Finding the integral curves of a vector field $x\partial y- y\partial x$ well, i know that the curves is (x,y) where 
$\displaystyle\frac{dx}{dt}=-y$
and 
$\displaystyle\frac{dy}{dt}=x$
i know that this is a PDE system, my professor said that if use second derivatires it may be turned into the ODE problem
$\displaystyle\frac{d^{2}x}{dt^2}=-x$
and its solution is a linear combination of $sin$ and $cos$, but this talk is too abstract for me, i would like if someone give me a exact solution of this problem. Is there another way, easier to solve this?
 A: Multiply $$\displaystyle\frac{dx}{dt}=-y$$ by $x$ to get $$\displaystyle x\frac{dx}{dt}=-xy$$
Multiply $$\displaystyle\frac{dy}{dt}=x$$ by $y$ to get $$\displaystyle y\frac{dy}{dt}=xy$$
Add them together to get $$ x \frac{dx}{dt} +y    \frac{dy}{dt}    =0$$
Integrate to get $$ x^2 + y^2 =C $$
That is the integral curve for your system.  
A: Perhaps it might be more clear if you write $y = y(t)$ and $x = x(t)$.  Then you get $x'(t) = -y(t)$ and $y'(t) = x(t)$.
As you wrote, substituting one of the equalities into the other gets us $x''(t) =  - y'(t) = -x(t)$.
Similarly, you have $y''(t) = x'(t) = -y(t)$.  Thus, you get the two equations:
$$x''(t) + x(t) = 0$$
$$y''(t) + y(t) = 0$$
The characteristic equation associated to these is $r^2 + 1 = 0$.  Let's just look at the differential equation associated with $x(t)$.  This has complex roots, so the solution should be of the form $x(t) = c_1 \cos(t) + c_2 \sin(t)$ for arbitrary constants $c_1$ and $c_2$.
So, $y(t) = c_1 \sin(t) - c_2 \cos(t)$.
Writing the curve as $\gamma(t)$, we have $\gamma(t) = (c_1 \cos(t) + c_2 \sin(t), c_1 \sin t - c_2 \cos t)$ which gives us circles except at the point $(0,0)$.
A: you can multiply the first equation by the second one
$$\frac {dx}{dt}=-y$$
$$\frac {dx}{dt}\frac {dt}{dy}=-\frac yx$$
$$\frac {dx}{dy}=-\frac yx$$
$$ x{dx}=-y{dy}$$
Integrate
$$x^2=-y^2+K$$
$$x^2+y^2=K$$
