If I have a potential given by: $$U=U_0\left[2\left(\frac xa\right)^2-\left(\frac xa\right)^4\right]$$ It says that at $t=0$, the particle is at the origin ($x=0$) and the velocity is positive and equal to the escape velocity, which I found to be $\sqrt {2U_0/m}$ I have the differential equation: $$m\ddot x=-\nabla U=-U_0\left[\frac {4x}{a^2}-\frac {4x^3}{a^4}\right]$$ EDIT::
So I have the initial $(x,\dot x)$. Now to find $x(t)$, I use conservation of energy.
$$E=K+U=\frac 12mv^2+U_0\left[2\left(\frac xa\right)^2-\left(\frac xa\right)^4\right]$$ The energy of the system is $U_0$, so I can change the above equation to: $$U_0=\frac 12 m \left (\frac {dx}{dt}\right)^2+U_0\left[2\left(\frac xa\right)^2-\left(\frac xa\right)^4\right]$$ $$\left(\frac{dx}{dt}\right)^2=\frac 2m U_0\left[1-2\left( \frac xa \right)^2+\left( \frac xa\right)^4 \right]$$
I found that the value in brackets reduces to $\frac{1}{a^2} (x^2-a^2)^2$
So: $$\frac {dx}{dt}=\sqrt{\frac{2U_0}{m}} \frac {x^2-a^2}{a^2}$$ So I want $$\int^{x}_{x_0} \frac{dx}{x^2-a^2}=\int^t_0 \sqrt{\frac {2U_0}{m}}\frac{dt}{a^2}$$ This ends up being a hyperbolic arctan function, which could potentially make sense, but am I going in the right direction?
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