# 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

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 two solution of a second order linear ODE, $$W(ϕ_1,ϕ_2)(x_0)=0$$ at one point implies $$W(ϕ_1,ϕ_2)(x)=0$$ everywhere and thus linear dependence.
For instance, consider the linear combinations $$\psi_1(x)=ϕ_1(x_0)ϕ_2(x)-ϕ_2(x_0)ϕ_2(x) ~\text{ and }~ ψ_2(x)=ϕ_1'(x_0)ϕ_2(x)−ϕ_2'(x_0)ϕ_2(x).$$ 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.