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If we assume that $p(t)$ and $q(t)$ are continuous , then why $\sin t^2$ can't be an answer of equation $$y'' + p(t)y'+q(t)y=0$$ on an interval which contains $t=0$?

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  • $\begingroup$ Hint: plug in $y = \sin(t^2)$, then evaluate both sides at $t=0$. $\endgroup$ Oct 25, 2020 at 10:22
  • $\begingroup$ @GregMartin thanks, but it just saying that 0 is in that interval, not 0 is root of the equation. can you explain more please $\endgroup$ Oct 25, 2020 at 10:38
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    $\begingroup$ Nowhere in their comment did they mention roots. What is $y'' + p(t)y'+ q(t)y$ if $y = \sin(t^2)$, and does it equal $0$ (which is the RHS of your equation) at $t = 0$? $\endgroup$
    – player3236
    Oct 25, 2020 at 10:42

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You get $y(0)=0$ and $y'(0)=0$. For any homogeneous linear equation with continuous coefficients (in explicit form), there is only one trivial solution for these initial conditions, and it is not the given function.

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