If $y_1(x)$ and $y_2(x)$ are two solutions of equation $y'' +P(x)y' +Q(x)y = 0$ on an interval $[a,b]$ and have a common zero , show linear dependence

If $y_1(x)$ and $y_2(x)$ are two solutions of equation $y'' +P(x)y' +Q(x)y = 0$ on an interval $[a,b]$ and have a common zero in this interval, show that one is a constant multiple of the 1other.

Suppose the initial values of the solutions $y_1$ and $y_2$ are defined as follows : $y_1(t_o)=0,y_1'(t_o)=c_1$ and

$y_2(t_o)=0, y_2'(t_o)=c_2,$ then,

$y_1-y_2$ is also a solution to the given differential equation which satisfies :

$(y_1 -y_2 )(t_o)=0 , (y_1~ -y_2)~'(t_o) = c_1-c_2$.

Since, all of $y_1~,y_2,~y_1-y_2$ are solutions to he given differential equation, as per the uniqueness theorem, there exist unique curves which satisfy the above initial conditions.

Hence, I do not completely understand why $y_1$, $y_2$ must be a constant multiple of the other . I found an answer to this problem here. However, I do not understand the answer quite much.

Could someone please give an explanation to my confusion above.

Thank you for the help!

Hint: instead of looking at $y_1-y_2$, consider $y_1 - \frac{c_1}{c_2} y_2$. Then you have $y_1(t_0) - \frac{c_1}{c_2} y_2(t_0) = 0$ and $y_1'(t_0) - \frac{c_1}{c_2} y'_2(t_0) = 0$. But the curve $y_3(t) := 0$ also satisfies these conditions. Now apply uniqueness.
the uniqueness of the initial value problem $y'' + py' + qy = 0, y(a) = 0, y'(a) = k$ shows that if two solutions $y_1, y_2$ are such that $y_1(a) = y_2(a) = 0,$ then $y_1$ is a multiple of $y_2.$ that is $y_1, y_2$ are linearly dependent. this can also be seen if you look at $y(x) = y_2'(a)y_1(x) - y_1'(a)y_2(x),$ you find that $y(a) = y'(a) = 0.$ by uniqueness,$y = 0$ for all $x.$ that is $y_1, y_2$ are linearly dependent.
A different would be to find the Wronskian of these two function. Since we know that Wronskian is either identically zero or non zero everywhere. Wronskian is identically zero if and only if $$y_1$$ and $$y_2$$ are linearly dependent.
Let $$a$$ denote the common zero then $$W(a)=0$$ which implies it is zero everywhere and which further implies the linear dependence.