When the switch is ON:

\( L \frac{di_L}{dt} = V_{in} -i_L (R_L +R_{DS,ON}) \)

\( C \frac{dV_C}{dt} = \frac{-V_{out}}{R + ESR} \)

\( C \frac{dV_C}{dt} = \frac{V_C - V_{out}}{ESR} \)

When the switch is off:

\( L \frac{di_L}{dt} = V_{in} -V_{out} -i_L(R_L + R_D) \)

\( C \frac{dV_C}{dt} = i_L -\frac{V_{out}}{R +ESR} \)

\( V_C = V_{out} -i_CESR \)

Then

\( \Delta i_{Lon} + \Delta i_{Loff} = 0 \)

But to get \( \Delta i_L \) terms I would have to integrate the equations with the inductor voltage, which both contain a term with \( i_L \). How can that be done? Thanks.