showing a function is not a solution to ODE

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?

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