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Studying for a test? Prepare with these 3 lessons on Second order linear equations.
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Let's do another problem with repeated roots. So let's say our differential equation is the second derivative of y minus the first derivative plus 0.25-- that's what's written here-- 0.25y is equal to 0. And they've actually given us some initial conditions. They said that y of 0 is equal to 2, and y prime of 0 is equal to 1/3. So like we've done in every one of these constant coefficient linear second order homogeneous differential equations, let's get the characteristic equation. So that's r squared minus r plus 0.25-- or we can even call it plus 1/4. So let's see, when I just inspected this, it always confuses me when I have fractions. So it becomes very hard to factor. So let's just do the quadratic formula. So the roots of this are going to be r is equal to negative b. Well, b is negative 1. So negative b is going to be 1. Plus or minus the square root of b squared. b is negative 1. So that squared is 1. Minus 4 times a, which is 1, times c. Well, 4 times 1 times 0.25, that's 1. Ah-ha. So notice that when you have a repeated root, this under the square root becomes 0. And that makes sense, because it's this plus or minus in the quadratic formula that gives you two roots, whether they be real or complex. But if the square root is 0, you're adding plus or minus 0 and you're only left with one root. Anyway, we're not done yet. What's the denominator of a quadratic equation? 2a. So a is 1, so over 2. So our one repeated root is 1 plus or minus 0 over 2, or it equals 1/2. And like we learned in the last video, you might just say, oh well, maybe the solution is just y is equal to ce to the 1/2 x. But like we pointed out last time, you have two initial conditions. And this solution is not general enough for two initial conditions. And then last time, we said, OK, if this isn't general enough, maybe some solution that was some function of x times e to the 1/2 x, maybe that would be our solution. And we said, it turns out it is. And so that more general solution that we found, that we figured out that v of x is actually equal to some constant plus x times some other constant. So our more general solution is y is equal to c1 times e to the 1/2 x soon. plus c2 times xe to the 1/2 x. I forgot the x here. Let me draw a line here so you don't get confused. Anyway, that's the reasoning. That's how we came up with this thing. And it is good to know. Because later on when you want to know more theory of differential equations-- and that's really the whole point about learning this if your whole goal isn't just to pass an exam-- it's good to know. But when you're actually solving these you could just kind of know the template. If I have a repeated root, well I just put that repeated root twice, and one of them gets an x in front of it, and they have two constants. Anyway, this is our general solution and now we can use our initial conditions to solve for c1 and c2. So let's just figure out the derivative of this first. So it becomes easy to substitute in for c2. So y prime is equal to 1/2 c1 e to the 1/2 x, plus-- now this becomes a little bit more complicated, we're going to have to use the product rule here-- so plus c2 times-- derivative of x is 1-- times e to the 1/2 x, that's the product rule. Plus the derivative of e to the 1/2 x times x. So that's 1/2 xe to the 1/2 x. Or we can write-- I don't want to lose this stuff up here-- we can write that it equals-- let's see, I have 1/2-- so I have c2 times e to the 1/2 x and I have 1/2 times c1 e to the 1/2 x. So I could say, it's equal to e to the 1/2 x times c1 over 2. That's that. Plus c2. That takes care of these two terms. Plus c2 over 2 xe to the 1/2 x. And now let's use our initial conditions. And let me actually clear up some space, because I think it's nice to have our initial conditions up here where we can see them. So let me delete all this stuff here. That, hopefully, makes sense to you by now. You know the characteristic equation. We figured out the general solu-- I don't want to erase our initial conditions-- we figured out the general solution was this. I'll keep our general solution there. And so, now we just substitute our initial conditions into our general solution and the derivative of the general solution, and hopefully we can get meaningful answers. So substituting into our general solution, y of 0 is equal to 2. So y is equal to 2 when x is equal to 0. So c1-- when x is equal to 0, all the e terms you become 1, right? This one will become 1. And then notice, we have an xe to the 0. So now this x is 0. So this whole term is going to be equal to 0. So we're done. c1 is equal to 2. That was pretty straightforward. This x actually made it a lot easier. So c1 is equal to 2. And now we can use the derivative. So let's see, this is the first derivative. And I'll substitute c1 in there so we can just solve for c2. So our first derivative is y prime is equal to-- let's see c1-- 1/2 plus c2-- so it's-- well I'll write this first-- it's equal to 2 over 2. So it's 1 plus c2 times e to the 1/2 x plus c2 over 2 times xe to the 1/2 x. There was an x here. So when x is equal to 0, y prime is equal to 1/3. So 1/3 is equal to-- well, x is equal to 0, this'll be 1-- so it's equal to 1 plus c2. And then this term, when x is equal to 0, this whole thing becomes 0, right? Because this x just cancels out the whole thing. You multiply by 0 you get 0. So then we get 1/3 is equal to 1 plus c2, or that c2 is equal to 1/3 of minus 1 is equal to minus 2/3. And now we have our particular solution. Let me write it down and put a box around it. So this is our general solution. Our particular solution, given these initial conditions for this repeated root problem, is y is equal to c1-- we figured that out to be 2 fairly quickly-- 2e to the 1/2 x plus c2. c2 is minus 2/3. So minus 2/3 xe to the 1/2 x. And we are done. There is our particular solution. So once again, kind of the proof of how do you get to this. Why is there this x in there? And it wasn't a proof, it was really more of just to show you the intuition of where that came from. And it did introduce you to a method called, reduction of order, to figure out what that function v was, which ended up just being c1 plus c2 times x. But all that can be pretty complicated. But you see that once you know the pattern, or once you know that this is going to be the general solution, they're pretty easy to solve. Characteristic equation. Get your general solution. Figure out the derivative of the general solution. And then substitute your initial conditions to solve for your constants. And you're done. Anyway, I'll see you in the next video. And actually, we're going to start solving non-homogeneous differential equations. See