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Laplace as linear operator and Laplace of derivatives Laplace transform Khan Academy.mp3 | So say some constant, c1 times my first function, f of t, plus some constant, c2, times my second function, g of t. Well, by the definition of the Laplace transform, this would be equal to the improper integral from 0 to infinity of e to the minus st times whatever our function that we're taking the Laplace transform o... |
Laplace as linear operator and Laplace of derivatives Laplace transform Khan Academy.mp3 | That is equal to c1 e to the minus st f of t plus c2 e to the minus st g of t. And all of that times dt. And just by the definition of how, or the properties of integrals work, we know that we can split this up into two integrals, right? If the integral of the sum of two functions is equal to the sum of their integrals... |
Laplace as linear operator and Laplace of derivatives Laplace transform Khan Academy.mp3 | And these are just constants. So this is going to be equal to c1 times the integral from 0 to infinity of e to the minus st times f of t, d of t, plus c2 times the integral from 0 to infinity of e to the minus st g of t, dt. And this was just a very long-winded way of saying, well, what is this? This is the Laplace tra... |
Laplace as linear operator and Laplace of derivatives Laplace transform Khan Academy.mp3 | This is the Laplace transform of f of t. This is the Laplace transform of g of t. So this is equal to c1 times the Laplace transform of f of t plus c2 times, this is the Laplace transform, the Laplace transform of g of t. And so we have just shown that the Laplace transform is a linear operator, right? The Laplace tran... |
Laplace as linear operator and Laplace of derivatives Laplace transform Khan Academy.mp3 | That's something useful to know, and you might have guessed that that was the case anyway, but now you know for sure. Now we'll do something which I consider even more interesting. And this is actually going to be a big clue as to why Laplace transforms are extremely useful for solving differential equations. So let's ... |
Laplace as linear operator and Laplace of derivatives Laplace transform Khan Academy.mp3 | So let's say I wanted to find the Laplace transform of f prime of t. So I have some f of t. I take its derivative, and then I want the Laplace transform of that. Let's see if we can find a relationship between the Laplace transform of the derivative of a function and the Laplace transform of the function. So we're goin... |
Laplace as linear operator and Laplace of derivatives Laplace transform Khan Academy.mp3 | So let's just put the integration. Well, actually, let me just say what this is. First of all, this is equal to the integral from 0 to infinity of e to the minus st times f prime of t dt. And to solve this, we're going to use integration by parts. Let me write it in the corner just so you remember what it is. So I thin... |
Laplace as linear operator and Laplace of derivatives Laplace transform Khan Academy.mp3 | And to solve this, we're going to use integration by parts. Let me write it in the corner just so you remember what it is. So I think I memorized it because I recorded that last video not too long ago. That the integral of u v prime, because that will match what we have up here better, is equal to both functions withou... |
Laplace as linear operator and Laplace of derivatives Laplace transform Khan Academy.mp3 | That the integral of u v prime, because that will match what we have up here better, is equal to both functions without the derivatives, u v, minus the integral of the opposite. So the opposite is u prime v. So here, the substitution is pretty clear, right? Because we want to end up with f of x, right? So let's make v ... |
Laplace as linear operator and Laplace of derivatives Laplace transform Khan Academy.mp3 | So let's make v prime is f prime. And let's make u e to the minus st. So let's do that. u is going to be e to the minus st. And v is going to equal what? v is going to equal f prime of t. And then u prime would be minus s e to the minus st. And then v prime is f prime of t. So v is just going to be equal to f of t. Hop... |
Laplace as linear operator and Laplace of derivatives Laplace transform Khan Academy.mp3 | u is going to be e to the minus st. And v is going to equal what? v is going to equal f prime of t. And then u prime would be minus s e to the minus st. And then v prime is f prime of t. So v is just going to be equal to f of t. Hope I didn't say that wrong the first time. But you see what I'm saying. This is u. That's... |
Laplace as linear operator and Laplace of derivatives Laplace transform Khan Academy.mp3 | This is u. That's u. And this is v prime. And if this is v prime, then if you take the antiderivative of both sides, then v is equal to f of t. So let's apply integration by parts. So this Laplace transform, which is this, is equal to u v, which is equal to e to the minus st. This is equal to e to the minus st times v.... |
