Any two solution of the equation $y''+p(x)y'+q(x)y=0$, $p(x)$ and $q(x)$ are continuous on $(a,b)$ and $x\in (a,b)$ are linearly dependent if
(a) they have common zero in $(a,b)$
(b) they have a maximum at some point in $(a,b)$
(c) they have a minimum at some point in $(a,b)$
(d) Their Wronskian doesn't vanish at some point $x_0\in (a,b)$
My attempt:-
Result:- If the function $\phi_1(x)$ and $\phi_2(x)$ are linearly dependent on an open interval $I$, the $W(\phi_1,\phi_2)(x)=0,\forall x\in I$
I know that $W(\phi_1,\phi_2)(x_0)=0,x_0$ is the common zeroes of the solutions. If $t_0$ is the extremum(minimum/maximum) the $\phi_1'(t_0)=\phi_2'(t_0)=0\implies W(\phi_1,\phi_2)(t_0)=0 $ I am not able to use the result. If the question is Any two solution of the equation $y''+p(x)y'+q(x)y=0$, $p(x)$ and $q(x)$ are continuous on $(a,b)$ and $x\in (a,b)$ are linearly dependent only if. Then I could have use the result. Can you please give direction? How to approach the solution?
For any two functions with the critical property of being solutions of the same second order linear ODE, the Wronskian value $W(ϕ_1,ϕ_2)(x_0)=0$ at one point implies $W(ϕ_1,ϕ_2)(x)=0$ everywhere and thus linear dependence.
For a short proof idea, consider the linear combinations $$\psi_1(x)=ϕ_1(x_0)ϕ_2(x)-ϕ_2(x_0)ϕ_2(x) ~\text{ and }~ ψ_2(x)=ϕ_1'(x_0)ϕ_2(x)−ϕ_2'(x_0)ϕ_2(x). $$ These are also solutions of the given linear DE. If $W(ϕ_1,ϕ_2)(x_0)=0$, then both $$ ψ_1(x_0)=ψ_2(x_0)=0 ~\text{ and }~ ψ_1'(x_0)=ψ_2'(x_0)=0. $$ This implies that both are the zero solution. At least one of both is a non-trivial linear combination in all your cases.