I am given a particular normal form for a system with a Bogdanov-Takens bifurcation, namely $\dot{\xi} = f(\beta_1, \beta_2, \xi)$ given by
\begin{align} \dot{\xi_1} &= \xi_2 \\ \dot{\xi_2} &= \beta_1 + \beta_2\xi_1 + \xi_1^2 - \xi_1\xi_2 \end{align}
For $\beta_2^2 > 4\beta_1$ we can apparently prove this system is orbitally equivalent to the (perturbed Hamiltonian) system $\dot{\zeta} = g(\gamma_1, \gamma_2, \zeta)$ given by
\begin{align} \dot{\zeta_1} &= \zeta_2 \\ \dot{\zeta_2} &= \zeta_1(\zeta_1 - 1) - (\gamma_1\zeta_2 + \gamma_2\zeta_1\zeta_2) \end{align}
with $\gamma_j = \gamma_j(\beta_1, \beta_2) \to 0$ for $(\beta_1, \beta_2) \to 0$.
I know that these two systems are orbitally equivalent if we can find a function $\mu: \mathbb{R}^2 \to \mathbb{R}$ such that $f(\beta_1, \beta_2, x) = \mu(x)g(\gamma_1, \gamma_2, x)$. Since we need $f_1 = \mu g_1$ we get that $\mu = 1$ is a constant function. To satisfy $f_2 = g_2$, our only tool left is writing $\gamma_1$ and $\gamma_2$ in terms of $\beta_1$ and $\beta_2$, but writing out my definition of orbitally equivalence all I get is that we need
$$\beta_1 + (\beta_2 + 1)x_1 + (\gamma_2 - 1)x_1x_2 + \gamma_1 x_2 = 0$$
which isn't true for arbitrary $x$ regardless of our choice of $\gamma_j$.
As you can see, I am completely stuck. Even if there is another definition of orbital equivalence we can work with, I don't see how we can ever satisfy the given definition. Could anyone point out what I am overlooking?