either $y_1(x)=y_2(x)$ or $\{x\in[a,b] \mid y_1(x)=y_2(x) \} $ is finite Let $ y_1 $ and $y_2$ be two solutions of $ y''+P(x)y'+Q(x)y=R(x)$ on $[a,b]$ where $P , O ,R $are continuous functions on $[a,b]$ . Prove that the either $y_1=y_2$ or the set $\{x\in[a,b] \mid y_1(x)=y_2(x) \} $ is finite .
Here is what i have tried -
Set $g(x)=y_1(x)-y_2(x)$ then the differential equation becomes $$g''(x)+P(x)g'(x)+Q(x)g(x)= 0 \tag 1 $$ 
and the question is equivalent to showing 
When $g(x)$ is a solution to the differential equation (1) then either $g(x)=0 \forall x\in[a,b] $ or $g(x)=0$ for finitely many $x\in[a,b] $ .
If $g(x)=0$ for all $x\in [a,b]$ is the trivial solution to (1) . So we must show that if $g(x)\neq 0$ for some $x\in [a,b]$ then there are only finitely many $x\in[a,b]$ satisfying $g(x)=0$ .
Assume to the contrary that $g(x)=0$ for infinitely many values $x$ . As $[a,b]$ is compact ,there exists a sequence $x_n \in [a,b]$ such that $ \lim x_n= \alpha$ , $x_n\neq\alpha$ and $g(x_n)=0$ . Hence $g(\alpha)=\lim(g(x_n))=\lim(0)=0$ .We also have $g'(\alpha)=\lim \frac{g(x_n)-g(\alpha)}{x_n-\alpha} = \lim \frac{0-0}{x_n-\alpha }=0$ . From Differential equation (1) we have $g''(\alpha)=0$ .I have no other idea more useful than this . Please help me with a hint . 

You got $g(α)=0$ and $g′(α)=0$.. With these initial values and a homogeneous linear differential equation, you can only get the zero solution.
