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Undetermined coefficients 3 Second order differential equations Khan Academy.mp3 | That's minus 13 over 2. So minus 13 over 2 is equal to 4c. Or c, divide both sides by 4, and then you get c is equal to minus 13 over 8. And I think I haven't made a careless mistake. So if I haven't, then our particular solution we now know. And actually, let's just write it in common. Well, let me write the whole sol... |
Undetermined coefficients 3 Second order differential equations Khan Academy.mp3 | And I think I haven't made a careless mistake. So if I haven't, then our particular solution we now know. And actually, let's just write it in common. Well, let me write the whole solution. So, and this is a nice stretch of horizontal real estate. So let's write our solution. Our solution is going to be equal to the pa... |
Undetermined coefficients 3 Second order differential equations Khan Academy.mp3 | Well, let me write the whole solution. So, and this is a nice stretch of horizontal real estate. So let's write our solution. Our solution is going to be equal to the particular solution, which is ax squared. So that's minus 1x squared. So minus 1x squared, right? ax squared plus bx plus 3 halves x plus c. Minus 13 ove... |
Undetermined coefficients 3 Second order differential equations Khan Academy.mp3 | Our solution is going to be equal to the particular solution, which is ax squared. So that's minus 1x squared. So minus 1x squared, right? ax squared plus bx plus 3 halves x plus c. Minus 13 over 8. So this is the particular solution. We solved for a, b, and c. We determined the undetermined coefficients. And now if we... |
Undetermined coefficients 3 Second order differential equations Khan Academy.mp3 | ax squared plus bx plus 3 halves x plus c. Minus 13 over 8. So this is the particular solution. We solved for a, b, and c. We determined the undetermined coefficients. And now if we want the general solution, we add to that the general solution of the homogenous equation. What was that? y prime minus 3y prime minus 4y ... |
Undetermined coefficients 3 Second order differential equations Khan Academy.mp3 | And now if we want the general solution, we add to that the general solution of the homogenous equation. What was that? y prime minus 3y prime minus 4y is equal to 0. And we've solved this multiple times. We know that the general solution of the homogenous equation is c1 e to the 4x plus c2 e to the minus x. Right? You... |
Undetermined coefficients 3 Second order differential equations Khan Academy.mp3 | And we've solved this multiple times. We know that the general solution of the homogenous equation is c1 e to the 4x plus c2 e to the minus x. Right? You just take the characteristic equation, r squared minus 3r minus 4. You get, what did you get? You get r minus 4 times r plus 1, and then that's how you get minus 1 an... |
Undetermined coefficients 3 Second order differential equations Khan Academy.mp3 | You just take the characteristic equation, r squared minus 3r minus 4. You get, what did you get? You get r minus 4 times r plus 1, and then that's how you get minus 1 and 4. Anyway, so this is the general solution to the homogenous equation. This is a particular solution to the non-homogenous equation. The general sol... |
Undetermined coefficients 3 Second order differential equations Khan Academy.mp3 | Anyway, so this is the general solution to the homogenous equation. This is a particular solution to the non-homogenous equation. The general solution to the homogenous equation is going to be the sum of the two. So let's add that. So plus c1 e to the 4x plus c2 e to the minus x. So there you go. I don't think that was... |
Undetermined coefficients 3 Second order differential equations Khan Academy.mp3 | So let's add that. So plus c1 e to the 4x plus c2 e to the minus x. So there you go. I don't think that was too painful. The most painful part was just making sure that you don't make a careless mistake with the algebra. But using a fairly straightforward, really algebraic technique, we were able to get a fairly fancy ... |
First order homogenous equations First order differential equations Khan Academy.mp3 | But the application here, at least I don't see the connection. Homogenous differential equation. And even within differential equations, we'll learn later, there's a different type of homogenous differential equation, and those are called homogenous linear differential equations, but they mean something actually quite ... |
First order homogenous equations First order differential equations Khan Academy.mp3 | But anyway, for this purposes, I'm going to show you homogenous differential equations, and what we're dealing with are going to be first order equations. And what does a homogenous differential equation mean? Well, let's say I had just a regular first order differential equation that could be written like this. So dy ... |
