I have the following equation: $m\dfrac{d^2\vec{r}(t)}{dt^2} = \vec{F}(\vec{r}(t))$.
I am asked to rewrite it into a differential equation of order 1.
I guesd I'd have to simplify the equation $\int m\dfrac{d^2\vec{r}(t)}{dt^2}dt = \int \vec{F}(\vec{r}(t))dt$
But I'm not sure how to do that.
Multiply the equation by $\frac{d \vec{r(t)}}{dt}$ on both sides and then you can use
$\frac{d}{dt}(\frac{d \vec{r(t)}}{dt}^2) = 2 \frac{d \vec{r(t)}}{dt} \frac{d^2 \vec{r(t)}}{dt^2}$.
Finally, integrate by time $dt$.
The Newton's law is $$m\ddot{r}+\nabla{V}(r)=0$$ Multiply by $\dot{r}$ you get $$m\ddot{r}\dot{r}+\nabla{V}(r)\dot{r}=\frac{d}{dt}(\frac{m}{2}\dot{r}^{2}+V(r))=0$$ Hence $$\frac{m}{2}\dot{r}^{2}+V(r)=E$$ If $V(r, t)$ is explicilty time dependent, then $$m\ddot{r}+\nabla{V}(r, t)=0$$ Multiply by $\dot{r}$ you get $$m\ddot{r}\dot{r}+\nabla{V}(r, t)\dot{r}=\frac{d}{dt}(\frac{m}{2}\dot{r}^{2}+V(r, t))-\frac{\partial{V}(r, t)}{\partial{t}}=0$$ Hence $$\frac{dE}{dt}=\frac{\partial{V}(r, t)}{\partial{t}}$$