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Show that if $y_1(x_0) =y_2(x_0) = 0$ then $y_1$ and $y_2$ cannot be the fundamental set of solutions for asecond order linear homogenous ODE on an interval $I$ with $x_0 \in I$.

How can one prove the above statement. I know by Wronskian that $W(y_1,y_2)(x_0) = y_1(x_0)y_2'-y_1'y_2(x_0) = 0$, but I don't have any idea what can be inferred from this equation.

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1 Answer 1

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There is some constant $c$ with $y_2'(x_0)=cy_1'(x_0)$. Use the uniqueness theorem to show that $z=cy_1-y_2$ must be the zero solution.


Or more compact, set $z(x)=y_1(x)y_2'(x_0)−y_1'(x_0)y_2(x)$ and show that it is the zero solution, proving linear dependence of $y_1,y_2$.

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  • $\begingroup$ Can you elaborate? I'm not sure how uniqueness theorem can be applied here because this is not IVP. $\endgroup$
    – Ted
    Mar 11, 2019 at 18:34
  • $\begingroup$ For $z$ you have an IVP with obviously $z=0$ as one solution. $\endgroup$ Mar 11, 2019 at 19:18

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