First order homogenous equations First order differential equations Khan Academy.mp3 | So dy dx is equal to some function of x and y. And let's say we try to do this, and it's not separable, and it's not exact. What we learn is that if it can be homogenous, if this is a homogenous differential equation, that we can make a variable substitution, and that variable substitution allows this equation to turn ... |
First order homogenous equations First order differential equations Khan Academy.mp3 | But before I need to show you that, I need to tell you, well what does it mean to be homogenous? Well, if I can algebraically manipulate this right side of this equation so that I can actually rewrite it, instead of a function of x and y, if I could rewrite this differential equation so that dy dx is equal to some func... |
First order homogenous equations First order differential equations Khan Academy.mp3 | Let me show you an example. So let's say that, and I'll just show you the example, show you it's homogenous, and then we'll just do the substitution. So let's say that my differential equation is d, the derivative of y with respect to x, is equal to x plus y over x. And you can, if you'd like, you can try to separate, ... |
First order homogenous equations First order differential equations Khan Academy.mp3 | And you can, if you'd like, you can try to separate, make this a separable, but it's not that trivial to solve, or at least I'm looking at an inspection, it doesn't seem that trivial to solve. And as we see right here, we have the derivative, it's equal to some function of x and y. And my question is to you, can I just... |
First order homogenous equations First order differential equations Khan Academy.mp3 | Well sure, if we just divide both of these top terms by x, right, this is the same thing as x over x plus y over x. This equation is the same thing as dy over dx is equal to this, which is the same thing as rewriting this whole equation, I'm going to switch colors arbitrarily, as this. dy over dx is equal to x divided ... |
First order homogenous equations First order differential equations Khan Academy.mp3 | So you're probably wondering, what did I mean by a function of y over x? Well you can see it here. When I just algebraically manipulated this equation, I got 1 plus y over x. So if I said that y over x is equal to some third variable, this is just a function of that third variable. And actually I'm going to do that rig... |
First order homogenous equations First order differential equations Khan Academy.mp3 | So if I said that y over x is equal to some third variable, this is just a function of that third variable. And actually I'm going to do that right now. So let's say, let's make a substitution for y over x. So let's say that v, and I'll do v in a different color, let's say that v is equal to y over x, or another way, i... |
First order homogenous equations First order differential equations Khan Academy.mp3 | So let's say that v, and I'll do v in a different color, let's say that v is equal to y over x, or another way, if you just multiply both sides by x, you could write that y is equal to xv. And we're going to substitute v for y over x, but we're also going to have to substitute dy over dx. So let's figure out what that ... |
First order homogenous equations First order differential equations Khan Academy.mp3 | Well if we assume that v is also a function of x, then we're just going to use the product rule. So the derivative of x is 1 times v, plus x times the derivative of v with respect to x. And now we can substitute this and this back into this equation, and we get. So dy over dx, that is equal to this. So we get v plus x ... |
First order homogenous equations First order differential equations Khan Academy.mp3 | So dy over dx, that is equal to this. So we get v plus x dv dx, derivative of v with respect to x, is equal to, that's just the left hand side, it's equal to 1 plus y over x, but we're making this substitution that v is equal to y over x, is equal to 1 plus v. And now this should be pretty straightforward. So let's see... |
First order homogenous equations First order differential equations Khan Academy.mp3 | And then what do we have left? We have x dv dx is equal to 1. Let's divide both sides by x, and we get the derivative of v with respect to x is equal to 1 over x. It should maybe start becoming a little bit clear what the solution here is, but let's just keep going forward. So if we multiply both sides by dx, we get dv... |
