How to find the first integral of a system of ODE

Question

Assume the problem $$ x'=x-2xy $$ $$ y'=2xy-y $$ to find the first integral, we solve

$$ \dfrac{dG}{dt}=\dfrac{\delta G}{\delta x}\dfrac{dx}{dt}+\dfrac{\delta G}{\delta y}\dfrac{dy}{dt} $$ How do we then find G? EDIT: According to the solution contributed below(thank u): $$ xye^(2(x-y))=c $$ it produces this graph: the solution curve

but that differs with the phase diagram: phase diagram with a specific trajectory

so what is the correlation between that solution and the phase plane and the G first integral(which to my understanding, has to follow the equation $$ \dfrac{dG}{dt}=\dfrac{\delta G}{\delta x}\dfrac{dx}{dt}+\dfrac{\delta G}{\delta y}\dfrac{dy}{dt} $$ )

Answer 1

Hint: You can eliminate $y$: $$y=\frac{x-x'}{2x}$$ and $$y'==\frac{(x'-x'')2x-(x-x')(2x')}{4x^2}$$ and you will get an equation only in $x,x'$ Plug this in the second equation.

Answer 2

Rewrite the system as $$ \frac{dx}{dt}=x-2xy, \quad \frac{dy}{dt}=2xy-y; $$ express $dt$ and separate the variables: $$ dt=\frac{dx}{x(1-2y)}=\frac{dy}{y(2x-1)} $$ $$ \frac{(2x-1) dx}{x}=\frac{(1-2y) dy}{y} $$ $$ (2-1/x)dx=(1/y-2) dy $$ $$ 2x-\ln |x|= \ln |y| -2y+\ln |C| $$ $$ \frac{e^{2x}e^{2y}}{xy}=C. $$

Answer 3

$$\begin{cases} \frac{dx}{dt}=x-2xy\\ \frac{dy}{dt}=2xy-y \end{cases} \qquad\to\qquad \frac{dy}{dx}=\frac{2xy-y}{x-2xy}=\frac{y(2x-1)}{x(1-2y)}$$ This is a separable ODE. $$\frac{1-2y}{y}dy=\frac{2x-1}{x}dx$$ $$\ln|y|-2y=2x-\ln|x|+\text{constant}$$ $$xy\;e^{-2(x+y)}=c$$ This is the implicite equation of the trajectory.

Finding analytically the position of the point $(x,y)$ on the trajectory as a function of time (that is finding the functions $x(t)$ and $y(t)$) is complicated. Better solve the problem thanks to numerical calculus.

NOTE :

The equation $\quad xy\;e^{-2(x+y)}=c\quad$ cannot be solved for $y(x)$ in terms of a finite number of elementary functions. If numerical calculus is not accepted, a special function is required. In the present case the Lambert W function : $$y(x)=-\frac12 \text{W}(X)\quad\text{with}\quad X=-\frac{2c}{x}e^{2x}$$ Similarly, solving for $x(y)$ lead to the inverse function : $$x(y)=-\frac12 \text{W}(Y)\quad\text{with}\quad Y=-\frac{2c}{y}e^{2y}$$ Theoretically putting the above functions $y(x)$ or $x(y)$ into the initial system of ODEs leads to separated ODEs with only one variable in each. But solving them analytically is doubtful.

On the other hand, the direct separation of variables from the two initial ODEs is easy. But the resulting second order ODE seems not solvable on closed form.

So, both methods after all lead to numerical calculus.