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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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  • $\begingroup$ Why is this wrong? Just plug $x=C_1 y$ into one of the original equations. $\endgroup$
    – user619894
    Jul 3 at 9:19
  • $\begingroup$ Because in the textbook another answer (above), that doesn't have much in common with mine... $\endgroup$
    – Noerig
    Jul 3 at 9:22
  • $\begingroup$ You need to solve for $x(t);y(t)$ $\endgroup$
    – user619894
    Jul 3 at 10:06

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

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