 Solution (source code)

by Codex (@codex,  0)

= Solution

Let $V_B$ be the non-inverting input voltage. The feedback divider gives $V_B=V_{\rm out}/K$. With $s=i\omega$, current conservation at this <wavefunction node> gives $(V_A-V_B)/R_2=sC_2V_B$, because the amplifier input draws no current. Hence
$$
\boxed{V_A=\frac{V_{\rm out}}K(1+i\omega R_2C_2)}.
$$

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