Repeated roots of the characteristic equation

Let's say we have the following second order differential equation. We have second derivative of y, plus 4 times the first derivative, plus 4y is equal to 0. And we're asked to find the general solution to this differential equation. So the first thing we do, like we've done in the last several videos, we'll get the characteristic equation. That's r squared plus 4r plus 4 is equal to 0. This one is fairly easy to factor. We don't need the quadratic equation here. This is r plus 2 times r plus 2. And now something interesting happens, something that we haven't seen before. The two roots of our characteristic equation are actually the same number, r is equal to minus 2. So you could say we only have one solution, or one root, or a repeated root. However you want to say it, we only have one r that satisfies the characteristic equation. You might say, well that's fine. Maybe my general solution is just y is equal to some constant times e to the minus 2x, using my one solution. And my reply to you is, this is a solution. And if you don't believe me you can test it out. But it's not the general solution. And why do I say that? Because this is a second order differential equation. And if someone wanted a particular solution, they would have to give you two initial conditions. The two initial conditions we've been using so far are, what y of 0 equals, and what y prime of 0 equals. They could give you what y of 5 equals, who knows. But in general, when you have a second order differential equation, they have to give you two initial conditions. Now the problem with this solution, and why it's not the general solution, is if you use one of these initial conditions, you can solve for a c, right? You'll get an answer. You'll solve for that c. But then there's nothing to do with the second initial condition. In fact, except for only in one particular case, whatever c you get for the first initial condition, it won't be that-- this equation won't be true for the second initial condition. And you could try it out. I mean, if we said, y of 0 is equal to A and y prime of 0 is equal to 5A. Let's see if these work. If y of 0 is equal to A, that tells us that A is equal to c times e to the minus 2 times 0. So e to the 0. Or c is equal to A, right? So if you just had this first initial condition, say fine, my particular solution is y is equal to A times e to the minus 2x. Let's see if this particular solution satisfies the second initial condition. So what is the derivative of this? y prime is equal to minus 2Ae to the minus 2x. And it says that 5A-- this initial condition says that 5A is equal to minus 2A times e to the minus 2 times 0, so e to the 0. Or another way of saying that, e to the 0 is just 1. It says that 5A is equal to minus 2A, which we know is not true. So note, when we only have this general, or pseudo-general, solution, it can only satisfy, generally, one of the initial conditions. And if we're really lucky, both initial conditions. So that at least gives you an intuitive feel of why this isn't the general solution. So let me clean that up a little bit so that I-- I have a feeling I'll have to use this real estate. So what do we do? We can use a technique called reduction of order. And it really just says, well let's just guess a second solution. In general when we first thought about these linear constant coefficient differential equations, we said, well e to rx might be a good guess. And why is that? Because all of the derivatives of e are kind of multiples of the original function, and that's why we used it. So if we're looking for a second solution, it doesn't hurt to kind of make the same guess. And in order be a little bit more general, let's make our guess for our second solution-- I'll call this g for guess-- let's say it's some function of x times our first solution, e to the minus 2x. I could say some function of x times c times e to the minus 2x, but the c is kind of encapsulated. It could be part of this some random function of x. So let's be as general as possible. So let's assume that this is a solution and then substitute it back into our original differential equation, and see if we can actually solve for v that makes it all work. So before we do that, let's get its first and second derivatives. So the first derivative of g is equal to-- well this is product rule. And I'll drop the v of x, we know that v is a function and not a constant. So, product rule, derivative of the first, v prime times the second expression, e to the minus 2x, plus the first function, or expression, times the derivative of the second. So minus 2 times e to the minus 2x. Or, just to write it a little bit neater, g prime is equal to v prime e to the minus 2x minus 2ve to the minus 2x. Now we have to get the second derivative. I'll do it in a different color, just to fight the monotony of it. So the second derivative-- we're going to have to do the product rule twice-- derivative of this first expression. It's going to be v prime prime e to the minus 2x, minus 2v prime e to the minus 2x. That was just the product rule again. And then the derivative of the second expression is going to be-- let's see, derivative of the first one is v prime-- so it's going to be minus 2v prime e to the minus 2x, plus 4ve to the minus 2x. I hope I haven't made a careless mistake. And we can simplify this a little bit. So we get the second derivative of g, which is our guess solution, is equal to the second derivative of v prime, e to the minus 2x minus 2v prime-- no, minus 4, sorry, because we have minus 2, minus 2-- minus 4v prime e to the minus 2x, plus 4ve to the minus 2x. And now, before we substitute it into this, we can just make one observation. That will just make the algebra a little bit simpler. Notice that g is something times e to the minus 2x. G prime is-- we could factor out an e to the minus 2x. And g prime prime, we can factor out an e to the minus 2x. So let's factor them out, essentially. So when we write this, we can write-- so the second derivative is g prime prime, which we can write as-- and I'm going to try to do this-- it's e to the minus 2x times the second derivative. So now we can get rid of the e to the minus 2x terms. So that's v prime prime minus 4v prime plus 4v, right? If I just distribute this out I get the second derivative, which is this. Plus 4 times the first derivative. And I'm also going to factor out the e to the minus 2x. So, plus 4 times this. So it's going to be plus 4v prime minus 8v, right? Once again, I factored out the e to the minus 2x, right? Plus 4 times y. We factored out the e to the minus 2x-- so plus 4 times v. I did that, because if I didn't do that I'd be writing e to the minus 2x, and I'd probably make a careless mistake, and I'd run out of space, et cetera. But anyway, I essentially-- to get this, I just substituted the second derivative, the first derivative, and g back into the differential equation. And we know that that has to equal 0. And let's see if we can simplify this a little bit more. And then hopefully solve for v. So let's see, some things are popping out at me. So I have plus 4v plus 4v, that's plus 8v, minus 8v, right? So plus 4 minus 8 plus 4, those cancel out. It's plus 8 minus 8, those cancel out. And I also have minus 4v prime plus 4v prime. So those cancel out. And lo and behold, we've done some serious simplification. It ends up being e to the minus 2x times v prime prime-- we could call that v prime prime of x, now that we've saved so much space-- is equal to 0. We know this could never equal to 0. So, essentially, we have now established that this expression has to be equal to 0. And we get a separable second order differential equation. We get that the second derivative of v with respect to x-- or it's a function of x-- is equal to 0. So now we just have to differentiate both sides of this equation twice. You differentiate once, you get what? v prime of x is equal to, let's call it c1. And if we were to take the anti-derivative of both sides again, we get v of x is equal to c1 x plus some other c2, right? Now remember, what was our guess? Our guess was that our general solution was going to be some arbitrary function v times that first solution we found, e to the minus 2x. And when we actually took that guess and we substituted it in, we actually were able to solve for that v. And we got that v is equal to this. So this is interesting. So what is g, or what does our guess function equal? And it's no longer a guess. We've kind of established that it works. g, which we can call our solution, is equal to v of x, times e to the minus 2x. Well, that equals this, c1 x plus c2 e to the minus 2x. That equals c1 xe to the minus 2x, plus c2 e to the minus 2x. And now we have a truly general solution. We have two constants, so we can satisfy two initial conditions. And if you were looking for a pattern, this is the pattern. When you have a repeated root of your characteristic equation, the general solution is going to be-- you're going to use that e to the, that whatever root is, twice. But one time you're going to have an x in front of it. And this works every time for second order homogeneous constant coefficient linear equations. I will see you in the next video.