I want to solve for the measurable function $c(x)$ for the following equation: $$c(2x)-2c(x)=2\alpha \int_0^{2x}e^{c(z)}\ z\ dz.$$ Where $x \in (0,\infty)$ and $\alpha \in [-1, 1].$
So I continued by assuming $c\in \mathcal{C}^2$ then: $$c'(2x)-c'(x)=4\alpha\ x\ e^{c(2x)}$$ Differentiating again: $$2c''(2x)-c''(x)=4\alpha\ \left(e^{c(2x)}+2x\ e^{c(2x)}\ c'(2x)\right)$$ Letting $w=2x$ and substituting $c'(2x)-c'(x)=4\alpha\ x\ e^{c(2x)}$ we get: $$2c''(2x)-c''(x)={c'(2x)-c'(x)\over x}+2c'(2x)(c'(2x)-c'(x))$$ $$wc''(w)-{w\over 2}c''({w\over 2})=c'({w})-c'({w\over 2})+wc'(w)^2-wc'(w)c'({w\over 2})$$ Any hints on how to solve the DDE given a boundary conditions of $c(0)=0$ and $c(\infty)=-2\alpha.$