Laplace as linear operator and Laplace of derivatives Laplace transform Khan Academy.mp3 | And if this is v prime, then if you take the antiderivative of both sides, then v is equal to f of t. So let's apply integration by parts. So this Laplace transform, which is this, is equal to u v, which is equal to e to the minus st. This is equal to e to the minus st times v. f of t minus the integral. And of course,... |
Laplace as linear operator and Laplace of derivatives Laplace transform Khan Academy.mp3 | And of course, we're going to have to evaluate this from 0 to infinity. I'll keep the improper integral with us the whole time, I won't switch back and forth between the definite and indefinite integral. So minus this part. So the integral from 0 to infinity of u prime. u prime is minus se to the minus st times v. v is... |
Laplace as linear operator and Laplace of derivatives Laplace transform Khan Academy.mp3 | So the integral from 0 to infinity of u prime. u prime is minus se to the minus st times v. v is f of t dt. Let's see, we have a minus and a minus. Let's make both of these pluses. This s is just a constant, so we can bring it out. So that is equal to e to the minus st f of t, evaluated from 0 to infinity, or as we app... |
Laplace as linear operator and Laplace of derivatives Laplace transform Khan Academy.mp3 | Let's make both of these pluses. This s is just a constant, so we can bring it out. So that is equal to e to the minus st f of t, evaluated from 0 to infinity, or as we approach infinity, plus s times the integral from 0 to infinity of e to the minus st f of t dt. And here we see, what is this? This is the Laplace tran... |
Laplace as linear operator and Laplace of derivatives Laplace transform Khan Academy.mp3 | And here we see, what is this? This is the Laplace transform of f of t, right? So this is equal to, let's evaluate this part. So when we evaluate at infinity, as we approach infinity, e to the minus infinity approaches 0. f of infinity, now this is an interesting question. f of infinity, I don't know, that could be lar... |
Laplace as linear operator and Laplace of derivatives Laplace transform Khan Academy.mp3 | So when we evaluate at infinity, as we approach infinity, e to the minus infinity approaches 0. f of infinity, now this is an interesting question. f of infinity, I don't know, that could be large, that could be small, that could approach some, that approaches some value, right? This approaches 0, so we're not sure. If... |
Laplace as linear operator and Laplace of derivatives Laplace transform Khan Academy.mp3 | If this increases faster than this approaches 0, then this will diverge. I won't go into the mathematics of whether this converges or diverges, but let's just say, in very rough terms, that this will converge to 0 if f of t grows slower than e to the minus st shrinks. And maybe later on we'll do some more rigorous defi... |
Laplace as linear operator and Laplace of derivatives Laplace transform Khan Academy.mp3 | But let's assume that f of t grows slower than e to the st, or it converges, or it grows slower than this, or it diverges slower than this converges is another way to view it, or this grows slower than this shrinks. So if this grows slower than this shrinks, then this whole expression will approach 0, and then you want... |
Laplace as linear operator and Laplace of derivatives Laplace transform Khan Academy.mp3 | So that's just f of 0 plus s times, we said this is the Laplace transform of f of t. That's our definition. So the Laplace transform of f of t. And now we have an interesting property. What was the left-hand side of everything we were doing? The Laplace transform of f prime of t. So let me just write it all over again.... |
Laplace as linear operator and Laplace of derivatives Laplace transform Khan Academy.mp3 | The Laplace transform of f prime of t. So let me just write it all over again. So we now know that, and I'll switch colors, the Laplace transform of f prime of t is equal to s times the Laplace transform of f of t minus f of 0. And now let's just extend this further. What is the Laplace transform? And this is a really ... |
Laplace as linear operator and Laplace of derivatives Laplace transform Khan Academy.mp3 | What is the Laplace transform? And this is a really useful thing to know. What is the Laplace transform of f prime of t? Well, we could do a little pattern matching here, right? That's going to be s times the Laplace transform of its antiderivative times the Laplace transform of f prime of t, right? This goes to this. ... |
Laplace as linear operator and Laplace of derivatives Laplace transform Khan Academy.mp3 | Well, we could do a little pattern matching here, right? That's going to be s times the Laplace transform of its antiderivative times the Laplace transform of f prime of t, right? This goes to this. That's an antiderivative. This goes to this. That's one antiderivative. Minus f prime of 0, right? |
Laplace as linear operator and Laplace of derivatives Laplace transform Khan Academy.mp3 | That's an antiderivative. This goes to this. That's one antiderivative. Minus f prime of 0, right? But then what's the Laplace transform of this? This is going to be equal to s times the Laplace transform of f prime of t, but what's that? That's this, right? |
