Finding the conserved energy function I have the differential system
\begin{eqnarray*}
x'(t) & = &  x^2(t)-y^2(t)  \\
y'(t) & = & 2xy \\
\end{eqnarray*}
I need to find the conservative energy function for it. Usually I would just divide the two expressions and then isolate $x$ and $y$ on either side and integrate. But I cant do that here. Any other methods to finding the conserved energy function? 
 A: Obviously, this is the real component notation for $z'=z^2$, $z(t)=x(t)+iy(t)$. Its solutions are given as $1/z=C-t$ or $z(t)=1/(C-t)$. 
Thus $1/z+t$ is constant along each solution, in real components $$x/(x^2+y^2)+t$$ and $$-y/(x^2+y^2).$$ The last one does not depend on $t$ and is thus a first integral. 
I would not call it energy, because this would imply that the system is obtained as the symplectic gradient of this function, Hamiltonian, $H$, i.e., $\dot x=\frac{\partial H}{\partial y}$ and $\dot y=-\frac{\partial H}{\partial x}$, which is impossible due to the second derivatives.
A: The answer given by LutzL is pretty good, but if you don't like complex numbers, here's what I would do
$$ \frac{dy}{dx} = \frac{2xy}{x^2 - y^2} = \frac{2\frac{y}{x}}{1-\frac{y^2}{x^2}}$$
Make the substitution $u = \frac{y}{x}$. Then $y = xu$ or $\frac{dy}{dx} = u + x\frac{du}{dx}$. The equation becomes
$$ u + x\frac{du}{dx} = \frac{2u}{1-u^2} $$
$$ x\frac{du}{dx} = \frac{2u}{1-u^2} - u $$ $$ x\frac{du}{dx} = \frac{u+u^3}{1-u^2} $$
Now you can separate the variables and integrate to get the answer
