Regarding common zeros of a pair of solutions to an ODE

Question

$\mathbf{Question}:$ Let $y_1$ and $y_2$ be the solutions to $y''x^2+y'+\sin(x)y=0$ which satisfy the boundary conditions $y_1(0)=0$ and $y_1'(1)=1$ and $y_2(0)=1$ and $y_2'(1)=0$ respectively. Then,

1. $y_1$ and $y_2$ do not have common zeros
2. $y_1$ and $y_2$ do have common zeros
3. either $y_1$ or $y_2$ has a zero of order $2$
4. both $y_1$ and $y_2$ have zeros of order $2$

$\mathbf{Attempt}:$ Each of the particular solutions can be written in terms of linear combinations that abide by the specified conditions. Let $A\phi_1(x)+B\phi_2(x)=y_1(x)$ and $C\phi_1(x)+D\phi_2(x)=y_2(x)$ be the solutions, with $\phi_1$ and $\phi_2$ being the fundamental ones.

Supposing that they indeed have at least one common zero, say $x_0$, then: $A\phi_1(x_0)+B\phi_2(x_0)=0$ and $C\phi_1(x_0)+D\phi_2(x_0)=0$. Now, we know from the Sturm Separation Theorem that the roots of the fundamental solutions alternate, hence, they cannot vanish together; implying that $\begin{vmatrix} A & B\\ C & D \end{vmatrix}=0$. This implies that $y_1(x)$ and $y_2(x)$ are linearly dependent, thus violating the boundary condition $y_1(0)=0 \neq y_2(0)$.

Is this the correct approach? Kindly Verify.