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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?

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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.

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