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In my differential equations course one of the problems asked the following:

Show that $sin(t^3)$ cannot be a solution of any equation of the form $$z''' + a_1(t)z'' + a_2(t)z' + a_3(t)z = 0$$ where a1(t), a2(t), and a3(t), are continuous over an open interval containing 0.

The solution was essentially that $z(t) = 0$ was unique by the basic existence and uniqueness theorem, therefore making it so $z=sin(t^3)$ cannot be a solution. However I don't see why the functions could be reversed to argue the opposite.

My professor's explanation was that "$z(t)=0$ is clearly a solution". How can this be answered more rigorously?

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    $\begingroup$ What do you mean by "the functions could be reversed"? In any case $z(t)=0$ is a solution, but if $z(t)=\sin t^3$ then $z(0)=z'(0)=z''(0)=0$ which by the uniqueness theorem is enough to determine $z$. $\endgroup$ – Lord Shark the Unknown Feb 22 '18 at 3:05
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    $\begingroup$ Why couldn't I say $z(t) = sin(t^3)$ is a solution and by the uniqueness theorem is unique and therefore $z(t) = 0$ is not a solution? $\endgroup$ – yerpderpington Feb 22 '18 at 4:01
  • $\begingroup$ I think Lord Shark's got the idea here. Once you realize that $z$ and its first $3$ derivatives at $0$ are $0$, $z=0$ becomes an obvious solution. $\endgroup$ – Mike Feb 22 '18 at 7:02

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