Solving differential equation for x I have a field $\phi(x,t)=\sin(t+|x|)(\frac{x}{|x|})$ where x is a point vector and t is the current time.
If this field describes the acceleration of a particle at a point in space and time: 
$$\frac{d^2x}{dt^2} = \phi(x,t)$$
Then surely the function that describes the path of the particle is just the second anti-derivative of $\phi$ with respect to time.
I haven't done much multivariable calculus, and I don't know how to do this. As far as I can tell,
$$\frac{dx}{dt}=\int \sin(t+|x|)(\frac{x}{|x|}) dt = -\cos(t+|x|)(\frac{x}{|x|})$$
$$x = \int-\cos(t+|x|)(\frac{x}{|x|})dt = -sin(t+|x|)(\frac{x}{|x|})$$
I'm not sure if this is correct, or how to solve this equation so that all the $x$ terms are on the left hand side.
Any help is appreciated.
 A: You make us terrified that as you force to solve the higher-order ODE with dependent variable by direct integration on the both sides! In fact this is a more serious wrong thing than applying separation of variables on the higher-order separable ODEs!
$\dfrac{d^2x}{dt^2}=\sin(t+|x|)\left(\dfrac{x}{|x|}\right)$
$\begin{cases}\dfrac{d^2x}{dt^2}=\sin(t+x)&\text{when}~x>0\\\dfrac{d^2x}{dt^2}=-\sin(t-x)&\text{when}~x<0\end{cases}$
Let $u=\begin{cases}t+x&\text{when}~x>0\\t-x&\text{when}~x<0\end{cases}$ ,
Then $x=\begin{cases}u-t&\text{when}~x>0\\t-u&\text{when}~x<0\end{cases}$
$\dfrac{dx}{dt}=\begin{cases}\dfrac{du}{dt}-1&\text{when}~x>0\\1-\dfrac{du}{dt}&\text{when}~x<0\end{cases}$
$\dfrac{d^2x}{dt^2}=\begin{cases}\dfrac{d^2u}{dt^2}&\text{when}~x>0\\-\dfrac{d^2u}{dt^2}&\text{when}~x<0\end{cases}$
$\therefore\begin{cases}\dfrac{d^2u}{dt^2}=\sin u&\text{when}~x>0\\-\dfrac{d^2u}{dt^2}=-\sin u&\text{when}~x<0\end{cases}$
$\dfrac{d^2u}{dt^2}=\sin u$
According to http://www.wolframalpha.com/input/?i=u%22%3Dsinu,
$u=\pm~2~\text{am}\biggl(\dfrac{\sqrt{C_1(t+C_2)^2}}{2}|-\dfrac{4}{C_1}\biggr)$
$t+|x|=\pm~2~\text{am}\biggl(\dfrac{\sqrt{C_1(t+C_2)^2}}{2}|-\dfrac{4}{C_1}\biggr)$
$|x|=\pm~2~\text{am}\biggl(\dfrac{\sqrt{C_1(t+C_2)^2}}{2}|-\dfrac{4}{C_1}\biggr)-t$
$x=\pm\biggl(\pm~2~\text{am}\biggl(\dfrac{\sqrt{C_1(t+C_2)^2}}{2}|-\dfrac{4}{C_1}\biggr)-t\biggr)$
