I'm reading through some notes from a past course of mine, where a system of ODE's of the form$$ \begin{array}{c} x'=h(x,y)f(x,y)\\ y'=h(x,y)g(x,y) \end{array} $$ appears, such that $x,y\in\mathbb{R}$ and $h(x,y)>0$ for all values of $x,y$. Now it says that one can divide the RHS by $h(x,y)$ (since it's positive) and the orbit of the system remain unchanged, so that the above system is equivalent to
$$ \begin{array}{c} x'=f(x,y)\\ y'=g(x,y) \end{array} $$Is this true ?
EDIT (due to Hans Lundmarks comment): Can someone please provide me with a reference or proof for this ?
Suppose that we have a solution $x(t),y(t)$ to $$x'=f(x,y)$$ $$y'=g(x,y).$$ Define $\tilde f(t)$ to satisfy the differential equation $$\tilde f'(t)=h(x(\tilde f(t)),y(\tilde f(t))).$$ Notice that, since we can regard $h$, $x$ and $y$ as known already, we can let, for some constant $c_1$: $$I(z)=\int_{c_1}^z\frac{d\lambda}{h(x(\lambda),y(\lambda))}$$ and notice that, from the differential equation $I(\tilde f(t))=t+c_2$ - meaning $\tilde f$ is the inverse function of $I$. Notice that $I$ is strictly increasing, since $h$ is positive, implying that $I$ is at least injective. To show surjectivity, suppose that $\lim_{z\rightarrow\infty}I(z)=k$ for some finite $k$. Then, it would follow, that the limit of the integrand would be $\lim_{z\rightarrow\infty}\frac{1}{h(x(z),y(z))}=0$ implying that $\lim_{(x,y)\rightarrow (x(k),y(k))}h(x,y)=\infty$, which would mean that $h$ would tend to infinity somewhere on the orbit of $(x,y)$ - impossible for a continuous function. A similar argument handles the case of $z\rightarrow-\infty$.
This implies that $I$ is bijective, and hence has a bijective inverse $\tilde f$ which solves the desired equation. Then, let $$x_f(t)=x(\tilde f(t))$$ $$y_f(t)=y(\tilde f(t))$$ we have that, from the chain rule $$x_f'(t)=\tilde f'(t)x'(\tilde f(t))$$ $$y_f'(t)=\tilde f'(t)y'(\tilde f(t))$$ but if we replace $x'$, $y'$, and $\tilde f'$ by the differential equations they solve and replace $x(\tilde f(t))$ with $x_f$, we get $$x_f'=h(x_f,y_f)f(x_f,y_f)$$ $$y_f'=h(x_f,y_f)g(x_f,y_f)$$ which means that $x_f$ and $y_f$ solve the modified differential equation, but since they are composition of $x$ and $y$ with a bijective $\tilde f$, they must have the same orbits.