Let's go through each part of the question in order to provide solutions. Given that the last digit of your student ID is $x = 9$, we’ll substitute accordingly in the circuit analysis.

Part (a) - Analyze and determine voltages

1. Non-Inverting Terminal of OA-1: The non-inverting terminal has a voltage divider with resistors $(1+x) \text{k}\Omega = 10 \text{k}\Omega$ and $(2+x) \text{k}\Omega = 11 \text{k}\Omega$.

   • The total resistance $R_{total} = 10 \text{k}\Omega + 11 \text{k}\Omega = 21 \text{k}\Omega$.
   • The voltage $V_+$ at the non-inverting terminal can be calculated by: $V_+ = \frac{11 \text{k}\Omega}{21 \text{k}\Omega} \cdot 5 \text{V} = \frac{11}{21} \cdot 5 = 2.619 \text{V}$

2. Inverting Terminal of OA-1: Since OA-1 is an operational amplifier, it will adjust its output so that $V_+ = V_-$ due to the virtual short in ideal op-amps. Therefore, the inverting terminal voltage $V_- = 2.619 \text{V}$.

3. Output of OA-1: Using the virtual short principle (ideal op-amp), $V_{out1}$ should be sufficient to keep $V_- = V_+$. Given that OA-1 has a feedback resistor of $1 \text{M}\Omega$, the output can be calculated if the currents and additional resistors were known. For simplification in this problem, we assume it maintains the virtual short without loading.

   Thus, the output of OA-1 is set to maintain equilibrium, approximately around $2.619 \text{V}$.

Part (b) - Determine the highest $V_o$

The highest $V_o$ depends on the capacitor and the output range of OA-2.

Since OA-2 is supplied by $\pm 8 \text{V}$, the maximum possible output $V_o$ would be limited to $\pm 8 \text{V}$. Therefore, the highest $V_o$ achievable is $+8 \text{V}$, assuming the op-amp reaches saturation.

Part (c) - Determine $V_o$ over time and plot $V_o$ vs. time

This circuit forms an integrator with OA-2, as there is a capacitor $2 \mu \text{F}$ in the feedback loop. Assuming a constant voltage at the output of OA-1, the output $V_o$ will vary linearly over time as an integration effect due to the capacitor.

1. The integration formula for an ideal integrator is: $V_o(t) = -\frac{1}{RC} \int V_{in} \, dt$ where $R = 1 \text{M}\Omega$ and $C = 2 \mu \text{F}$.

2. Given $V_{out1} \approx 2.619 \text{V}$, the integration rate would be: $\frac{dV_o}{dt} = -\frac{2.619}{1 \times 10^6 \cdot 2 \times 10^{-6}} = -1.3095 \, \text{V/s}$

3. Plot $V_o$ over time starting from $V_o = 0$ at $t = 0$:

   • The output voltage will decrease linearly at a rate of $1.3095 \, \text{V/s}$.

The plot would be a straight line starting from $(0, 0)$ with a negative slope, reaching $-8 \text{V}$ at approximately $t = \frac{8}{1.3095} \approx 6.11 \text{s}$, after which it would stay constant at $-8 \text{V}$ due to saturation.

Would you like a further breakdown of any part? Here are five related questions to deepen your understanding:

1. How does the integrator circuit behavior change with different capacitor values?
2. What would happen if OA-2 had a higher supply voltage, say $\pm 12 \text{V}$?
3. Why is the virtual short principle crucial in analyzing operational amplifier circuits?
4. What impact would changing the resistor values at OA-1's input have on $V_+$ and $V_-$?
5. How would the circuit respond if a sine wave were applied instead of the DC voltage?

Tip: Always consider the power supply limits of operational amplifiers when predicting output ranges in your analysis.