The ODE I am working with is, $y''+p(x)y'+q(x)y=0$. Let $y_1(x)$ and $y_2(x)$ be its solutions. Then, $W(x)=W(y_1,y_2)(x)$.
While trying to prove the above, I assume that in an interval $I$ containing $x_0, x_1, x_2$, $W(x_1) =0$ and $W(x_2)=0$ and $W(x)\neq0$ for any $x_1<x<x_2$.
Then the function should be continuous at end points and differentiable there too. If that is possible, then I cannot see any further contradiction as to why such a $W(x)$ cannot exist.
(At least on the basis of the fact, that if $W(x)\neq0$, $W(x)=W(x_0)e^{\int^x_{x_0} p(t)dt}$ )
Now using the conditions at end points, we get
$$\lim_{x\to x_1} W(x_0)e^{\int^x_{x_0} p(t)dt}=0$$ and $$\lim_{x\to x_1} p(x)W(x_0)e^{\int^x_{x_0} p(t)dt}=0$$ Since $W'(x)=-p(x)W(x)$.
And similarly for $x_2$.
Now if first condition is proved, the second one follows because $p(x)$ is assumed to be continuous function on $I$. So $p(x_1)$ must exist. Also, due to the same fact, $p(x)$ must be bounded on the closed interval $x_1<x<x_2$.
For the second condition, $-|x-x_0||M|<\int^x_{x_0} p(t)dt<|x-x_0||M|$. So $0<e^{-|x-x_0||M|}<e^{\int^x_{x_0} p(t)dt}<e^{|x-x_0||M|}$.
So the limit cannot be zero.
Does this proof seem correct?
Also, can anyone suggest a simpler proof?