First order homogenous equations First order differential equations Khan Academy.mp3 | It should maybe start becoming a little bit clear what the solution here is, but let's just keep going forward. So if we multiply both sides by dx, we get dv is equal to 1 over x times dx. Now we can take the antiderivative of both sides, integrate both sides, and we're left with v is equal to the natural log of the ab... |
First order homogenous equations First order differential equations Khan Academy.mp3 | So let's do that. What was v? We made the substitution that v is equal to y over x, so let's reverse substitute it now, or unsubstitute it. So we get y over x is equal to the natural log of x plus c, some constant. Multiply both sides times x, and you get y is equal to x times the natural log of x plus c. And we're don... |
First order homogenous equations First order differential equations Khan Academy.mp3 | So we get y over x is equal to the natural log of x plus c, some constant. Multiply both sides times x, and you get y is equal to x times the natural log of x plus c. And we're done. We solved that seemingly inseparable differential equation by recognizing that it was homogenous, and making that variable substitution v... |
First order homogenous equations First order differential equations Khan Academy.mp3 | That turned it into a separable equation in terms of v, and then we solved it, and then we unsubstituted it back, and we got the solution to the differential equation. You can verify it for yourself that y is equal to the x natural log of the absolute value of x plus c. Oh, actually I made a mistake. y over x is equal ... |
The convolution and the laplace transform Laplace transform Khan Academy.mp3 | Now that you've had a little bit of exposure to what a convolution is, I can introduce you to the convolution theorem, or at least the convolution theorem, or at least in the context of, there may be other convolution theorems, but we're talking about differential equations in Laplace transforms. So this is the convolu... |
The convolution and the laplace transform Laplace transform Khan Academy.mp3 | And if I have another function, g of t, and I take its Laplace transform, that of course is capital G of s, that if we were to convolute these two functions, so if I were to take f and I were to convolute it with g, which is going to be another function of t, and we already saw this, we saw that in the last video, I co... |
The convolution and the laplace transform Laplace transform Khan Academy.mp3 | Actually, let me write one more thing. If this is true, then we could also do it the other way. We could also say that F, and I'll just do it all in yellow, it takes me too much time to keep switching colors, that the convolution of F and G, which is just a function of t, I could just say it's the inverse Laplace trans... |
The convolution and the laplace transform Laplace transform Khan Academy.mp3 | F of s times G of s. There you go. Now, what good does all of this do? Well, we can take inverse Laplace transforms. Let's just say that I had, let me write it down here. Let's say I gave you, I told you that the following expression or function, let's say H of s is equal to 2s over s squared plus 1. We did this long d... |
The convolution and the laplace transform Laplace transform Khan Academy.mp3 | Let's just say that I had, let me write it down here. Let's say I gave you, I told you that the following expression or function, let's say H of s is equal to 2s over s squared plus 1. We did this long differential equation at the end, we end up with this thing, and we have to take the inverse Laplace transform of it. ... |
The convolution and the laplace transform Laplace transform Khan Academy.mp3 | We want to figure out the inverse Laplace transform of H of s, or the inverse Laplace transform of this thing right there. We want to figure out the inverse Laplace transform of this expression right here, 2s over s squared plus 1 squared. Don't want to lose that right there. Now, can we write this as a product of two ... |
The convolution and the laplace transform Laplace transform Khan Academy.mp3 | Now, can we write this as a product of two Laplace transforms that we do know? Let's try to do it. We can rewrite this, so this is the inverse Laplace transform. Let me rewrite this expression down here. I can rewrite 2s over s squared plus 1 squared. This is the same thing as, let me write it this way, 2 times 1 over ... |
The convolution and the laplace transform Laplace transform Khan Academy.mp3 | Let me rewrite this expression down here. I can rewrite 2s over s squared plus 1 squared. This is the same thing as, let me write it this way, 2 times 1 over s squared plus 1, times s over s squared plus 1. I just kind of broke it up. If you multiply the numerators here, you get 2 times 1 times s, or 2s. If you multipl... |
