The topic of this problem is energy storage elements and operational amplifiers. The problem is to find V0(t) for the circuit shown below. The circuit has a couple resistors, a 20k and a 100k ohm resistor. And they have, also, a 2 microfarad capacitor in the circuit. Vin is attached to the left-hand side of the circuit and to the inverting input of the op-amp. So to solve this problem, we're going to use two things. We're going to use the properties on ideal op-amp, and we're going to use our prior knowledge of nodal analysis. So we'll start with the circuit symbol for the ideal op-amp. It has two inputs, an inverting input and a non-inverting input. There are currents associated with both these inputs, and also voltages associated with both of these inputs. And the output is on the right-hand side of the circuit symbol. We know that the ideal op-amp has certain properties that help us in solving linear circuits. The first of which is that the currents into the inputs of the op-amp are always 0. A second of which is that the voltages at the inverting and non-inverting terminals of the op-amp are also equal to each other. So we can use these two properties as a way of helping us solve our circuit. So if we look at our circuit, we have the op-amp with the non-inverting input tied to ground. So the voltage at the non-inverting input is 0 volts. Through our properties of the ideal op-amp, we know the voltage at the inverting input is also 0 volts. So that helps us in finding the Vout in the circuit. Now we can use Kirchhoff's Current Law at this upper node, let's call it Node 1. And if we do that, if we sum the currents into or out of, our choice, Node 1, then we can find a relationship between the input voltage and the output voltage. In our case, we're going to choose a Vin(t), which is equal to 0 for t < 0. And it's going to be 12 cosine of 100t for t greater than or equal to 0. And we're also going to have the initial condition that the voltage at the output at time t is equal to 0, is equal to 0. So that's our initial condition. So let's go and solve our problem. If we take the, Node 1, and look at the currents flowing into it, we have a current which is flowing from left to right through a 20 kilo resistor. That current is Vin- 0 divided by 20k. We also have a current flowing right to left through the 2 microfarad capacitor. We know that the IV characteristics for a capacitor tell us that the current is equal to the capacitance, which is 2 microfarads, Times the time derivative of the voltage across a capacitor. In our case, the voltage across a capacitor is Vout- 0. And we take the time derivative of that. And the third current into this upper node, Node 1, is a current out of the inverting input of the op-amp. And we know that that current is 0 through our properties of the ideal op-amp. And so the sum of those is equal to 0. So now we have a relationship that relates Vout to Vin. Namely, if we go back and we revise this, To put it in a form where we're looking in for Vout in terms of Vin, then it's -1 over RC times the integral of Vin(tau)d tau + our initial condition, which we know is 0 for this problem. So if we continue with this and I put in our values for R and C, you have -1 divided by (20x10 to the 3rd)(2x10 to the -6). And again, we have the integral from 0 to t of 12 cosine of 100t, 100 tau d tau. So now what we need to do is we need to take our integral, And solve the integral for the cosine of 100 tau d tau. So if we do that, we end up with a -25 times 12 sine of 100 tau divided by 100, and that's evaluated from 0 to t. So if we continue with that evaluation, then we end up with -3 sine of 100t. So, our output voltage, Vout(t) = -3 times the sine of 100t.