Solving an implicit equation for a cdf $M$: $M(c)= a \int_{-\infty}^c M^\prime(z) M(c^b z^{1-b}) dz + (1-a) G(z)$ I'm trying to analytically characterize a cumulative distribution function $M$ that takes a particular implicit form
$$ M(c) = a \int_{0}^c M^\prime(z) M(c^b z^{1-b}) dz + (1-a) G(c)$$
and has support on $c \in [0,\infty)$. The parameters $a \in [0,1]$ and $b \in [1,\infty)$ are both given, as is some exogenous c.d.f. $G$. I've characterized some of the limiting cases. If $a = 0$ or $b \rightarrow +\infty$ it naturally collapses to $M(c) = G(c)$. If $b = 1$, then the problem becomes
$$ a M(c)^2 - M(c) + (1-a) G(c) = 0$$
which implies a characterization of $M$ via the quadratic formula. The trouble for me is the interior case, in which the integral becomes non-trivial.
I have not been able to prove existence given a generic c.d.f. $G$, though solving it computationally seems to converge. I would therefore also accept any answer that imposes restrictions on $G$ to generate the result (i.e., "if $G$ is normal then $M = \ldots$"). Thanks.

Supposing everything differentiable, maybe you can try this:
$$M ' (c) =a M'(c) M(c) + (1+a) G'(c)$$
and solve the ODE. Rewrting as
$$- \frac{(a + 1) G'(c)}{a} + \left(-\frac{1}{a} + M(c)\right)  M'(c) = 0$$
is an exact equation and with this is possible to find out the expression for M(c) and thus study wich kind of $G(c)$ is more suitable for your needs.
