You know the current through R4:
\$i_4 = \dfrac{A}{R_4}\$
Thus, you know the current through R3:
\$i_3 = i_4 \$
Thus, you know the output voltage of the 2nd op-amp:
\$v_{O2} = i_4(R_4 + R_3) = A(1 + \dfrac{R_3}{R_4})\$
Thus, you know the voltage across R2:
\$v_{R2} = A - v_{O2} = -A\dfrac{R_3}{R_4} \$
Thus, you know the current through R2 which is identical to the current through the capacitor:
\$i_C = i_{R2} = -\dfrac{A}{R_2}\dfrac{R_3}{R_4}\$
Now recall:
\$i_C = C \dfrac{dv_C}{dt}\$
Can you take it from here?
I can't find an error in my calcs, so I think I'm just
misinterpreting what Vs and Is actually are.
Switching to the phasor domain, we have:
\$I_c = -\dfrac{A}{R_2}\dfrac{R_3}{R_4} = j \omega C V_c\$
or
\$V_c = -\dfrac{A}{j \omega}\dfrac{R_3}{R_2R_4C}\$
Thus, the output voltage of the first op-amp is:
\$V_{o1} = A + V_c = A - \dfrac{A}{j \omega}\dfrac{R_3}{R_2R_4C}\$
And the voltage across R1 is:
\$V_{r1} = A - V_{o1} = \dfrac{A}{j \omega}\dfrac{R_3}{R_2R_4C}\$
Finally, the current through R1 is:
\$I_{r1} = \dfrac{V_{r1}}{R_1} = \dfrac{A}{j \omega}\dfrac{R_3}{R_1R_2R_4C} \$
But the source current is identical to \$I_{r1}\$, thus:
\$\dfrac{V_s}{I_s} = \dfrac{A}{I_{r1}} = j\omega \dfrac{R_1R_2R_4C}{R_3} = j \omega L_{eq} \$
Best Answer
The gyrator simulates only some of the properties of the inductor, e.g. its frequency characteristics and Q factor, but not all of them, e.g. not energy storage.