How to find an orbit in implicit form for a first order non-linear system of differential equations?

Question

How to find an orbit in implicit form for a first order non-linear system of differential equations? Say $x'= x - xy$, $y'= y - 2xy$ is our system. How do we find an orbit of it in an implicit form?

Answer 1

$$\frac{d x(t)}{dt}=x(t) (1-y(t))\tag{1}$$ $$\frac{d y(t)}{dt}=y(t) (1-2x(t))\tag{2}$$

(2)/(1) leads to:

$$\frac{\frac{d y(t)}{dt}}{\frac{d x(t)}{dt}}=\frac{y(t) (1-2x(t))}{x(t) (1-y(t))}\tag{3}$$

or

$$\frac{d y}{d x}=\frac{y (1-2x)}{x (1-y)}\tag{4}$$

or

$$\frac{(1-y)}{y}d y=\frac{(1-2x)}{x }dx\tag{5}$$

The solution is

$$y(x)=\frac{x}{x-c e^{2x}}$$

Answer 2

Assume that $(x(t),y(t))$ belongs to the curve of equation $$H(x)=K(y),$$ for every $t$, that is, that, for every $t$, $$H(x(t))=K(y(t)).$$Then $$\frac{\mathrm d}{\mathrm dt}H(x(t))=\frac{\mathrm d}{\mathrm dt}K(y(t)),$$ that is, $$H'(x(t))x'(t)=K'(y(t))y'(t).$$ Thus, a sufficient conditio, is that, for every $(x,y)$, $$(x-xy)H'(x)=(y-2xy)K'(y),$$ that is, $$\frac{x}{1-2x}H'(x)=\frac{y}{1-y}K'(y).$$ The LHS does not depend on $y$ and the RHS does not depend on $x$, hence both are constant, without loss of generality, $$H'(x)=\frac{1-2x}x,\qquad K'(y)=\frac{1-y}y,$$ that is, $$H(x)=\log x-2x+h,\qquad K(y)=\log y-y+k.$$ Finally, the orbits are defined by the condition that $(\log y-y)-(\log x-2x)$ is constant or, equivalently, by the condition that, for some constant $C$, $$y\mathrm e^{-y}=Cx\mathrm e^{-2x}.$$