I do not think that the particle would ever start moving, if $\ddot{x} = x^2$ is truly its equation of motion. You could imagine this as the particle being trapped at exactly the top of some potential well (e.g a ball exactly at the top of a hill). If the particle was slightly "left or right" of x=0 (e.g x=0.00001), you would see motion. This is clear if we look at the potential (ball on a hill) picture too, the particle is clearly in an unstable equilibrium.

Your solution for the differential equation $$\ddot{x}=x^2\tag{1}$$ is not well defined at $t=0$ as $\frac{6}{0^2}$ is probelmatic! If you require the position at $t=0$ to be $0$, you must apply this as a boundary condition when solving the differential equation (1). However, the solution to (1) is not easy, I believe it is the Weierstrass Functions which is probably not useful for a dynamics problem. I believe that your solution ($x=6/t^2$) is actually the first term of the expansion of this function. While you do correctly recover $a(x) = x^2$ from this solution, im not sure if it has any physical meaning at $t=0$. Also, the particle would only every have $x=0$ at $t=\infty$ according to this solution!

I do not think that the particle would ever start moving if $\ddot{x} = x^2$ is truly its equation of motion. You could imagine this as the particle being trapped at exactly the top of some potential well (e.g. a ball exactly at the top of a hill). You would see movement if the particle were slightly left or right of x=0 (e.g. x=0.00001). This is clear if we look at the potential (ball on a hill) picture too, the particle is clearly in an unstable equilibrium.

Your solution for the differential equation $$\ddot{x}=x^2\tag{1}$$ is not well defined at $t=0$ as $\frac{6}{0^2}$ is problematic! If you require the position at $t=0$ to be $0$, you must apply this as a boundary condition when solving the differential equation (1). However, the solution to (1) is not easy, I believe it is the Weierstrass Functions which is probably not useful for a dynamics problem. I believe that your solution ($x=6/t^2$) is actually the first term of the expansion of this function. While you do correctly recover $a(x) = x^2$ from this solution, I'm not sure if it has any physical meaning at $t=0$. Also, the particle would only ever have $x=0$ at $t=\infty$ according to this solution!

I do not think that the particle would ever start moving, if $\ddot{x} = x^2$ is truly its equation of motion. You could imagine this as the particle being trapped at exactly the top of some potential well (e.g a ball exactly at the top of a hill). If the particle was slightly "left or right" of x=0 (e.g x=0.00001), you would see motion. This is clear if we look at the potential (ball on a hill) picture too, the particle is clearly in an unstable equilibrium.

Your solution for the differential equation $$\ddot{x}=x^2\tag{1}$$ is not well defined at $t=0$ as $\frac{6}{0^2}$ is probelmatic! If you require the position at $t=0$ to be $0$, you must apply this as a boundary condition when solving the differential equation (1). However, the solution to (1) is not easy, I believe it is the Weierstrass Functions which is probably not useful for a dynamics problem. I believe that your solution ($x=6/t^2$) is actually the first term of the expansion of this function. While you do correctly recover $a(x) = x^2$ from this solution, im not sure if it has any physical meaning at $t=0$.

I do not think that the particle would ever start moving, if $\ddot{x} = x^2$ is truly its equation of motion. You could imagine this as the particle being trapped at exactly the top of some potential well (e.g a ball exactly at the top of a hill). If the particle was slightly "left or right" of x=0 (e.g x=0.00001), you would see motion. This is clear if we look at the potential (ball on a hill) picture too, the particle is clearly in an unstable equilibrium.

Your solution for the differential equation $$\ddot{x}=x^2\tag{1}$$ is not well defined at $t=0$ as $\frac{6}{0^2}$ is probelmatic! If you require the position at $t=0$ to be $0$, you must apply this as a boundary condition when solving the differential equation (1). However, the solution to (1) is not easy, I believe it is the Weierstrass Functions which is probably not useful for a dynamics problem. I believe that your solution ($x=6/t^2$) is actually the first term of the expansion of this function. While you do correctly recover $a(x) = x^2$ from this solution, im not sure if it has any physical meaning at $t=0$. Also, the particle would only every have $x=0$ at $t=\infty$ according to this solution!