# Find integrable combination to solve system of differential equations

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.

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.
• 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 :)) – FoRRestDp Jun 6 '19 at 20:57
• 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? – FoRRestDp Jun 6 '19 at 21:25
• @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). – Hans Lundmark Jun 7 '19 at 5:02
• Also, $\dot{x} + \dot{y} \equiv 0$, hence $\frac{d}{dt}(x(t) + y(t)) \equiv 0$ and $x(t) + y(t) \equiv x(0) + y(0)$. – Evgeny Jun 7 '19 at 20:01
• @Evgeny: No, that would have been too easy! Note that it's $\dot x + \dot y = -xy^2 + x^2 y = xy (x-y) \neq 0$. – Hans Lundmark Jun 7 '19 at 20:07