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

c_{1}\vec{v_{1}}(t)+c_{2}\vec{v_{2}}(t)+...+c_{n}\ vec{v_{n}}(t) \equiv \vec{0}$

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

Thanks