Suppose that $a(t)$, $b(t）$ is continuous over $[0,1]$. Proof:any solutions satisfy: $$x"+a(t)x'+b(t)x=0$$ and $$x(0)=0$$ is linear dependent.
Obviously the Wronski determinant is zero, but it only means the matric $\left(\begin{matrix}\phi(t)&q(t)\\ \phi(t)&q(t)\end{matrix}\right)$ does not have an inverse, but for the linear dependent, we should prove that it is a left zero divisor, so we wonder does the matrix ring have this property: $u$ is unit$\leftrightarrow$ $u$ is not a zero divisor.
Take two solutions, $x_1(t)$ and $x_2(t)$, there are $\alpha_1$ and $\alpha_2$ at least one of them is nonzero such that $\alpha_1x_1'(0)+\alpha_2x_2'(0)=0$. Now consider the linear combination $\alpha_1x_1(t)+\alpha_2x_2(t)$. It makes a solution of the problem $$ x''+a(t)x'+b(t)x=0,\quad x(0)=0,\quad x'(0)=0. $$ On the other hand, $x(t)\equiv0$ is a solution as well. And because of the uniqueness of the solution, we can conclude that $\alpha_1x_1(t)+\alpha_2x_2(t)\equiv0$, and $x_1(t)$, $x_2(t)$ are linearly dependent.