your functions to be linear combinations of the other derivatives of your function.

 

ie. for one of the rows of the wronskian to be a linear combination of the others.

You probably want to look up "Why Does the Wronskian Work?" by Mark Krusemeyer. The American Mathematical Monthly Vol 95, pp. 46-49, 1988.