$y''+p(x)y'+q(x)y=0$ , $p,q$ are continuous functions in $[a,b]$ and $y_1(x), y_2(x)$ are the solutions.
$a\leq x_0\leq b$
Prove\disprove:
1.If $y_1(x_0)=y_2(x_0)=0 \implies y_1,y_2$ are linearly dependent.
2.If $y_1,y_2$ get an extremum in $x_0 \implies y_1,y_2$ are linearly dependent.
3.If $y_1,y_2$ get an inflection point in $x_0$, and $\forall x\in [a,b] q(x)\neq0 $or$ p(x)\neq0 $(or both) $\implies y_1,y_2$ are linearly dependent.
I'd be grateful for your feedback for 1,2 and some help for 3
My solution :
$1.W[y_{1,}y_{2}](x_{0})=\left|\begin{matrix}y_{1}(x_{0}) & y_{2}(x_{0})\\ y'_{1}(x_{0}) & y'_{2}(x_{0}) \end{matrix}\right|=\left|\begin{matrix}0 & 0\\ y'_{1}(x_{0}) & y'_{2}(x_{0}) \end{matrix}\right|=\implies W[y_{1,}y_{2}](x_{0})=0$
$p,q$ are continuous functions, since the ode exists the Uniqueness and Existence theorem then $\forall x\in [a,b]$ exists $W[y_{1,}y_{2}](x)=0 \implies y_1,y_2$ are linearly dependent.
$2.W[y_{1,}y_{2}](x_{0})=\left|\begin{matrix}y_{1}(x_{0}) & y_{2}(x_{0})\\ y'_{1}(x_{0}) & y'_{2}(x_{0}) \end{matrix}\right|=\left|\begin{matrix}y_{1}(x_{0}) & y_{2}(x_{0})\\ 0 & 0 \end{matrix}\right|=\implies W[y_{1,}y_{2}](x_{0})=0$
$p,q$ are continuous functions, since the ode exists the Uniqueness and Existence theorem then $\forall x\in [a,b]$ exists $W[y_{1,}y_{2}](x)=0 \implies y_1,y_2$ are linearly dependent.
$3.W[y_{1,}y_{2}](x_{0})=\left|\begin{matrix}y_{1}(x_{0}) & y_{2}(x_{0})\\ y'_{1}(x_{0}) & y'_{2}(x_{0}) \end{matrix}\right|=y_1y'_2-y_2y'_1$
I tried to derivative this for getting $0$.
$(y_1y'_2-y_2y'_1)'=y'_1y'_2+y_1y''_2-y'_2y'_1-y_2y''_1=0$
I get stuck and I don't know how to solve $3$ and why the given $\forall x\in [a,b] q(x)\neq0,p(x)\neq0$ are important.
Thank you !