For simplicity, let's assume initially  that the support of $\mu$ is the whole real line, so that $F$ is a homeomorphism $ \mathbb{R} \to (0,1) $. Let's denote $b:=F(0)=\mu(-\infty,0]=1-\mu[0,+\infty)\in(0,1)$.  

The relevant function is the strictly convex function $\Phi :(0,1)\to\mathbb{R}$ defined as
$$\Phi(t):=\int_{b}^t F^{-1}(s)\, ds  $$
Finally, let's denote $\Psi_{+}:=(\Phi_{|[b,1)})^{-1}$ and $\Psi_-:=(\Phi_{|(0,b]})^{-1}$, and consider the solution $u:[0,+\infty)\to \mathbb{R}$ of the first order autonomous ODE 

$$u'(t) = \frac{1- \Psi_+(u )+\Psi_-(u )}{2},\quad t\ge0$$
with $u(0)=0$, given  explicitly by
$$u^{-1}(x)=2\int_0^x\frac{dy}{1-\Psi_+(y)+\Psi_-(y)}$$
(note that the integrand is well defined, positive and increasing for $y\ge0$. 


Then it is easy to check that $\phi_{\pm}:=F^{-1}\circ\Psi_\pm\circ u$ solve the initial problem written as
$$ 2\phi_+ F'(\phi_+)\phi_+'=2\phi_- F'(\phi_-)\phi_-'=1-F(\phi_+)+F(\phi_-)$$
that we can write equivalently, because $\phi_\pm=F^{-1}\circ F\circ \phi_\pm=\Phi'\circ F\circ \phi_\pm$, in the form
$$ 2(\Phi\circ F\circ \phi_+)'=2(\Phi\circ F\circ \phi_-)'=1-\Psi_+\circ\Phi\circ F\circ \phi_+  +\Psi_-\circ\Phi\circ F\circ \phi_- .$$




**Rmk.** The assumption on the support of $\mu$ is not crucial; if it is not the whole real line, $F$ is constant on some open set, and $F^{-1}$ has jumps and it is only defined as a left inverse of $F$; $\Phi$ is however well defined. It may also be defined as a Legendre transform of an antiderivative of $F$.