Travel vector field Lets say I have a vector field that looks like this:

Where the x-component of each vector is x=x and the y-component is y=y.
Each of these vectors represents acceleration.
Now I "drop" a fresh particle (p) into the field at a certain position (p) which already has a velocity (v). 
How can I calculate the new position (p2) and velocity (v2) of the particle after a specific time (t)?
From my basic understanding it has something to do with Line Integrals? However this doesn't help me much since it was years since I even solved simple integrals.
I'm doing this for a program I'm writing and I'm not well versed in the world of mathematics so I would really appreciate if you would keep the mathematic notation simple!
Hopefully I didn't miss any important piece of information.
Thanks, Andreas.
 A: An acceleration has units of length per time squared (or inverse length in natural units), and the vector field you describe has units of length, so there needs to be some factor of proportionality connecting the two. Let's say the acceleration is
$$\ddot{\vec{r}}=\lambda^2 \vec{r}\;,$$
where $\vec{r}$ is the position vector with components $x$ and $y$, a dot indicates differentiation with respect to time, and $\lambda^2$ is some positive proportionality constant which I've written as a square for later convenience. This is a second-order linear ordinary differential equation for $\vec{r}$, whose solution is
$$\vec{r}(t)=\vec{r}_1\mathrm e^{\lambda t}+\vec{r}_2\mathrm e^{-\lambda t}$$
with arbitrary vectors $\vec{r}_1$ and $\vec{r}_2$ which need to be determined using the initial conditions. Substituting your initial values $\vec{r}(0)=\vec{p}$ and $\dot{\vec{r}}(0)=\vec{v}$ yields
$$
\begin{eqnarray}
\vec{r}_1+\vec{r}_2&=&\vec{p}\;,\\
\vec{r}_1-\vec{r}_2&=&\frac{\vec{v}}{\lambda}\;,
\end{eqnarray}
$$
with solution
$$
\begin{eqnarray}
\vec{r}_1&=&\frac{1}{2}\left(\vec{p}+\frac{\vec{v}}{\lambda}\right)\;,\\
\vec{r}_2&=&\frac{1}{2}\left(\vec{p}-\frac{\vec{v}}{\lambda}\right)\;,
\end{eqnarray}
$$
so the overall solution is
$$\vec{r}(t)=\frac{1}{2}\left[
\left(\vec{p}+\frac{\vec{v}}{\lambda}\right)\mathrm e^{\lambda t}
+
\left(\vec{p}-\frac{\vec{v}}{\lambda}\right)\mathrm e^{-\lambda t}
\right]\;.
$$
This gives you the position at time $t$. To obtain the velocity at time $t$, you can differentiate with respect to time:
$$\dot{\vec{r}}(t)=\frac{1}{2}\left[
\left(\lambda\vec{p}+\vec{v}\right)\mathrm e^{\lambda t}
-
\left(\lambda\vec{p}-\vec{v}\right)\mathrm e^{-\lambda t}
\right]\;.
$$
To determine $\lambda$, take the square root of $\lambda^2$. Your units appear to be such that $\lambda^2=1$; in that case, you'd have $\lambda=1$.
[Edit:] I just realized that the vector field in your image doesn't look like the vector field you defined -- the vectors seem to all have the same length. If this is what you want the acceleration would be
$$\ddot{\vec{r}}=\lambda^2 \frac{\vec{r}}{|\vec{r}|}\;.$$
This can also be solved in closed form, but it's quite a bit more complicated, so I'll only work it out if that's what you need.