Laplace as linear operator and Laplace of derivatives Laplace transform Khan Academy.mp3 | Minus f prime of 0, right? But then what's the Laplace transform of this? This is going to be equal to s times the Laplace transform of f prime of t, but what's that? That's this, right? That's s times the Laplace transform of f of t minus f of 0, right? I just substituted this with this. Minus f prime of 0. |
Laplace as linear operator and Laplace of derivatives Laplace transform Khan Academy.mp3 | That's this, right? That's s times the Laplace transform of f of t minus f of 0, right? I just substituted this with this. Minus f prime of 0. And we get the Laplace transform of the second derivative is equal to s squared times the Laplace transform of our function, f of t, minus s times f of 0 minus f prime of 0. And... |
Laplace as linear operator and Laplace of derivatives Laplace transform Khan Academy.mp3 | Minus f prime of 0. And we get the Laplace transform of the second derivative is equal to s squared times the Laplace transform of our function, f of t, minus s times f of 0 minus f prime of 0. And I think you're starting to see a pattern here. This is the Laplace transform of f prime prime of t. And I think you're sta... |
Exact equations example 3 First order differential equations Khan Academy.mp3 | One, trying to figure out whether the equations are exact. And then if you know they're exact, how do you figure out the xi and figure out the solution of the differential equation? So the next one in my book is 3x squared minus 2xy plus 2 times dx plus 6y squared minus x squared plus 3 times dy is equal to 0. So just ... |
Exact equations example 3 First order differential equations Khan Academy.mp3 | So just the way it was written, this isn't superficially in that form that we want, right? What's the form that we want? We want some function of x and y plus another function of x and y times y prime or dy dx is equal to 0. We're close. How could we get this equation into this form? We just divide both sides of this e... |
Exact equations example 3 First order differential equations Khan Academy.mp3 | We're close. How could we get this equation into this form? We just divide both sides of this equation by dx, right? And then we get 3x squared minus 2xy plus 2. We're dividing by dx so that dx just becomes a 1. Plus 6y squared minus x squared plus 3. And then we're dividing by dx so that becomes dy dx is equal to what... |
Exact equations example 3 First order differential equations Khan Academy.mp3 | And then we get 3x squared minus 2xy plus 2. We're dividing by dx so that dx just becomes a 1. Plus 6y squared minus x squared plus 3. And then we're dividing by dx so that becomes dy dx is equal to what's 0 divided by dx? Well, it's just 0. And there we have it. We have written this in the form that we need, in this f... |
Exact equations example 3 First order differential equations Khan Academy.mp3 | And then we're dividing by dx so that becomes dy dx is equal to what's 0 divided by dx? Well, it's just 0. And there we have it. We have written this in the form that we need, in this form, and now we need to prove to ourselves that this is an exact equation. So let's do that. So what's the partial of m? This is the m ... |
Exact equations example 3 First order differential equations Khan Academy.mp3 | We have written this in the form that we need, in this form, and now we need to prove to ourselves that this is an exact equation. So let's do that. So what's the partial of m? This is the m function, right? This is the m function. What's the, this was a plus here. What's the partial of this with respect to y? |
Exact equations example 3 First order differential equations Khan Academy.mp3 | This is the m function, right? This is the m function. What's the, this was a plus here. What's the partial of this with respect to y? This would be 0. This would be minus 2x and then just a 2. So that's the partial of this with respect to y is minus 2x. |
Exact equations example 3 First order differential equations Khan Academy.mp3 | What's the partial of this with respect to y? This would be 0. This would be minus 2x and then just a 2. So that's the partial of this with respect to y is minus 2x. What's the partial of n with respect to x? This would be 0. This would be minus 2x. |
Exact equations example 3 First order differential equations Khan Academy.mp3 | So that's the partial of this with respect to y is minus 2x. What's the partial of n with respect to x? This would be 0. This would be minus 2x. So there you have it. The partial of m with respect to y is equal to the partial of n with respect to x. My is equal to nx. |
Exact equations example 3 First order differential equations Khan Academy.mp3 | This would be minus 2x. So there you have it. The partial of m with respect to y is equal to the partial of n with respect to x. My is equal to nx. So we are dealing with an exact equation. So now we have to find xi. The partial of xi with respect to x is equal to m, which is equal to 3x squared minus 2xy plus 2. |
