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Let's say we have a linear ODE with polynomial coefficients $p_j(x)$: $$ p_n(x) y^{(n)}(x)+\dots+p_1(x)y'(x)+p_0(x)y=0 $$ and let's say $x_0$ is a root of $p_n(x)$. What can be said about uniqueness of the IVP for this ODE at $x_0$? In particular, if I succeeded to prove that an analytic solution $y_0(x)$ satisfies $y_0(x_0)=y'(x_0)=\dots=y^{(n-1)}(x_0)=0$, does it necessarily follow that $y_0(x)\equiv 0$?

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up vote 3 down vote accepted

No. Consider Cauchy problem $xy'-y=0\;$, $y(0)=0\;$. It has solutions $y=Cx\;$, $\ C\in\mathbb R\;$.

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