The convolution and the laplace transform Laplace transform Khan Academy.mp3 | I just kind of broke it up. If you multiply the numerators here, you get 2 times 1 times s, or 2s. If you multiply the denominators here, s squared plus 1 times s squared plus 1, that's just s squared plus 1 squared. This is the same thing. If we want to take the inverse Laplace transform of this, it's the same thing a... |
The convolution and the laplace transform Laplace transform Khan Academy.mp3 | This is the same thing. If we want to take the inverse Laplace transform of this, it's the same thing as taking the inverse Laplace transform of this right here. Now, this something should hopefully start popping out at you. If these were separate transforms, if they were on their own, we know what this is. If we call ... |
The convolution and the laplace transform Laplace transform Khan Academy.mp3 | If these were separate transforms, if they were on their own, we know what this is. If we call this f of s, if we say that this is a Laplace transform of some function, we know what that function is. This piece right here, I'm just doing a little dotted line around it, this is the Laplace transform of sine of t. Then, ... |
The convolution and the laplace transform Laplace transform Khan Academy.mp3 | This implies that g of t, if we define this as the Laplace transform of g, this means that g of t is equal to cosine of t. Of course, when you take the inverse Laplace transform, you could take the 2s out. You could say, now what can we say? We can now say that the inverse, or we could actually have a better thing to d... |
The convolution and the laplace transform Laplace transform Khan Academy.mp3 | I didn't want to leave that out and confuse the issue. I wanted a very pure f of s times g of s. This expression right here is the product of the Laplace transform of 2 sine of t and the Laplace transform of cosine of t. Now, our convolution theorem told us this right here, that if we want to take the inverse Laplace t... |
The convolution and the laplace transform Laplace transform Khan Academy.mp3 | This right here is the Laplace transform of 2 sine of t. This is the Laplace transform of cosine of t. We just wrote that as g of s and f of s. If I have an expression written like this, I can take the inverse Laplace transform and it will be equal to the convolution of the original functions. It will be equal to the c... |
The convolution and the laplace transform Laplace transform Khan Academy.mp3 | We know that f of t is equal to the inverse Laplace transform of f of s. We know that g of t is equal to the inverse Laplace transform of g of s. We could rewrite the convolution theorem as the inverse Laplace transform of, I'm going to try to stay true to the colors, of f of s times g of s is equal to, I'm just restat... |
The convolution and the laplace transform Laplace transform Khan Academy.mp3 | That's f of s in our example. The g of s was s over s squared plus 1. All I got that from is I just broke this up into two things that I recognize. If I multiply this together, I get back to my original thing that I was trying to take the inverse Laplace transform of. The convolution theorem just says that the inverse ... |
The convolution and the laplace transform Laplace transform Khan Academy.mp3 | If I multiply this together, I get back to my original thing that I was trying to take the inverse Laplace transform of. The convolution theorem just says that the inverse Laplace transform of this is equal to the inverse Laplace transform of 2 over s squared plus 1 convoluted with the inverse Laplace transform of our ... |
The convolution and the laplace transform Laplace transform Khan Academy.mp3 | I already told them to you, but they should be somewhat second nature now. This is 2 times sine of t. Take the Laplace transform of sine of t, you get 1 over s squared plus 1, and then you multiply it by 2, you get the 2 up there. You're going to have to convolute that with the inverse Laplace transform of this thing h... |
The convolution and the laplace transform Laplace transform Khan Academy.mp3 | We already went over this. This is cosine of t. Our result so far, let me be very clear, it's always good to take a step back and just think about what we're doing, much less why we're doing it. Let's see, the Laplace transform, the inverse Laplace transform of this thing up in this top left corner of 2s over s squared... |
The convolution and the laplace transform Laplace transform Khan Academy.mp3 | Actually, this should be a curly bracket right here, but you get the idea. It's equal to this. It's equal to 2 sine of t convoluted with cosine of t. You're like, Sal, throughout this whole process, I've already forgotten what it means to convolute two functions, so let's convolute them. I'll just write the definition,... |
