Solving differential equations using circuits

To make it possible to follow the solution, I have labeled intermediate quantities in the diagram. Recall the two "golden rules" of op amps in the inverting feedback configuration:

1) The voltages at the inputs are equal
2) The currents into the inputs are zero

Proceed to solve for the quantities:

$I_1 = C \dot{V} \\ V_1 = 0 - R I_1 = - R C \dot{V} \\ I_2 = C \dot{V}_1 = - R C^2 \ddot{V} \\ V_2 = 0 - R I_2 = R^2 C^2 \ddot{V} \\ I_4 = \frac{V_2}{12 R} = \frac{R C^2 \ddot{V}}{12} \\ I_3 = \frac{V_1}{R} = -C \dot{V} \\ V_3 = 0 - R I_3 = R C \dot{V} \\ I_5 = \frac{V_3}{1.5 R} = \frac{C \dot{V}}{1.5} \\ I_6 = I_4 + I_5 = \frac{R C^2 \ddot{V}}{12} + \frac{C \dot{V}}{1.5}$

Having determined the intermediate quantities, we can now write a final expression for $V$ .

$V = 0 - 6 R I_6 = - 6 R \Big(\frac{R C^2 \ddot{V}}{12} + \frac{C \dot{V}}{1.5} \Big) \\ V = - \frac{R^2 C^2 \ddot{V}}{2} - 4 R C \dot{V}$

Simplifying a bit more, and substituting in $RC = 1$ , results in:

$\ddot{V} + 8 \dot{V} + 2 V = 0$