Question 10.SP.11: Let Rx be the inherent resistance of the inductor L for the ......

Figure 10-12(a) represents the circuit of Fig. 10-4 with Q ON and D OFF from which KVL gives
L \frac{d i_{L}}{d t} + R_{x}i_{L} = V_{1} \qquad 0 \leq t \leq D T_{s}           (1)

The equivalent circuit of Fig. 10-12(b) is valid for Q OFF and D ON, yielding

L \frac{d i_{L}}{d t} + R_{x}I_{L} = V_{1}  –  V_{2} \qquad D T_{s} \leq t \leq T_{s}             (2)

Integrate both (1) and (2) over their time regions of validity to give
L\int_{i_{L}(0)}^{i_{L}(D T_s)}d i_{L} + R_{x}\int_{0}^{D T_s}d t = V_{1}\int_{0}^{D T_s}d t                      (3)
L\int_{i_{L}(D T_{s})}^{i_{L}(T_{s})}d i_{L} + R_{x}\int_{D T_{s}}^{T_{s}}d t = V_{1}\left.\int_{D T_{s}}^{T_{s}}d t  –  V_{2}\int_{D T_{s}}^{T_{s}}d t\right.                    (4)

Add (3) and (4) and divide by T_s to find
{\frac{L}{T_{s}}}\int_{i_{L}(0)}^{i_{L}(T_{s})}\,d i_{L} + R_{x}\,{\frac{1}{T_{s}}}\int_{0}^{T_{s}}\,i_{L}\,d t = {\frac{V_{1}}{T_{s}}}\int_{0}^{D T_{s}}\,d t  –  {\frac{V_{2}}{T_{s}}}\int_{D T_{s}}^{T_{s}}\,d t               (5)

If i_L is periodic, i_L(0) = i_L(T_s).   Hence, the first term of (5) has a value of zero. The second term is R_x 〈i_L〉 = R_xI_L.   Thus, (5) can be written as
R_{x}I_{L} = V_{1}  –  (1  –  D)V_{2}              (6)

From the waveform sketch of Fig. 10-5,
I_{2} = {\frac{1}{T_{s}}}\int_{D T_{s}}^{T_{s}}i_{L}\,d t

Since i_L is described by straight line segments, it follows that

or
I_{L} = {\frac{I_{2}}{1  –  D}} = {\frac{V_{2}}{R_{L}(1  –  D)}}           (7)

Substitute (7) into (6) and rearrange to yield
G_{V}^{\prime} = {\frac{V_{2}}{V_{1}}} = {\frac{(1  –  D)R_{L}}{R_{x}  +  R_{L}(1  –  D)^{2}}}                (8)