Solve $\frac{dx}{dt} = \frac{1}{y}$ and $\frac{dy}{dt} = \frac{1}{x}$ by integrable combinations method? I have a such system to solve using integrable combinations method:
$$
\frac{dx}{dt} = \frac{1}{y}
$$
$$
\frac{dy}{dt} = \frac{1}{x}
$$
And the right answer for it is:
$$
C_1 x^2 = 2t + C_2
$$
$$
y^2 = C_1 (2t + C_2)
$$
And I really didn't understand from what this answer goes, because then I divide these to equations I get for the first $x = C_1 y$ and for the second $y = C_2 x$, that's definitely wrong.

\begin{align*}
\frac{dx}{dt}  & =\frac{1}{y}\tag{1}\\
\frac{dy}{dt}  & =\frac{1}{x}\tag{2}
\end{align*}
From (1) $dt=ydx$ and from (2) $dt=xdy$. Hence $ydx=xdy$ or $ydx-xdy=0$. But
$d\left(  \frac{x}{y}\right)  =\frac{ydx-x dy}{y^{2}}$. This shows that
$d\left(  \frac{x}{y}\right)  =0$ or $\frac{x}{y}=c$. Where $c$ is arbitrary
constant. Hence
$$
x=cy
$$
(2) now becomes
\begin{align*}
\frac{dy}{dt}  & =\frac{1}{cy}\\
ydy  & =\frac{1}{c}dt\\
\frac{1}{2}y^{2}  & =\frac{1}{c}t+c_{2}\\
y^{2}  & =\frac{2}{c}t+2c_{2}%
\end{align*}
Hence
$$
y=\pm\sqrt{\frac{2}{c}t+2c_{2}}
$$
