Signal Flow Graph for electrical circuits
Let me label the intermediate nodes in the circuit using the letters A, B and C as shown below.



The nodal equations at the nodes A,B and C can be re-arranged to get the three equations given below.

$$\begin{align}U_AS_A &amp;= CsU_1 + CsU_B\tag1\\
U_BS_B &amp;= \frac{G}{2}U_1 + GU_2+CsU_A\tag2\\
U_CS_C &amp;=  GU_1 + G'U_2\tag3\end{align}$$ 

Where, \$G=\frac{1}{R}\$, \$G'=\frac{1}{K}\$. \$U_A, U_B\$ and \$U_C\$ are the potential at the nodes A, B and C respectively. And \$S_A, S_B\$ and \$S_C\$ are defined as follows:

$$\begin{align}S_A &amp;= G+2Cs\\
S_B &amp;= \frac{3}{2}G + Cs\\
S_C &amp;=G+G'\end{align}$$

Let the gain of the operational amplifier be \$A_{op}\$ and \$A_{op}\rightarrow \infty\$. Now the output voltage of op-amp can be written as:

$$U_2 = A_{op}(U_B-U_C)|_{A_{op}\rightarrow\infty}\tag4$$

From the equations (1) to (3) potential at nodes can be written as:
$$\begin{align}U_A &amp;= \frac{Cs}{S_A}U_1+\frac{Cs}{S_A}U_B\tag5\\
U_B &amp;= \frac{G}{2S_B}U_1+\frac{G}{S_B}U_2+\frac{Cs}{S_B}U_A\tag6\\
U_C &amp;= \frac{G}{S_C}U_1+\frac{G'}{S_C}U_2\tag7\end{align}$$

The signal flow graph can be drawn using the equations from (4) to (7) as given below:



You can apply the limit \$A_{op}\rightarrow \infty\$ while simplifying the calculations.