Exact equations example 3 First order differential equations Khan Academy.mp3 | My is equal to nx. So we are dealing with an exact equation. So now we have to find xi. The partial of xi with respect to x is equal to m, which is equal to 3x squared minus 2xy plus 2. Take the antiderivative with respect to x on both sides and you get xi is equal to x to the third minus x squared y, because y is just... |
Exact equations example 3 First order differential equations Khan Academy.mp3 | The partial of xi with respect to x is equal to m, which is equal to 3x squared minus 2xy plus 2. Take the antiderivative with respect to x on both sides and you get xi is equal to x to the third minus x squared y, because y is just a constant, plus 2x plus some function of y, right? Because we know xi is a function of... |
Exact equations example 3 First order differential equations Khan Academy.mp3 | So when you take a derivative, when you take a partial with respect to just x, a pure function of just y would get lost. So it's like the constant when we first learned taking antiderivatives. And now to figure out xi, we just have to solve for h of y. And how do we do that? Well, let's take the partial of xi with resp... |
Exact equations example 3 First order differential equations Khan Academy.mp3 | And how do we do that? Well, let's take the partial of xi with respect to y. And that's going to be equal to this right here. So the partial of xi with respect to y, this is 0, this is minus x squared, so it's minus x squared, this is 0, plus h prime of y is going to be equal to what? That's going to be equal to our n ... |
Exact equations example 3 First order differential equations Khan Academy.mp3 | So the partial of xi with respect to y, this is 0, this is minus x squared, so it's minus x squared, this is 0, plus h prime of y is going to be equal to what? That's going to be equal to our n of xy. It's going to be equal to this. And then we can solve for this. So that's going to be equal to 6y squared minus x squar... |
Exact equations example 3 First order differential equations Khan Academy.mp3 | And then we can solve for this. So that's going to be equal to 6y squared minus x squared plus 3. Can add x squared to both sides to get rid of this and this. And then we're left with h prime of y is equal to 6y squared plus 3. Antiderivative. So h of y is equal to 2y cubed plus 3y. And you could put a plus c there, bu... |
Exact equations example 3 First order differential equations Khan Academy.mp3 | And then we're left with h prime of y is equal to 6y squared plus 3. Antiderivative. So h of y is equal to 2y cubed plus 3y. And you could put a plus c there, but the plus c merges later on when we solve the differential equation, so you don't have to worry about it too much. So what is our function xi? I'll write it i... |
Exact equations example 3 First order differential equations Khan Academy.mp3 | And you could put a plus c there, but the plus c merges later on when we solve the differential equation, so you don't have to worry about it too much. So what is our function xi? I'll write it in a new color. Our function xi as a function of x and y is equal to x to the third minus x squared y plus 2x plus h of y, whi... |
Exact equations example 3 First order differential equations Khan Academy.mp3 | Our function xi as a function of x and y is equal to x to the third minus x squared y plus 2x plus h of y, which we just solved for. So h of y is plus 2y to the third plus 3y. And then there could be a plus c there, but you'll see that it doesn't matter much. Actually, I want to do something a little bit different. I'm... |
Exact equations example 3 First order differential equations Khan Academy.mp3 | Actually, I want to do something a little bit different. I'm not just going to chug through the problem. I want to kind of go back to the intuition, because I don't want this to be completely mechanical. Let me just show you what the derivative, using what we knew before you even learned anything about the partial deri... |
Exact equations example 3 First order differential equations Khan Academy.mp3 | Let me just show you what the derivative, using what we knew before you even learned anything about the partial derivative chain rule. What is the derivative of xi with respect to x? Here we just use our implicit differentiation skills. So the derivative of this, I'll do it in a new color, is 3x squared minus, now we'r... |
Exact equations example 3 First order differential equations Khan Academy.mp3 | So the derivative of this, I'll do it in a new color, is 3x squared minus, now we're going to have to use the chain rule here, so the derivative of the first expression with respect to x is, well, let me just put the minus sign and I can put like that, so it's 2x times y plus the first function, x squared times the der... |
Exact equations example 3 First order differential equations Khan Academy.mp3 | Fair enough. Plus, the derivative of this with respect to x is easy, 2. Plus, the derivative of this with respect to x, well, let's take the derivative of this with respect to y first. We're just doing implicit differentiation in the chain rule. So this is plus 6y squared, and then we're using the chain rule, so we too... |
