Difference Amplifier with 3 Op. Amp. How to analyze it?
Who doesn't like a good old \$Y - \Delta \$ transform: https://en.wikipedia.org/wiki/Y-Δ_transform ? [EDIT after solving the question: might not be necessary].

The computed equivalent values for the \$ \Delta \$ are:

\$R_{eq1} = \frac{R_1R_1+2R_3R_1}{4R_3}\$

\$R_{eq3} = \frac{R_1+2R_3}{2}\$

The op-amps 1 &amp; 2 have negative feedback so we can assume they are not in saturation and for them \$v_+ = v_- \$.

The current \$i_2\$ across \$R_{2top}\$ is equal to:

\$i_2 = \frac{v_o-v_2}{R_2} = \frac{v_2-v_1}{R_{eq1}}+\frac{v_2-v_{oa}}{R_{eq3}}\$ [1]

The current \$i_1\$ across \$R_{2bot}\$ is equal to:

\$i_1 = \frac{v_1-v_y}{R_2} = \frac{v_2-v_1}{R_{eq1}}+\frac{v_{oa}-v_{1}}{R_{eq3}}\$ [2]

From [1] we derive:

\$ \frac{v_{oa}}{R_{eq3}} = \frac{v_2-v_o}{R_2} + \frac{v_2-v_1}{R_{eq1}}+\frac{v_2}{R_{eq3}}\$ [3]

From [2] we derive:

\$ \frac{v_{oa}}{R_{eq3}} = \frac{v_1-v_y}{R_2} + \frac{v_1-v_2}{R_{eq1}}+\frac{v_{1}}{R_{eq3}}\$ [4]

Isolating \$ v_{oa} \$

\$ v_{oa} =\frac{R_{eq3}(v_1-v_y)}{R_2}+\frac{R_{eq3}(v_1-v_2)}{R_{eq1}}+v_{1}\$

\$ v_y \$ &amp; \$ v_{oa} \$ have opposite signs in this equation! So it turns out there is negative feedback from \$ v_y \$ on \$ v_{oa} \$. This explains the stability of the circuit.

So we can assume \$ v_y = 0 \$ (this is the big trick of this exercise - you can indeed verify that v_y is 0 in simulations). So now it is easier from [3] = [4] and setting  \$ v_y = 0 \$ we get:

\$ \frac{v_2-v_o}{R_2} + \frac{v_2-v_1}{R_{eq1}}+\frac{v_2}{R_{eq3}} = \frac{v_1}{R_2} + \frac{v_1-v_2}{R_{eq1}}+\frac{v_{1}}{R_{eq3}}\$

\$ \frac{v_o}{R_2} = \frac{v_2-v_1}{R_2} + \frac{v_2-v_1}{R_{eq1}} + \frac{v_2-v_1}{R_{eq3}}\$

\$ v_o = (v_2-v_1) \left(1 + \frac{R_2}{R_{eq1}} + \frac{R_2}{R_{eq3}}\right)\$

\$ v_o = (v_2-v_1) \left(1 + \frac{4R_3R_2}{R_1R_1+2R_3R_1} + \frac{2R_1R_2}{R_1R_1+2R_3R_1}\right)\$

\$ v_o = (v_2-v_1) \left(1 + \frac{2R2(2R_3+R_1)}{R_1(R_1+2R_3)}\right)\$

\$ v_o = (v_2-v_1) \left(1 + \frac{2R2}{R_1}\right)\$
Hence the answer.







simulate this circuit &ndash; Schematic created using CircuitLab