The convolution and the laplace transform Laplace transform Khan Academy.mp3 | I'll just write the definition, or the definition we're using of the convolution, that f convoluted with g, it's going to be a function of t, I'll just write this shorthand, is equal to the integral from 0 to t of f of t minus tau times g of tau d tau. So 2 sine of t convoluted with cosine of t is equal to, so this is ... |
The convolution and the laplace transform Laplace transform Khan Academy.mp3 | If we take the 2 out, we get 2 times the integral from 0 to t of sine of t minus tau times the cosine of tau. I actually solved this in the previous video. This right here, this is a convolution of sine of t and cosine of t. It's sine of t convoluted with cosine of t. I show you in the previous video, just watch that v... |
The convolution and the laplace transform Laplace transform Khan Academy.mp3 | Those cancel out, so it equals t sine of t. So all of this mess, and once you get the hang of it, you won't have to go through all of these steps, but the key is to recognize that this could be broken down as a product of 2 Laplace transforms that you recognize. This could be broken down as a product of 2 Laplace trans... |
The convolution and the laplace transform Laplace transform Khan Academy.mp3 | And if you watched the previous video, you'll realize that actually calculating that convolution was no simple task, but it can be done. So you actually can get an integral form. Even if it can't be done, you can get your answer at least in terms of some integral. I haven't proven the convolution theorem to you just ye... |
Integrating factors 2 First order differential equations Khan Academy.mp3 | It was different than n with respect to x. We said, oh boy, it's not exact. But we said, what if we could multiply both sides of this equation by some function that would make it exact? And we called that mu. And in the last video, we actually solved for mu. We said, well, if we multiply both sides of this equation by ... |
Integrating factors 2 First order differential equations Khan Academy.mp3 | And we called that mu. And in the last video, we actually solved for mu. We said, well, if we multiply both sides of this equation by mu of x is equal to x, it should make this into an exact differential equation. It's important to note, there might have been a function of y that if I multiplied by both sides, it would... |
Integrating factors 2 First order differential equations Khan Academy.mp3 | It's important to note, there might have been a function of y that if I multiplied by both sides, it would also make it exact. There might have been a function of x and y that would have done the trick. But really, our whole goal is just to make this exact. So it doesn't matter which one we pick, which integrating fact... |
Integrating factors 2 First order differential equations Khan Academy.mp3 | So it doesn't matter which one we pick, which integrating factor, and this is called the integrating factor. Which integrating factor we pick. So anyway, let's do it now. Let's solve the problem. So let's multiply both sides of this equation by mu, and mu of x is just x. So we multiply both sides by x. So we get, see, ... |
Integrating factors 2 First order differential equations Khan Academy.mp3 | Let's solve the problem. So let's multiply both sides of this equation by mu, and mu of x is just x. So we multiply both sides by x. So we get, see, if you multiply this term by x, you get 3x squared y plus xy squared plus, we're multiplying these terms by x now, plus x to the third plus x squared y, y prime is equal t... |
Integrating factors 2 First order differential equations Khan Academy.mp3 | So we get, see, if you multiply this term by x, you get 3x squared y plus xy squared plus, we're multiplying these terms by x now, plus x to the third plus x squared y, y prime is equal to 0. Well now, first of all, just as a reality check, let's make sure that this is now an exact equation. So what's the partial of th... |
Integrating factors 2 First order differential equations Khan Academy.mp3 | Well, it's 3x squared, that's just a kind of constant coefficient of y, plus 2xy, that's the partial with respect to y of that expression. Now let's take the partial of this with respect to x. So we get 3x squared plus 2xy. And there we have it. The partial of this with respect to y is equal to the partial of this with... |