Exact equations example 3 First order differential equations Khan Academy.mp3 | We're just doing implicit differentiation in the chain rule. So this is plus 6y squared, and then we're using the chain rule, so we took the derivative with respect to y, then you have to multiply that times the derivative of y with respect to x, which is just y prime. Plus, the derivative of this with respect to y is ... |
Exact equations example 3 First order differential equations Khan Academy.mp3 | And let's try to see if we can simplify this. So we get this is equal to 3x squared minus 2xy plus 2, so that's this term, this term, and this term, plus, let's say, let's just put the y prime outside, y prime times, see, I have a negative sign out here, minus x squared plus 6y squared plus 3. So this is the derivative... |
Exact equations example 3 First order differential equations Khan Academy.mp3 | And notice, look at this closely, and notice that that is the same, hopefully it's the same, as our original problem. What was our original problem that we started working with? The original problem was 3x squared minus 2xy plus 2 plus 6y squared minus x squared plus 3 times y prime is equal to 0. So this was our origi... |
Exact equations example 3 First order differential equations Khan Academy.mp3 | So this was our original problem. And notice that the derivative of xi with respect to x, just using implicit differentiation, is exactly this. So hopefully this gives you a little intuition of why we can just rewrite this equation as the derivative with respect to x of xi, which is a function of x and y, is equal to 0... |
Exact equations example 3 First order differential equations Khan Academy.mp3 | Because this is the derivative of xi with respect to x. I wrote it out here. It's the same thing. It's right here. So that equals 0. So if we take the antiderivative of both sides, we know that the solution of this differential equation is that xi of x and y is equal to c is the solution. And we know what xi is, so we ... |
Exact equations example 3 First order differential equations Khan Academy.mp3 | So that equals 0. So if we take the antiderivative of both sides, we know that the solution of this differential equation is that xi of x and y is equal to c is the solution. And we know what xi is, so we just set that equal to c, and we have the implicit, we have a solution to the differential equation, all this defin... |
Exact equations example 3 First order differential equations Khan Academy.mp3 | So the solution, you don't have to do this every time. This step right here, you wouldn't have to do if you were taking a test, unless the teacher explicitly asked for it. I just wanted to kind of make sure that you know what you're doing, that you're not just doing things completely mechanically. That you really see t... |
Exact equations example 3 First order differential equations Khan Academy.mp3 | That you really see that the derivative of xi really does give you, like we solved for xi, and I just want to show that the derivative of xi with respect to x, just using implicit differentiation and our standard chain rule, actually gives you the left-hand side of the differential equation, which was our original prob... |
Repeated roots of the characteristic equations part 2 Khan Academy.mp3 | That's what's written here. 0.25 y 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 third. So like we've done in every one of these constant coefficient linear second order homogenous differential equations, let's get the character... |
Repeated roots of the characteristic equations part 2 Khan Academy.mp3 | They said that y of 0 is equal to 2, and y prime of 0 is equal to 1 third. So like we've done in every one of these constant coefficient linear second order homogenous differential equations, let's get the characteristic equation. So that's r squared minus r plus 0.25, or we could even call it plus 1 fourth. So let's s... |
Repeated roots of the characteristic equations part 2 Khan Academy.mp3 | So let's see. When I'm just inspecting 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 squ... |
Repeated roots of the characteristic equations part 2 Khan Academy.mp3 | 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. So notice that when you have a repeated root... |
Repeated roots of the characteristic equations part 2 Khan Academy.mp3 | 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. So notice that when you have a repeated root, this under the square root becomes 0. That makes sense, because this plus or minus in the quadratic formula gives you two roots, whether they be real or comple... |
Repeated roots of the characteristic equations part 2 Khan Academy.mp3 | That makes sense, because this plus or minus in the quadratic formula 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 the quadratic equation? 2a. So a is 1 over 2. ... |
Repeated roots of the characteristic equations part 2 Khan Academy.mp3 | 2a. So a is 1 over 2. So our one repeated root is 1 plus or minus 0 over 2, or it equals 1 half. And like we learned in the last video, you might just say, oh, well, maybe the solution is just y is equal to c e to the 1 half x. But like we pointed out last time, you have two initial conditions, and this solution is not... |
