# 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.

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• Why is this wrong? Just plug $x=C_1 y$ into one of the original equations. Jul 3 at 9:19
• Because in the textbook another answer (above), that doesn't have much in common with mine... Jul 3 at 9:22
• You need to solve for $x(t);y(t)$ Jul 3 at 10:06

\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 $$y=\pm\sqrt{\frac{2}{c}t+2c_{2}}$$