There is the system:

$$ \left\{ \begin{array}{c} \dot{x} = -xy^2+x+y \\ \dot{y} = -x-y+x^2y \end{array} \right. $$

the way it should be solved is to find an integrable combination there is the description (page 349 of pdf document).

I've already tried multiplying the first equation by $x$, the second one by $y$ then adding first to second so I got: $xdx+ydy = x^2 - y^2$, which I have no idea how to integrate.

Also I tried multiplying the first by $y$, the second by $x$, also adding first to second so I got: $ydx+xdy=(xy - 1)(x^2 - y^2)$, which I also don't know how to integrate.

Could you plese provide any integrating combinations or ideas how to deal with equations I got.


1 Answer 1


That's a good start, except that you should write $\dot x$ and $\dot y$ instead of $dx$ and $dy$.

Anyway, you have $$ (\tfrac12 (x^2+y^2))\dot{} = x \dot x + y \dot y = x^2-y^2 $$ and $$ (xy)\dot{} = \dot x y + x \dot y = (xy-1)(x^2-y^2) . $$ These can be combined to give $$ (xy)\dot{} = (xy-1) (\tfrac12 (x^2+y^2))\dot{} $$ so that either $xy-1=0$ identically or $$ \frac{(xy)\dot{}}{xy-1} = (\tfrac12 (x^2+y^2))\dot{} $$ where both sides can be integrated to find a constant of motion.

Can you take it from there, or do you need further hints?

  • $\begingroup$ Thank you for the answer. I used your steps and got $ce^{x^2+y^2}=(xy-1)^2$. As I know after that we should express $x$ or $y$ from there and substitute the result to one of the equations so we get $x(t)$ or $y(t)$. Am I right or there is some another way get the answer (cause mine seems to be too complicated :)) $\endgroup$
    – FoRRestDp
    Commented Jun 6, 2019 at 20:57
  • $\begingroup$ There also was Cauchy problem in the task so I've tried to use to simplify my task. It is: $x(2)=1$, $y(2)=1$. I substituted it to the result above and got that $c=0$ hence $1=xy$ hence $x=\frac{1}{y}$. Then I substituted this to the second equation and got $\frac{dy}{y} = -dt$. Am I right doing this? $\endgroup$
    – FoRRestDp
    Commented Jun 6, 2019 at 21:25
  • 1
    $\begingroup$ @ЕгорПономарёв: Looks fine! I doubt it's possible to find expressions for $x(t)$ and $y(t)$ in the general case (in practice, that is; it's always possible in principle once you have a constant of motion). $\endgroup$ Commented Jun 7, 2019 at 5:02
  • $\begingroup$ Also, $\dot{x} + \dot{y} \equiv 0$, hence $\frac{d}{dt}(x(t) + y(t)) \equiv 0$ and $x(t) + y(t) \equiv x(0) + y(0)$. $\endgroup$
    – Evgeny
    Commented Jun 7, 2019 at 20:01
  • $\begingroup$ @Evgeny: No, that would have been too easy! Note that it's $\dot x + \dot y = -xy^2 + x^2 y = xy (x-y) \neq 0$. $\endgroup$ Commented Jun 7, 2019 at 20:07

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