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 of the first order autonomous ODE 

$$u' = \frac{1- \Psi_+(u )+\Psi_-(u )}{2}$$
with $u(0)=0$.
Then it is easy to check that $\phi_{\pm}:=F^{-1}\circ\Psi_\pm\circ u$ solve uniquely the initial problem.

**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 transformation of an antiderivative of $F$.