Let $\frac{\mathrm{d}x}{\mathrm{d}t}=\dot{x}=v$ and $\frac{\mathrm{d}^2x}{\mathrm{d}t^2}=\ddot{x}=\dot{v}$. The original equation can now be written as $\dot{v}=x^3-x^5-v$. In consequence, the second order differential equation has been replaced by a system of two first order differential equations, namely
$$\left\\{\begin{array}{l l} \dot{x}=v\\\ \dot{v}=x^3-x^5-v \end{array}\right.$$
Note that the derivatives depend only on $x$ and $v$.