Let be $y'(x)=A(x)y(x)$ an ODE and $y_1(x),y_2(x)\cdots,y_n(x)$ some solutions. Let be $Y(x)$ the Wronskian of the those solutions.
We have proven the statement that if we find an $x_0$ such that rank$(Y(x_0))=n$, then all $y_1(x),y_2(x)\cdots,y_n(x)$ are linearly independent.
Our professor added that this means that we only have to check one value, i.e. plug $x_0$ into the Wronskian $Y(x)$, in order to determine if $y_1(x),y_2(x)\cdots,y_n(x)$ are linearly independent or not.
However, I don't see why this statement allows us to conclude that rank$(Y(x_0))<n$ means linear dependence of $y_1(x),y_2(x)\cdots,y_n(x)$?
Just assume that $\mathrm{rank}(Y(x_0)) < n$. This means that $Y(x_0)$ is a family of linearly dependent vectors. This implies that there exists a vector $(\lambda_1, ..., \lambda_n) \neq 0$ such that $$ \sum_{j = 1}^n \lambda_j y_j(x_0) = 0. $$ Note that solutions are unique for these kinds of linear ODEs and that $$ g(x) := \sum_{j = 1}^n \lambda_j y_j(x) $$ is also a solution. It fulfils the initial condition $g(x_0) = 0$ due to assumption. Because of uniqueness and since $0$ is also a solution, we must have $$ 0 = g(x) = \sum_{j = 1}^n \lambda_j y_j(x) $$ for all $x$ and thus linear dependence.