Integrating factors 2 First order differential equations Khan Academy.mp3 | And there we have it. The partial of this with respect to y is equal to the partial of this with respect to n. So we now have an exact equation whose solution should be the same as this. All we did is we multiplied both sides of this equation by x. So it really shouldn't change the solution of that equation or that dif... |
Integrating factors 2 First order differential equations Khan Academy.mp3 | So it really shouldn't change the solution of that equation or that differential equation. So it's exact. Let's solve it. So how do we do that? Well, what we say is, since we've shown this exact, we know that there's some function xi where the partial derivative of xi with respect to x is equal to this expression right... |
Integrating factors 2 First order differential equations Khan Academy.mp3 | So how do we do that? Well, what we say is, since we've shown this exact, we know that there's some function xi where the partial derivative of xi with respect to x is equal to this expression right here. So it's equal to 3x squared y plus xy squared. Let's take the antiderivative of both sides with respect to x, and w... |
Integrating factors 2 First order differential equations Khan Academy.mp3 | Let's take the antiderivative of both sides with respect to x, and we'll get xi is equal to what? Let's see, it's x to the third y plus, let's see, x, we could write 1 half x squared y squared. And of course, this xi is a function of x and y, so when you take the partial with respect to x, when you go that way, you mig... |
Integrating factors 2 First order differential equations Khan Academy.mp3 | So instead of a plus c here, there could have been a whole function of y that we lost. So we'll add that back when we take the antiderivative. So this is our xi, but we're not completely done yet because we have to somehow figure out what this function of y is. And the way we figure that out is we use the information t... |
Integrating factors 2 First order differential equations Khan Academy.mp3 | And the way we figure that out is we use the information that the partial of this with respect to y should be equal to this. So let's set that up. So what's the partial of this expression with respect to y? So it would be, so I could write xi, the partial of xi with respect to y is equal to, let's see, x to the third p... |
Integrating factors 2 First order differential equations Khan Academy.mp3 | So it would be, so I could write xi, the partial of xi with respect to y is equal to, let's see, x to the third plus, let's see, 2 times 1 half, so it's just x squared y plus h prime of y. That's the partial of a function purely of y with respect to y. And then that has to equal our new n, or the new expression we got ... |
Integrating factors 2 First order differential equations Khan Academy.mp3 | So that's going to be equal to this right here. This will hopefully be making sense to you at this point. So that should be equal to x to the third plus x squared y. And interesting enough, both of these terms are on this side, so let's subtract both of those terms from both sides. So x to the third, x to the third, x ... |
Integrating factors 2 First order differential equations Khan Academy.mp3 | And interesting enough, both of these terms are on this side, so let's subtract both of those terms from both sides. So x to the third, x to the third, x squared y, x squared y, and we're left with h prime of y is equal to 0. Or you could say that h of y is equal to some constant. And so there's really no y, I guess, e... |
Integrating factors 2 First order differential equations Khan Academy.mp3 | And so there's really no y, I guess, extra function of y. There's just some constant left over. So for our purposes, we can just say that xi is equal to this, because this is just a constant. We're going to take the antiderivative anyway and get a constant on the right-hand side. And in the previous videos, the constan... |
Integrating factors 2 First order differential equations Khan Academy.mp3 | We're going to take the antiderivative anyway and get a constant on the right-hand side. And in the previous videos, the constants all merged together. So we'll just assume that that is our xi. And we know that this differential equation up here can be rewritten as the derivative of xi with respect to x. And that just ... |
Integrating factors 2 First order differential equations Khan Academy.mp3 | And we know that this differential equation up here can be rewritten as the derivative of xi with respect to x. And that just falls out of the partial derivative chain rule. The derivative of xi with respect to x is equal to 0. If you took the derivative of xi with respect to x, it should be equal to this whole thing, ... |
