In the AC modeling of converters, what is considered is the average over one period of a certain quantity (e.g. \$\langle v(t)\rangle=\frac{1}{T}\int_0^Tv(t)\mathrm{dt}\$), in order to maintain the slowly varying behaviour of electric quantities, while eliminating high frequency variations and ripples.
Using this approach to model the inductor in the circuit you showed, it is
\begin{equation}
\langle v_L(t)\rangle=d(t)\langle v_g(t)\rangle + [1-d(t)]\langle v(t)\rangle = L\frac{d}{dt}\langle i(t)\rangle
\end{equation}
where \$d(t)\$ is the duty cycle in the period that was considered, and the other quantities are named as per the circuit you provided.
If you now consider all quantities as being a DC value + an AC small variation/perturbation (e.g. \$\langle v_L(t)\rangle=V_L + \hat{v}_L(t)\$), you obtain
\begin{equation}
L\frac{d}{dt}[I_L + \hat{i}_L(t)]=[D + \hat{d}(t)][V + \hat{v}(t)] + [1-D-\hat{d}(t)][V+\hat{v}(t)].
\end{equation}
If you now compute all the products and neglect all DC (e.g. \$DV\$) and second order terms (e.g. \$\hat{d}(t)\hat{v}(t)\$), you obtain the very expression shown in the picture you uploaded (where \$D'=1-D\$).
Now, consider which quantities are you using to control your converter: you obviously use the duty-cycle, and \$\hat{d}(t)\$ is the perturbation on it that causes the circuit to steer away from the precise working point you wish to have; thus \$(V_g - V)\hat{d}(t)\$ is your independent source, since it is due to your control circuit.
What about the other two sources? They are the effect on the inductor of variations/perturbations happening on input and output voltage, which you do not control directly, since they are due to the power supply you are exploiting and to the correct work of your circuit, respectively.
