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The steady state solution for first stage is disconnected from outside by C1 and C2. The non inverting input has 3V on it (6 * (R4 / (R3 + R4))). The inverting input simply has the output connected around to it. For DC, R2 does nothing because of C1.
The second stage recognizes that the LM358 does not know how to pull its output below one or two transistor junctions above V(-). Or it has applied the amplifier gain to the LM358's input offset voltage.
The circuit does not have a signal input and the voltages do not say they are DC but they are probably DC.
1) The first opamp is a "follower" at DC with a gain of 1.
The non-inverting input is 3.0V.
The output voltage is 3V x 1= 3V.
2) The second opamp has a DC gain of 1+ (R10/R9)= 221.
Its inputs are at 0V.
It might have an input offset voltage of +3.17mV so its output is 3.17mV x 221= 0.7V.
I think the point that everyone was making, is that you haven't actually asked a question. You presented us with a schematic and a couple of formulae. What is it that you want to know?
One thing that would help you a lot i think is to do a simulation point for pin 2 and another for pin 3 of the second op amp shown in your schematic. That would tell you the voltages there and help to understand what exactly is happening here. The input offset voltage appears to be the cause of the 0.7v output, and doing the calculations it works out to a typical value for this type of op amp so i would believe that is what is happening here.
Once you assume the 0.7v output, you can use a simple resistive divider formula to get the answer to the input pin 2. The formula is:
V2=V1*R2/(R1+R2)
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