Integrating factors 2 First order differential equations Khan Academy.mp3 | If you took the derivative of xi with respect to x, it should be equal to this whole thing, just using the partial derivative chain rule. Well, we know what xi is, so we can write. Or actually, we could use this fact to say, well, if we integrate both sides, that the solution of this differential equation is that xi is... |
Integrating factors 2 First order differential equations Khan Academy.mp3 | So a solution to the differential equation is xi is equal to c. So xi is equal to x to the third y plus 1 half x squared y squared. And we could have said plus c here, but we know the solution is that xi is equal to c, so we'll just write that there. I could have written a plus c here, but then you have a plus c here, ... |
Integrating factors 2 First order differential equations Khan Academy.mp3 | But anyway, there we have it. We had a differential equation that at least superficially looked exact. But then when we tested the exactness of it, it was not exact, but we multiplied it by an integrating factor. In the previous video, we figured out that a possible integrating factor is that we could just multiply bot... |
Integrating factors 2 First order differential equations Khan Academy.mp3 | In the previous video, we figured out that a possible integrating factor is that we could just multiply both sides by x. And when we did that, we tested it, and true enough, it was exact. And so given that it was exact, we knew that a xi would exist where the derivative of xi with respect to x would be equal to this en... |
Integrating factors 2 First order differential equations Khan Academy.mp3 | So we could rewrite our differential equation like this, and we'd know that a solution is xi is equal to c. And to solve for xi, we just say, OK, the partial derivative of xi with respect to x is going to be this thing, antiderivative of both sides. There's some constant h of y, not constant, there's some function of y... |
Worked example identifying separable equations AP Calculus AB Khan Academy.mp3 | which of the differential equations are separable. And I encourage you to pause this video and see which of these are actually separable. Now, the way that I approach this is I try to solve for the derivative, and if when I solve for the derivative, if I get dy dx is equal to some function of y times some other functio... |
Worked example identifying separable equations AP Calculus AB Khan Academy.mp3 | You would go from this first equation to the second equation just by dividing both sides by g of y and multiplying both sides by dx, and then it's clear you have a separable equation, you can integrate both sides. But the key is let's solve for the derivative and see if we can put this in a form where we have the produ... |
Worked example identifying separable equations AP Calculus AB Khan Academy.mp3 | So let's see, if I subtract y from both sides, I'm just trying to solve for the derivative of y with respect to x, I'm gonna get x times, I'll write y prime as the derivative of y with respect to x is equal to three minus y. So I subtracted y from both sides. Now, let's see, if I divide both sides by x, I'm gonna get t... |
Worked example identifying separable equations AP Calculus AB Khan Academy.mp3 | And so it's clear, I'm able to write the derivative as the product of a function of y and a function of x. So this indeed is separable. And I could show you, I can multiply both sides by dx, and I can divide both sides by three minus y now, and I would get one over three minus y dy is equal to one over x dx. So clearly... |
Worked example identifying separable equations AP Calculus AB Khan Academy.mp3 | So clearly, this one right over here is separable. Now let's do the second one. And I'm gonna just do the same technique, I'll do it in a different color so we don't get all of our math all jumbled together. So in this second one, let's see, if I subtract the two x, the two y from both sides, so actually, let me just d... |
Worked example identifying separable equations AP Calculus AB Khan Academy.mp3 | So in this second one, let's see, if I subtract the two x, the two y from both sides, so actually, let me just do, whoops, let me do a couple things at once. I'm gonna subtract two x from both sides. I am going to subtract two y from both sides. So I'm gonna subtract two y from both sides. I'm gonna add one to both sid... |
Worked example identifying separable equations AP Calculus AB Khan Academy.mp3 | So I'm gonna subtract two y from both sides. I'm gonna add one to both sides. So I'm gonna add one to both sides. And then what am I going to get if I do that? This is gonna be zero, this is gonna be zero, this is gonna be zero. I'm gonna have two times the derivative of y with respect to x is equal to negative two x m... |
