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let $y_1(x)=\sin(x^2)$, $y_2(x)=\cos(x^2)$.

  • Prove that the function $y_2$ can't be a solution for a linear homogeneous second order equation which fulfills the conditions of Existence and Uniqueness Theorem.
  • Prove that the function $y_1$ can't be a solution for a linear homogeneous second order equation which fulfills the conditions of Existence and Uniqueness Theorem.

My solution attempt: I thought about using the Wronskian property for the proof. but I didn't have and direction for how to do it. any kind of help would be appreciated.

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  • $\begingroup$ Do you know about the behaviour of Wronskian in the interval of given ODE? $\endgroup$ – Nitin Uniyal Nov 30 '15 at 18:06
  • $\begingroup$ I know a lot of stuff about the Wronskian, can you clarify your question please ? $\endgroup$ – F1sargyan Nov 30 '15 at 18:11
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You do not need to use the Wronskian for $y_1$. Observe that $y_1(0)=y_1'(0)=0$. Now think uniqueness. Using the Wronskian, it is enough to observe that for any $y$, $W(y,y_1)$ vanishes at $x=0$.

For $y_2$ I have not found such a simple argument. Let $z=y_2'=-2\,x\sin x^2$. If $y_2$ satisfies second order homogeneous ODE, then $z$ satisfies a third order one. Since $z(0)=z'(0)=z''(0)=0$, again a uniqueness argument shows that it is not possible.

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  • $\begingroup$ I know that usually when the Wronskian is zero then the $y_1$ and $y$ are linearly dependent. (but not in all cases).and that the two solutions can cross each other as long as they don't have and same initial points, as well as their derivatives(it might happen that only part of them are equal, but not all of them). that all I can think about right now and im not seeing that this is helping me figuring out why getting zero Wronskian prevents $y_1$ from being a solution for a linear homogeneous second order differential equation. $\endgroup$ – F1sargyan Dec 1 '15 at 16:57
  • $\begingroup$ I am assuming that the equation and the solutions are defined on all $\mathbb R$. Then if the Wronskian vanishes at some point, it vanishes on all of $\mathbb R$. If not, $y_1$ is a solution of the equation $$y''-\Bigl(\frac1x+2\,x\tan x^2\Bigr)y'=0$$ on $(0,\pi/2)$. $\endgroup$ – Julián Aguirre Dec 1 '15 at 17:05
  • $\begingroup$ I get this, because your equation doesn't fulfill the conditions of Existence and Uniqueness Theorem, because $\frac{1}{x}$ is not continuous. but i'm talking about your first explanation, how does Uniqueness plays a part in the solution, I can't see that yet. $\endgroup$ – F1sargyan Dec 1 '15 at 17:22
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    $\begingroup$ $y=0$ is always a solution. You would have two solutions with $y(0)=y'(0)=0$. $\endgroup$ – Julián Aguirre Dec 1 '15 at 17:23
  • $\begingroup$ Turns out that $y_2$ has a counterexample. (which I'm not aware of at the moment, but still thinking about it), if u find it, it would be better if you edit your answer. thanks. $\endgroup$ – F1sargyan Dec 4 '15 at 9:33

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