Coupled differential equation arising in flow line. So, I ran (certainly not literally) across these two coupled differential equation given by:

$$x'(t)=\left(x(t)\right)^2-\left(y(t)\right)^2   $$
   $$ y'(t)=2x(t)y(t)$$

These equation occurred when I was trying to figure flow lines, to a vector field given by. 
$\vec F:\mathbb{R}^2\to \mathbb{R}^2\;\; \mid \; \vec{F}(x,y)=(x^2-y^2,2xy) $
And the flow line is given by $\vec{r}'(t)=\vec{F}(\vec{r}(t))$  , where $t$ is any introduced parameter. And obviously $\vec{r}(t)=(x(t),y(t))$ , Right?
Here's further background if it might be useful , http://ocw.mit.edu/courses/mathematics/18-022-calculus-of-several-variables-fall-2010/lecture-notes/MIT18_022F10_l_17.pdf. Example 17.10
And now for what have I tried.
I differentiated equation in terms of $x'(t)$, to get,
$$x''(t)=2x(t)x'(t)-2y(t)y'(t)$$
From original equation we have $y(t)=\pm \sqrt{(x(t))^2-x'(t)}$
Using this and above, to eliminate $y'(t)$ and $y(t)$ from the equation $y'(t)=2x(t)y(t)$, yet not really yielding good expression. 
Any help, to make the solution bit easier?

Alternatively, observe that
$$x'(t) + i y'(t) = (x(t)+iy(t))^2$$
$$x'(t)-iy'(t)=(x(t)-iy(t))^2$$
So,
$$u'(t) = [u(t)]^2$$
$$v'(t)=[v(t)]^2$$
Where $x(t)=\dfrac{1}{2}(u(t)+v(t)), y(t)=\dfrac{1}{2i}(u(t)-v(t))$ are easily found.