Repeated roots of the characteristic equations part 2 Khan Academy.mp3 | And like we learned in the last video, you might just say, oh, well, maybe the solution is just y is equal to c e to the 1 half 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... |
Repeated roots of the characteristic equations part 2 Khan Academy.mp3 | 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 half x plus c2 times x e to the 1 half x. I forgot the x here. Let me draw ... |
Repeated roots of the characteristic equations part 2 Khan Academy.mp3 | 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 ex... |
Repeated roots of the characteristic equations part 2 Khan Academy.mp3 | 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 rep... |
Repeated roots of the characteristic equations part 2 Khan Academy.mp3 | 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. But 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 eas... |
Repeated roots of the characteristic equations part 2 Khan Academy.mp3 | So y prime is equal to 1 half c1 e to the 1 half x plus. Now this becomes a little bit more complicated. We're going to have to use the product rule here. So let's see. Plus c2 times derivative of x is 1 times e to the 1 half x. That's the product rule. Plus the derivative of e to the 1 half x times x. |
Repeated roots of the characteristic equations part 2 Khan Academy.mp3 | So let's see. Plus c2 times derivative of x is 1 times e to the 1 half x. That's the product rule. Plus the derivative of e to the 1 half x times x. So that's 1 half x e to the 1 half 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 half, so I have c2 times e... |
Repeated roots of the characteristic equations part 2 Khan Academy.mp3 | Plus the derivative of e to the 1 half x times x. So that's 1 half x e to the 1 half 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 half, so I have c2 times e to the 1 half x, and I have 1 half times c1 e to the 1 half x. So I could say it's equal to e to t... |
Repeated roots of the characteristic equations part 2 Khan Academy.mp3 | I have 1 half, so I have c2 times e to the 1 half x, and I have 1 half times c1 e to the 1 half x. So I could say it's equal to e to the 1 half x times c1 over 2, that's that, plus c2, that takes care of these two terms, plus c2 over 2 x e to the 1 half x. And now let's use our initial conditions. Let me actually clear... |
Repeated roots of the characteristic equations part 2 Khan Academy.mp3 | Let me actually clear up some space, because I think it's nice to have our initial conditions up here, so we can see them. So let me delete all this stuff here that hopefully makes sense to you by now. We know the characteristic equation, we figured out the general solution. I don't want to erase our initial conditions... |
Repeated roots of the characteristic equations part 2 Khan Academy.mp3 | 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 o... |
Repeated roots of the characteristic equations part 2 Khan Academy.mp3 | 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 b... |
Repeated roots of the characteristic equations part 2 Khan Academy.mp3 | So c1, when x is equal to 0, all the e terms become 1, right? This one will become 1. And then notice, we have an x e 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. |
Repeated roots of the characteristic equations part 2 Khan Academy.mp3 | 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. |
Repeated roots of the characteristic equations part 2 Khan Academy.mp3 | 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. So let's see. And I'll substitute c1 in there so we can just solve for c2. |
Repeated roots of the characteristic equations part 2 Khan Academy.mp3 | And now we can use the derivative. So let's see, this is the first derivative. So let's see. 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 half 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 ... |
Repeated roots of the characteristic equations part 2 Khan Academy.mp3 | 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 half 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 half x plus c2 over 2 times x e to the 1 half x. There was an x here. So when x is e... |
Repeated roots of the characteristic equations part 2 Khan Academy.mp3 | It's equal to 2 over 2, so it's 1 plus c2 times e to the 1 half x plus c2 over 2 times x e to the 1 half x. There was an x here. So when x is equal to 0, y prime is equal to 1 third. So 1 third is equal to, well, x is equal to 0, this will be 1, so it's equal to 1 plus c2. And then this term, when x is equal to 0, this... |
Repeated roots of the characteristic equations part 2 Khan Academy.mp3 | So 1 third is equal to, well, x is equal to 0, this will 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 it by 0, you get 0. So then we get 1 third is equal to 1 plus c2, or that c2 is equal to 1... |
