Dear friends,

I have a problem with the solution representation formula for first-order (delay) differential equations.
Let t_{0}\in\mathbb{R}, A,f\in C([t_{0},\infty),\mathbb{R}) and \alpha\in C([t_{0},\infty),\mathbb{R}) be a function such that \alpha(t)\leq t for all t\geq t_{0} and \lim\nolimits_{t\to\infty}\alpha(t)=\infty.
Let t_{-1}:=\min\{\alpha(t):t\in[t_{0},\infty)\}, and \varphi\in C([t_{-1},t_{0}],\mathbb{R}).
Consider the following differential equation:
\begin{cases}<br />
x^{\prime}(t)+A(t)x(\alpha(t))=f(t)\quad\text{for}  \  t\in[t_{0},\infty)\\<br />
x(t)=\varphi(t)\quad\text{for}\ t\in[t_{-1},t_{0}].<br />
Then, x can be represented in the following form uniquely:
x(t)=\varphi(t_{0})\mathcal{X}(t,t_{0})+\int_{t_{0  }}^{t}\mathcal{X}(t,\eta)\big[f(\eta)-A(\eta)\varphi(\alpha(\eta))\big]\mathrm{d}\eta\rule{2cm}{0cm}(2)
for t\geq t_{0}, where f,A,\varphi are assumed to be identically zero out of their domains, and the fundamental solution \mathcal{X}=\mathcal{X}(t,s) is the solution of the following differential equation:
\begin{cases}<br />
x^{\prime}(t)+A(t)x(\alpha(t))=0\quad\text{for}\ t\in[s,\infty)\\<br />
x(s)=1\quad\text{and}\quad x(t)\equiv0\quad\text{for}\ t\in(-\infty,s).<br />
In (3), A,\alpha are same as in (1).
I would be glad if some one can help me with the proof of the representation formula (2).