Welcome back to Linear Circuits this is Dr Weitnauer. This lesson is about the direct method. Our objective is for a second-order circuit, learn how to apply the direct method to determine the differential equation for the desired circuit variable. This builds on that a second order circuit is modeled by a second order differential equation, the IV relationships for capacitors and inductors in KCL and KVL. Here's a summary of the direct method. Suppose the inductor currents and/or the capacitor voltages are x1 and x2. Write a circuit equation for dx1 dt in terms of x1 and x2. Write a second circuit equation for dx2 dt in terms of x1 and x2. Then substitute to achieve one equation in terms of the desired circuit variable. In other words, eliminate the unwanted circuit variable, using derivatives only. So we'll do two examples. Here's the first, the parallel RLC circuit. And our objective is to find differential equation for i. I emphasize, this lesson is not about solving for the complete response. It's only about getting the differential equation. So let's let i be x1 and let v, the voltage across the capacitor, be x2. We seek an equation that uses di dt and i and/or v. Well conveniently because the capacitor is in parallel with the inductor and the voltage v is also across the inductor, we can use v = L di dt. All right, that takes care of one of the equations. Continuing this example, we next seek an equation that relates dv dt to v and/or i. It's helpful to define the current i sub C. Then we can write, i sub C = C dv dt. But that is not enough, because this equation does not relate dv dt to v and/or i. But to get that, we can use KCL at the top node. So here's the KCL equation. i s is equal to the sum of the three branch currents, v over R + i + C dv dt. Now we've achieved relating dv dt to i and v in the same equation, and we have two equations that do that. Our desired circuit variable is i so we want to eliminate v from the equation. So we're going to substitute v = L di dt in for v in this equation. And here I've done that directly. And notice that this becomes a second order derivative of i, so I just rewrite this a little bit more compactly. And the final step is to put this in standard form which means that you lead on the left with your highest order derivative, and you make it have a unity coefficient. Then you have the other terms that involve i, and then all the other terms are on the right-hand side. And then we're finished for this because we have the differential equation for i. Here's another example. In this case we have two inductors. They're not in parallel or in series so this is second order circuit. And our objective is to find the differential equation for i2 which is the current In the top inductor. We want to exploit v = L di dt, the vi characteristic for inductors. So we should try KVL. It will be helpful to define this current i, which by KCL is going to be equal to i1 + i2. Now we can do KVL around the left loop. So starting here we have- vs. Then we have i times R1. And then we have the voltage across the inductor which is L1 di1 dt = 0. And what I've done here is I have just substituted in for i. I've substituted i1 + i2 right there. Now let's do KVL around the right loop. Starting here we have -L1 di1 dt + L2, di2 dt, and then R2 times i2 = 0. We want to eliminate i1 so that we can end up with a differential equation for i2. So we look in our two equations and ask which one has the fewest i1 terms? It's this one. This one has just one i1 term. So I can solve for this term and then try to substitute it into this equation. But if I solve for this derivative, I can't directly substitute it into this equation, because I have i1 here. To make that possible, I can just simply differentiate this entire equation, and that's what I've done here. That's all I've done, see dvs dt and then the respective derivatives of i1 and i2. And then finally, this last term is a second-order derivative, but I've written it like this to preserve the di1 dt. So you can see where you have to substitute that. So we solved this equation for di1 dt, and we substituted in here and here. And it gives us this big equation. And it remains to put this differential equation because now we have the differential equations in terms of i2, but we need to put it into standard form, and here it is. So, to summarize this lesson, for a second order circuit, write a circuit equation for each capacitor voltage and for each inductor current. For an inductor, try to exploit v = L di dt by using KVL. For a capacitor, try to exploit i = C dv dt by using KCL. And then, if necessary, differentiate one of the circuit equations to enable substitution. Thank you. [MUSIC]