Repeated roots of the characteristic equations part 2 Khan Academy.mp3 | You multiply it by 0, you get 0. So then we get 1 third is equal to 1 plus c2, or that c2 is equal to 1 third minus 1 is equal to minus 2 thirds. And now we have our particular solution. Let me write it down and put a box around it. So this was our general solution, our particular solution, given these initial conditio... |
Repeated roots of the characteristic equations part 2 Khan Academy.mp3 | Let me write it down and put a box around it. So this was 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. 2 e to the 1 half x plus c2. c2 is minus 2 thirds. So minus 2 thirds x e to the 1 half ... |
Repeated roots of the characteristic equations part 2 Khan Academy.mp3 | 2 e to the 1 half x plus c2. c2 is minus 2 thirds. So minus 2 thirds x e to the 1 half 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. |
Repeated roots of the characteristic equations part 2 Khan Academy.mp3 | 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 w... |
Repeated roots of the characteristic equations part 2 Khan Academy.mp3 | 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 of that can be pretty complicated. But you see that once you k... |
2nd Order Linear Homogeneous Differential Equations 4 Khan Academy.mp3 | Let's solve another second order linear homogenous differential equation. And this one, well, I won't give you the details before I actually write it down. So the differential equation is 4 times the second derivative of y with respect to x minus 8 times the first derivative plus 3 times the function times y is equal t... |
2nd Order Linear Homogeneous Differential Equations 4 Khan Academy.mp3 | And we have our initial conditions. y of 0 is equal to 2. And we have y prime of 0 is equal to 1 half. Now, I could go into the whole thing, y is equal to e to the rx as a solution, substitute it in, then factor out e to the rx and have the characteristic equation. And if you want to see all of that over again, you mig... |
2nd Order Linear Homogeneous Differential Equations 4 Khan Academy.mp3 | Now, I could go into the whole thing, y is equal to e to the rx as a solution, substitute it in, then factor out e to the rx and have the characteristic equation. And if you want to see all of that over again, you might want to watch the previous video just to see where that characteristic equation comes from. But in t... |
2nd Order Linear Homogeneous Differential Equations 4 Khan Academy.mp3 | So if this is our original differential equation, the characteristic equation is going to be, and I'll do this in a different color, 4r squared minus 8r plus 3r is equal to 0. And watch the previous video if you don't know where this characteristic equation comes from. But if you want to do these problems really quick,... |
2nd Order Linear Homogeneous Differential Equations 4 Khan Academy.mp3 | And then the coefficient on the original function is just the constant, right? I think you see what I did. Second derivative, r squared, first derivative, r, no derivative, you could say that's r to the 0 or just 1, but this is our characteristic equation. And now we can just figure out its roots. This is not a trivial... |
2nd Order Linear Homogeneous Differential Equations 4 Khan Academy.mp3 | And now we can just figure out its roots. This is not a trivial one for me to factor, so if it's not trivial, you can use the quadratic equation. So we could say the solution of this is r is equal to negative b, well, b is negative 8, so it's positive 8, 8 plus or minus the square root of b squared, so that's 64, minus... |
2nd Order Linear Homogeneous Differential Equations 4 Khan Academy.mp3 | 2 times 4 is 8. That equals 8 plus or minus square root of 64 minus, what's 16 times 3 minus 48, all of that over 8. What's 64 minus 48? Let's see. 12, it's 16, right? 10 plus 48 is 58, then another, so it's 16. So we have r is equal to 8 plus or minus the square root of 16 over 8 is equal to 8 plus or minus 4 over 8. |
2nd Order Linear Homogeneous Differential Equations 4 Khan Academy.mp3 | Let's see. 12, it's 16, right? 10 plus 48 is 58, then another, so it's 16. So we have r is equal to 8 plus or minus the square root of 16 over 8 is equal to 8 plus or minus 4 over 8. That equals 1 plus or minus 1 half. So the two solutions of this characteristic equation, and ignore that, let me scratch that out in bla... |
2nd Order Linear Homogeneous Differential Equations 4 Khan Academy.mp3 | So we have r is equal to 8 plus or minus the square root of 16 over 8 is equal to 8 plus or minus 4 over 8. That equals 1 plus or minus 1 half. So the two solutions of this characteristic equation, and ignore that, let me scratch that out in black so that you know that that's not like a 30 or something. The two solutio... |
2nd Order Linear Homogeneous Differential Equations 4 Khan Academy.mp3 | The two solutions of this characteristic equation are r is equal to, well, 1 plus 1 half is equal to 3 halves, and r is equal to 1 minus 1 half is equal to 1 half. So we know our two r's, and we know that from previous experience in the last video that y is equal to c times e to the rx is a solution. So the general sol... |
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