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.

  • $\begingroup$ Welcome to MSE. Your question is phrased as an isolated problem, without any further information or context. This does not match many users' quality standards, so it may attract downvotes, or be closed. To prevent that, please edit the question. This will help you recognize and resolve the issues. Concretely: please provide context, and include your work and thoughts on the problem. These changes can help in formulating more appropriate answers. $\endgroup$ Jul 3 at 9:17
  • $\begingroup$ 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

1 Answer 1


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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.