 Welcome back. We're at lecture 36. We should finish up the supplement today, so I don't know if you want to continue to carry this around, whether you have this or the copy that I made, but we'll be back in the textbook tomorrow in chapter 8, so we should finish these today. Remind me when we start. We're going to do a couple of kind of sample problems or example problems that we kind of put a clock on how long it takes us to do that problem. That'll give me a clue about do I include one of these all the way through on the test or you know if it takes us 27 minutes to do one problem it probably is not a test worthy kind of problem and just remind me when we start that we kind of put a stopwatch on it and see how long it takes us. We were in the middle of a problem but it was a question that Nicole asked yesterday. She and I worked after class and got that figured out. Is there anybody that needed to go beyond where we were in that problem? We had the first particular solution which was linear, negative half x minus a fourth I think was the right. It was the way that went and then we were about to embark on the particular solution for sine of 2x or something like that. Procedure is kind of the same as we found out yesterday. It's very easy to when you're doubling or tripling equations and putting an equation with C1 and C2 together with another equation with C1 and C2 to make some arithmetic errors. So you do have to be kind of careful when you're trying to solve after you found the homogeneous solution, the first particular solution, the second particular solution and then you go back to the initial conditions and try to solve for C1 and C2. It can be a little bit cumbersome arithmetically. Anybody need to see that second part of the particular solution? All right well let's go to electric circuits as we encountered before. Some first-order linear equations with trying to model electric circuits. I confess to you that I'm not an expert by any means. I can once we get it kind of translated from the electric circuit problem to the second-order differential equation problem I feel myself giving a sigh of relief that we now have it into a form that I feel comfortable with. So it's a matter of kind of translating what Kirchhoff's voltage drops law says. Add them up. Set them equal to the electromotive force or the voltage that's being supplied and then as I said before breathe the sigh of relief and then we've got it in a form that we have already worked with so it should be the same thing we've been doing. So we're going to have these kind of things on our electric circuit. These things right here are the voltage drops across this particular circuit and we will take the sum of the voltage drops and set them equal to the basically the electromotive force whatever it is. The voltage whatever it is in the problem. So we're going to have things such as resistors. The unit will be in ohms. Looks like a capital omega. We'll have an inductor. The unit will be in Henry's. We'll have a capacitor. The unit will be in Faraday's. You can probably kind of assign some of this to who did some research and came up with a lot of this stuff in electricity. The charge in fact we want a second order differential equation in terms of charge will be Q. We don't have that here yet. We have Q in this term. We don't have Q in this term and we don't have Q in this term so we're going to have to kind of doctor it up slightly from the kind of taking the voltage drops across this particular circuit and get it in terms of Q which is the charge. So the charge is Q and the current is I. So when you see I it's a current. Now how are current and charge related? Current is derivative of Q with respect to T and that gets us from what appears to be kind of a first order differential equation now to a second order. So these are the three things we want to deal with in this particular circuit. We want to add them up. There's a reason why this term was written first because the derivative of I with respect to T while I itself is the current it's already the derivative of Q the charge with respect to T. So if I is already a first derivative what's the derivative of this first derivative? That's the second derivative of Q so DI over DT since I is already DQ over DT derivative of I would be second derivative of Q with respect to T. I which is right there is a first derivative. It is derivative of Q with respect to T and if you'll factor out this division by C bring it out in front as 1 over C then that will be our Q. So here's our second derivative of Q term. Here's our first derivative of Q term and here's our Q term itself. So this is what our equation is going to look like. Now once we get to that it's the same old stuff that we've been doing in 7.8 and 7.9 thus far but we have a second order linear differential equation in terms of Q which is charge. So I'm sure that seems a lot more comfortable to a lot of you in here than it does to me but I am relieved when we get everything plugged in and we get to this point in an electric circuit problem. So this thing on the right is what they refer to as the electromotive force, battery, generator, whatever it happens to be it goes on the right side then we no longer have a homogeneous differential equation. We have a non-homogeneous differential equation. So let's look at an example from the supplement and another example that is one of the web-assigned problems. I think it's web-assigned problem 5 from this section and I kind of want to see how long it takes us to get through all this with the initial conditions as well as see what these problems are like. Once you get them set up they're not that different from what we've already done. This example from the supplement the R value is 40 ohms. The L value, let me go ahead and put the symbolism here, is 1 that makes our lead coefficient 1 which is kind of helpful on this problem and our C value is 16 times 10 to the negative fourth. Now it's not Faraday's, it's Farad's, Farad's right but that comes from Faraday and the E of t which is going to be out there on the right side the so-called electromotive force is that. So you're going to be supplied these things if we put them in the right places as coefficients in the equation then we're in business we can go from there. On this example problem the initial charge is zero, charge is q so the q of zero we're going to have an equation q of t when we're done so the q of zero is zero the initial current is also zero so either right at that way or the q prime of zero is zero same thing so the q of zero is zero the q prime of zero is zero well those will be at the end of the problem we do have one additional thing in this problem that will analyze what happens as t approaches infinity so we get a so-called steady state what happens way out to the right as t approaches zero you'll see one of the terms kind of eventually disappear all right so our equation is l one second derivative of q with respect to t that worked out nicely what's our next one what's our first derivative what's our q prime or dq 40 dq over dd somebody start a watch here so we can kind of keep track now that we know the numbers how long a problem like this is going to take us uh the last term is q over c or if you bring it out in front it's one over c so 16 times 10 to the negative fourth ought to be that does that seem odd on the right side that that's our so-called electromotive force is that could that be i have a question about the left side can you like for visual whatever can you just put q double yes i will on the next yes is that i mean does anything really kind of electrically look like that alternating currents right do in fact have equations that look like that so we're going to our second example is just 12 volts and we'll throw a 12 over there but this is actually an alternating current an oscillating current all right so one so we've got q double prime we've got 40 q prime one over 0.0016 is what 625 is that right all right there's my sigh of relief i'm at a place where i feel real comfortable on this problem now because we're kind of putting the entries and the ferrets and the ohms behind us and now we have a second order differential equation non-homogeneous what's the path from here right homogeneous solution so q sub h we've got an equation in terms of q double prime q prime and q so we'll solve it for a q of t so our homogeneous solution what's the characteristic equation look like all right factors probably not so negative b plus or minus b squared minus four times a times c all over twice a all right what do we have under the radical there 1600 minus 2600 2500 so what do we have under there negative 900 square root of negative 900 30 i so what do we have negative 20 plus or minus 15 i does that look good so our homogeneous solution what's it look like two c one kind of 15 t so that's going to be part of our answer second part particular solution it's going to be a joyful looking solution in it with 15 t's and now what 10 t's how are we going to generate a certain amount of in this case 100 cosine of 10 t's well we don't know how many to start with so it's just a we don't know how many cosine of 10 t's to start with right and then b sine of 10 t so we need sines and cosines in the q position so that we can generate cosines and sines in the q prime position and we can generate back to our original sines and cosines in the q double prime and we want the what sine of 10 t's to eventually drop out right we want to be left with 100 cosine of 10 t's first derivative plus 10 b derivative again all right we plug those things in to the left side and then we set it equal to cos 100 cosine of 10 t so for q double prime anything ridiculously difficult to this point in time in the problem just kind of cumbersome and lengthy right i think anything's outrageously difficult so there is q double prime we had that in there once we have what 40 q primes and we have 625 of the original q things and when we do that with our particular solution one of those 40 of those 625 of those we should get 100 cosine of 10 t so what are the different kinds of terms that we're going to generate on the left side cosine of 10 t's and you can do this other ways i'm just going to kind of gather up what we have so let's take our q double prime term which we have one of how many cosine of 10 t's negative 100 a minus 100 b of those is that correct how many from the q prime term what is that negative 400 a of them and 400 b of these and now the other one what 625 a of those and 625 b of those so our coefficients for cosine of 10 t what is our kind of final coefficient there equals 100 equals 100 is that correct because that's how many we have over here we have a hundred of them and our other coefficient is what minus 100 no 400 a that's our coefficient of sine we don't have any sine of 10 t's over here so that must be zero all right any recommendations there to eliminate yeah let's give substitution rather than work with those coefficients so if we said from the second equation 520 we're not going to get a very delightful answer on this in fact none of the answers are things that we would like to be working with but it's kind of what we're handed 525 b equals 400 a divide by 400 so we can substitute that in for a in this equation so 525 a becomes 525 times so this might be time to switch to the calculator to get our a and b values i don't know if i have these written down or not i don't get coefficients of b add those coefficients together anybody else zero nine two i don't remember that value but that's possible that i don't remember that value okay they didn't give them in decimals they gave them in fractions with a denominator of 697 okay so we'll stick with decimals yeah 64 over 697 decimals are fine because that's kind of annoying at that point to try to get everything with the common denominator and go from there all right now we have b we have an equation that interrelates a and b so if we take 525 over 400 times b that should give us a 1205 thank you so there's a and b so let's write our solution and then we'll go to the initial conditions so q of t what was the homogeneous part so we add to that the particular solution which is what what was a the coefficient of cosine 10 t and b was the coefficient of sine of 10 t everybody okay to that point where are we on the time on the clock 15 minutes okay now what do we have to do we have initial conditions so q of zero this ought to go pretty quick so the initial charge is zero so q of zero is zero so we get a zero on the left side what do we get out here in front when t is zero okay that's one cosine of 15 zero is cosine of zero so that's one just c1 this has sine of zero in it so that's gone point 1205 cosine of zero which would be and this has sine of zero which is gone right so that's not bad c1 is negative point 1205 that'd be nice if the next one went that quickly it won't because it is what first derivative so it's q prime of where is the information q prime of zero is zero so that's the other initial condition so we need q prime of t that looks like fun so where's our q of t it is the first piece the homogeneous piece is a product so it is first times derivative of second what's derivative of c1 cosine 15 t and derivative of c2 sine 15 t all right so that is first times derivative of second now we need second times derivative of first what's derivative of e to the negative 20 t all right so that's the only product rule we have the rest are individual terms what's the derivative of point 1205 cosine 10 t and what's the derivative of point 092 sine of 10 t everybody feel reasonably confident with all those numbers and coefficients so we're supposed to put in zero for every t and it's supposed to kick out zero on the left side so e to the negative 20 times zero is just one so what do we get out of this one if t is zero zero out of that one because it's sine of zero 15 c2 out of that one out in front of the next term i've actually got it at the end we're going to have a negative 20 times one is that right and then what for this one c1 and what for this one zero this one has a sine of zero in it so it's gone this one has a cosine of zero so that's point nine two okay so we already have c1 right what do we get for c1 negative point 1205 so take all the numerical stuff to the left side keep all the c2 stuff on the right side divide by the coefficient of c2 negative 1.28 got a couple of those positive well we've got this was negative and this was also negative so this was positive but when we moved it over here it became negative right and then we've got a positive point nine two which got moved to negative so we've got a negative divided by positive so it should be negative point two two yeah repeating which would be what two nights we needed it okay so let's write our final solution what 20 minutes later so q of t equals e to the negative 20t c1 cosine 15t plus c2 what was c2 negative point two two sine of 15t plus point one two oh five cosine 10t yeah let me see how those check out because some of those in the supplement here we're given as fractions our 10s and 15s look good what's negative 84 divided by 697 negative 205 okay and what is negative 464 divided by 2091 negative 222 okay so we're okay because they gave exact fractions we're okay with these decimals all right now i said we're going to do one more thing that we normally don't do at the end of this problem is that is look at what happens as t approaches infinity with our final answer so what what happens to this kind of lengthy equation with these odd coefficients and all that's present as t approaches infinity can you tell do we kind of lose part of this equation the first part why do we lose the first part because doesn't e to the negative 20t as t approaches infinity this piece gets smaller and smaller and smaller right because it's one over e to the 20t and if t gets larger that denominator gets huge in a hurry so this term right here goes to zero which eliminates everything that it's multiplied by as t approaches infinity so the steady state as t approaches infinity q of t takes on equation because the first part kind of eventually disappears for large values of t all right well that tells me something that 20 minutes on one problem even though it's our first problem is probably not a fair test question so what i could do which um could have you do the homogeneous part the particular part on one problem similar to this and possibly on another problem give you those give you some initial conditions that maybe are a little bit kinder than these decimal values and have you take it from there to figure out the final result i'll have to mess with that but 20 minutes on one problem is probably too long all right second example this is a web assign problem so you might want to jot down at least kind of the different stages that we get to from the supplement this is problem 12 but as far as web assign this is your problem five on the web assign a series circuit contains a resistor with r equal 24 ohms an inductor with l equals two inries and a capacitor with c equal point 005 ferrids and a 12 volt battery so it's a little different on the right side the initial charge is point 001 and the initial current is zero find the charge and the current by the way i think we didn't completely finish that other problem because the directions said to find the charge and the current at time t and we finished with q of t which is the charge right what would we need to do with that q of t to get the current take its derivative so we had the answer so we take its derivative and we have the other piece to the solution all right so let's fill in our numbers um r is 24 l is two that's our lead coefficient r is 20 no 24 and c which we really want one over c one over that is 200 now this is a little different because we just have 12 volts so we're just going to put a 12 over there now the nice thing about the 12 over there is the particular solution because we have a constant therefore what is our particular solution it's a constant some random constant a and it's going to make that piece really easy so a particular solution of this will go real quickly all right so two q double prime plus 24 q prime plus that was 200 equals 12 so the homogeneous part could reduce everything by two right you might think of factors of 100 that are going to give us minus 12 or plus 12 in the middle i think we need to waste our time on that there's probably a good reason of why we didn't need to waste our time on that what's under the radical what's 144 minus 416 minus so the square root of negative 256 is 16 i so we get negative 6 plus or minus 8 i alpha is negative 6 beta is 8 sorry that's not right yes so what's our homogeneous solution look like okay we look at the right side what do we have to generate we have to generate a constant so we start with some generic constant it's about as generic as you can get not really a lot of real interesting stuff here the derivative first derivative of the unknown constant is zero the second derivative is also zero so the left side of the equation is two q double prime plus 24 q prime plus 200 q so that works out nicely with the 12 volts on the right side of the equation 200 a equals 12 so what is a so our final solution is the homogeneous part i guess not final but at least the two pieces thus far right that's it it's a constant so we just kind of it's not like it's a coefficient of anything it's just some arbitrary constant um let's see this is problem 12 here the initial charge q equals point 001 let's see what that tells us so we're going to use our solution so on the left side we're going to get point 001 everywhere we see a t on the right side we're going to put in zero so that is one that's going to be c1 that's going to be gone right and that's going to be plus point 06 so c1 is whatever that arithmetic gives and our other information that's given the initial current which is derivative so q prime of zero is zero so we need q prime so the first piece is a product first times derivative of second that should be the derivative of the second is that look right plus second times derivative of first and then the derivative of point 06 the derivative of the constant is zero so left side we're going to get zero right side every time we see a t we put in zero so that's e to the zero which is one negative eight c1 sign that's zero this is going to be what eight c2 that look right if t is zero here we've got negative six e to the zero so that's just negative six here we have c1 and this one is gone we already know that c1 is that so move that to the other side zero four four two five that all right so i think now we have a final answer q of t e to the negative six t c1 cosine of eight t plus c2 sine of eight t plus our constant that we figured out earlier that one was a little less time q of p was a little quite a bit simpler what was that 10 minutes okay so probably it's about a 15 minute problem a normal problem that we go through again all right so that should finish us from this you do have some web assign problems from this there's one of them already done for you we will continue back in the regular textbook starting tomorrow have a good afternoon