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We are asked whether it's possible that $x(t) = t^2$ is a solution to the ODE $\ddot{x}+a(x,\dot{x})\dot{x}+b(x,\dot{x})x = 0$ where $a,b$ are smooth function, in the interval $[-1,1]$.

I believe the answer is no. Notice that $x(0) = 0, \dot{x}(0) = 0$, and from existence and uniqueness theorem if I am not mistaken, $\begin{cases}\ddot{x}+a(x,\dot{x})\dot{x}+b(x,\dot{x})x = 0\\x(0) = 0 \\ \dot{x} (0) = 0\end{cases}$ admits a unique solution, and we can see that $x_*(t) = 0$ is that solution.

Is this reasoning correct?

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  • $\begingroup$ By "smooth" you mean "of class $C^1$"? Yes, it is correct. $\endgroup$ – user539887 Feb 13 at 7:52

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