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Complex roots of the characteristic equations 3

Video transcript
Let's do a couple of problems where the roots of the characteristic equation are complex. And just as a little bit of a review, and we'll put this up here in the corner so that it's useful for us. We learned that if the roots of our characteristic equation are r is equal to lambda plus or minus mu i, that the general solution for our differential equation is y is equal to e to the lambda x times c1 with some constant cosine of mu x, plus c2 times sine of mu x. And with that said, let's do some problems. So let's see, this first one that our differential equation-- I'll do this one in blue-- our differential equation is the second derivative, y prime prime plus 4y prime plus 5y is equal to 0. And they actually give us some initial conditions. They say y of 0 is equal to 1, and y prime of 0 is equal to 0. So now we'll actually be able to figure out a particular solution, or the particular solution, for this differential equation. So let's write down the characteristic equation. So it's r squared plus 4r plus 5 is equal to 0. Let's break out our quadratic formula. The roots of this are going to be negative b. So minus 4 plus or minus the square root of b squared. So that's 16. Minus 4 times a-- well that's 1-- times 1, times c times 5. All of that over 2 times a. a is 1 so all of that over 2. And see this simplifies. This equals minus 4 plus or minus 16-- this is 20, right? 4 times 1 times 5 is 20. So 16 minus 20 is minus 4. Minus 4 over 2. And that equals minus 4 plus or minus 2i, right? Square root of minus 4 is 2i. All that over 2. And so our roots to the characteristic equation are minus 2-- just dividing both by 2-- minus 2 plus or minus, we could say i or 1i, right? So if we wanted to do some pattern match, if we just wanted to do it really fast, what's our lambda? Our lambda is just minus 1. Let me write that down. That's our lambda. What's our mu? Well mu is the coefficient on the i, so mu is 1. mu is equal to 1. And now we're ready to write down our general solution. So the general solution to this differential equation is y is equal to e to the lambda x-- well lambda is minus 2-- minus 2x times c1 cosine of mu x-- but mu is just 1-- so c1 cosine of x, plus c2 sine of mu x, when mu is 1, so sine of x. Fair enough. Now let's use our initial conditions to find the particular solution, or a particular solution. So, when x is 0, y is equal to 1. So y is equal to 1 when x is 0. So 1 is equal to-- let's substitute x is 0 here. So e to the minus 2 times 0, that's just 1. So this whole thing becomes 1, so we could just ignore it. It's just 1 times this thing. So I'll write that down. e to the 0 is 1 times c1 times cosine of 0, plus c2 times sine of 0. Now what's sine of 0? Sine of 0 is 0. So this whole term is going to be 0. Cosine of 0 is 1. So there, we already solved for c1. We get this, this is 1. So 1 times c1 times 1 is equal to 1. So we get our first coefficient. c1 is equal to 1. Now let's take the derivative of our general solution. And we could even substitute c1 in here, just so that we have to stop writing c1 all the time. And we can solve for c2. So right now we know that our general solution is y-- we could call this our pseudo-general solution, because we already solved for c1-- y is equal to e to the minus 2x times c1, but we know that c1 is 1, so I'll write times cosine of x, plus c2 times sine of x. Now let's take the derivative of this, so that we can use the second initial condition. So y prime is equal to-- we're going to have to do a little bit of product rule here. So what's the derivative of the first expression? It is minus 2e to the minus 2x. And we multiply that times the second expression. Cosine of x plus c2 sine of x. And then we add that to just the regular first expression. So plus e to the minus 2x times the derivative of the second expression. So what's the derivative of cosine of x? It's minus sine of x. And then what's the derivative of c2 sine of x? Well , it's plus c2 cosine of x. And let's see if we can do any kind of simplification here. Well actually, the easiest way, instead of trying to simplify it algebraically and everything, let's just use our initial condition. Our initial condition is y prime of 0 is equal to 0. Let me write that down. Second initial condition was y prime of 0 is equal to 0. So y prime, when x is equal to 0, is equal to 0. And let's substitute x is equal to 0 into this thing. We could have simplified this more, but let's not worry about that right now. So if x is 0, this is going to be 1, right? E to the 0. e to the 0 is 1, so we're left with just minus 2, right? Minus 2 times e to the 0, times cosine of 0, that's 1, plus c2 times sine of 0. Sine of 0 is 0. So that's just 1 plus 0, plus e to the minus 2 times 0. That's just 1. Times minus sine of 0. Sine of 0 is just 0. Plus c2 times cosine of 0. Cosine of 0 is 1. So plus c2. That simplified things, didn't it? So let's see, we get 0 is equal to-- this is just 1-- minus 2 plus c2. Or we get c2 is equal to 2. Add 2 to both sides. c2 is equal to 2. And then we have our particular solution. I know it's c2 is equal to 2, c1 is equal to 1. Actually, let me erase some of this, just so that we can go from our general solution to our particular solution. So we had figured out, you can remember, c1 is 1 and c2 is 2. That's easy to memorize. So I'll just delete all of this. I'll write it nice and big. So our particular solution, given these initial conditions, were, or are, or is y of x is equal to-- this was a general solution-- e to the minus 2x times-- we solved for c1, it's equal to 1. So we can just write cosine of x. And then we solved for c2. We figured out that that was 2. Plus 2 sine of x. And there you go. We have our particular solution to this-- sorry where did I write it-- to this differential equation with these initial conditions. And what's neat is, when we originally kind of proved this formula-- when we originally showed this formula-- we had all of these i's and we simplified. We said c2, it was a combination of some other constants times some i's. And we said, oh we don't know whether they're imaginary or not, so let's just merge them into some constant. But what's interesting is this particular solution has no i's anywhere in it. And so, that tells us a couple of neat things. One that, if we had kept this c2 in terms of some multiple of i's, and our constants actually would have had i's and they would have canceled out, et cetera. And it also tells us that this formula is useful beyond formulas that just involve imaginary numbers. For example, this differential equation, I don't see an i anywhere here. I don't see an i anywhere here. And I don't see an i anywhere here. But given this differential equation, to get to the solution, we had to use imaginary numbers in between. And I think this is the first time-- if I'm remembering all my playlists correctly-- this is the first time that we used imaginary numbers for something useful. We used it as an intermediary tool where we got a real, a non-imaginary solution, to a real problem, a non-imaginary problem. But we used imaginary numbers as a tool to solve it. So, hopefully, you found that slightly interesting. And I'll see you in the next video.