Worked example identifying separable equations AP Calculus AB Khan Academy.mp3 | And then what am I going to get if I do that? This is gonna be zero, this is gonna be zero, this is gonna be zero. I'm gonna have two times the derivative of y with respect to x is equal to negative two x minus two y plus one. And now let's see, I can divide everything by two. I would get the derivative of y with respe... |
Worked example identifying separable equations AP Calculus AB Khan Academy.mp3 | And now let's see, I can divide everything by two. I would get the derivative of y with respect to x is equal to, and actually, yeah, I would get, I'm just gonna divide by two. So I'm gonna get negative x minus y and then I'm going to get plus 1 1 2. So it's not obvious to me how I can write this as a product of a func... |
Worked example identifying separable equations AP Calculus AB Khan Academy.mp3 | So it's not obvious to me how I can write this as a product of a function of x and a function of y. So this one does not feel, this one right over here is not separable. I don't know how to write this as a function of x times a function of y. So this one I'm gonna say is not separable. Now this one, they've already wri... |
Worked example identifying separable equations AP Calculus AB Khan Academy.mp3 | So this one I'm gonna say is not separable. Now this one, they've already written it for us as a function of x times a function of y. So this one is clearly separable right over here and if you want me to do the separating, I can rewrite this as, well, this is dy dx. If I multiply both sides by dx and divide both sides... |
Worked example identifying separable equations AP Calculus AB Khan Academy.mp3 | If I multiply both sides by dx and divide both sides by this right over here, I would get one over y squared plus y. Dy is equal to x squared plus x dx. So clearly separable. Alright, now this last choice, this is interesting. They've essentially distributed the derivative right over here. So let's see, if we were to u... |
Worked example identifying separable equations AP Calculus AB Khan Academy.mp3 | They've essentially distributed the derivative right over here. So let's see, if we were to unfactor the derivative, I'm just gonna solve for dy dx. So I'm gonna factor it out. I'm gonna get dy dx times x plus y, x plus y is equal to x. Now if I were to divide both sides by x plus y, I'm gonna get dy dx is equal to x o... |
Laplace transform of the unit step function Laplace transform Khan Academy.mp3 | I know everything we've done so far has really just been a toolkit of being able to solve them. But the whole reason is that because differential equations can describe a lot of systems and that we can actually model them that way. And we know that in the real world, everything isn't these nice, continuous functions. S... |
Laplace transform of the unit step function Laplace transform Khan Academy.mp3 | So over the next couple of videos, we're going to talk about functions that are a little bit more discontinuous than what you might be used to even in kind of your traditional calculus or traditional pre-calculus class. And the first one is the unit step function. Let's write it as u, and I'll put a little subscript c ... |
Laplace transform of the unit step function Laplace transform Khan Academy.mp3 | Let me draw my x-axis right here. I'll do a little thicker line. That's my x-axis right there. This is my y-axis right there. And when we talk about Laplace transforms, which we'll talk about shortly, we only care about t is greater than 0, because we saw our definition of a Laplace transform, we're always taking the i... |
Laplace transform of the unit step function Laplace transform Khan Academy.mp3 | This is my y-axis right there. And when we talk about Laplace transforms, which we'll talk about shortly, we only care about t is greater than 0, because we saw our definition of a Laplace transform, we're always taking the integral from 0 to infinity. So we're only dealing with the positive x-axis. But anyway, by this... |
Laplace transform of the unit step function Laplace transform Khan Academy.mp3 | But anyway, by this definition, it would be 0 all the way until you get to some value c. So you'd be 0 until you get to c. And then at c, you jump, and the point c is included. x is equal to c here. So it's included, so I'll put a dot there, because it's greater than or equal to c. You're at 1. So this is 1 right here.... |
Laplace transform of the unit step function Laplace transform Khan Academy.mp3 | So this is 1 right here. And then you go forward for all of time. And you're like, Sal, you just said that differential equations, we're going to model things. Why is this type of a function useful? