Is my intuition correct about vector field? Let us take a $\mathbb{R}^2 $ coordinate system and in it let us create a vector field of acceleration/force - the vector field will be: $$\vec{r(x,y)} =  0 \boldsymbol{\hat{\textbf{i}}}  -9.8 \boldsymbol{\hat{\textbf{j}}}$$ 
I have plotted this on Wolfram Mathematica, to get: Mathematica Vector Plot
As we see it can be thought as gravitation field, or something physical. That said, since we can assume it is a gravitational field, then the vector field is a acceleration field so then if say, I wanted to calculate the velocity of a particle after $t$ seconds, or position of it after a interval. How could I find it out? 
That might be quite simplistic but say I had a more complicated vector field like: 
$$\vec{r(x,y)} =  xy \boldsymbol{\hat{\textbf{i}}}  -9.8 \boldsymbol{\hat{\textbf{j}}}$$ 
Then how can I work out the position vector or velocity equation of the acceleration field, its because of the fact that the there are 2 variables I need to find the values of to compute it and since each value is changing then how could I do it? Differential Equation? Or am I confusing my self here? 
Regards,
Rohan 
 A: Suppose we have a force field $\mathbf{F}(\mathbf{x})$. Recall that force equals mass times acceleration, so if a particle has mass $m$ its acceleration at the point $\mathbf{x}$ is $\frac{1}{m}\mathbf{F}(\mathbf{x})$. Acceleration is the rate of change of velocity, and velocity is the rate of change of position, so we get $\mathbf{x}''(t)=\frac{1}{m}\mathbf{F}(\mathbf{x}(t))$, a differential equation, subject to some initial conditions $\mathbf{x}(0)=\mathbf{x}_0$, $\mathbf{x}'(0)=\mathbf{v}_0$ for some initial position $\mathbf{x}_0$ and initial velocity $\mathbf{v}_0$. Solving such a differential equation is not very easy in full generality.
If we're assuming Earth is flat, there is no wind resistance, the object has mass $1$, and that there is a gravitational force exists pulling downward with a constant force of $-9.8$ at all points in space, this means $\mathbf{F}=-9.8\mathbf{j}$ is a constant vector, which we can integrate to get $\mathbf{x}'(t)=-9.8t\mathbf{j}+\mathbf{v}_0$, which we may integrate again with respect to time to get
$$\mathbf{x}(t)=-4.9t^2\mathbf{j}+t\mathbf{v}_0+\mathbf{x}_0. $$
Suppose we consider $\mathbf{F}(x,y)=xy\mathbf{i}-9.8\mathbf{j}$ next. This gives the system of differential equations
$$\begin{cases} x''=xy \\ y''=-9.8 \end{cases} $$
Write $\mathbf{x}_0=x_0\mathbf{i}+y_0\mathbf{j}$ and $\mathbf{v}_0=u_0\mathbf{i}+v_0\mathbf{j}$. The second differential equation is easy to solve: integrate to get $y'=-9.8t+v_0$ and integrate again to get $y=-4.9t^2+v_0t+y_0$. Then we can substitute this expression in for $y$ in the first differential equation to obtain an ordinary differential equation $x''=(-4.9t^2+v_0t+y_0)x$. Wolfram|Alpha expresses the solution to the general ODE $x''=(at^2+bt+c)x$ and the specific ODE $x''=at^2 x$ in terms of a special function called the parabolic cylinder function, which essentially means they gave the solution to a certain differential equation its own name possibly because it couldn't be expressed in terms of elementary functions, so this one doesn't seem tractable symbolically.
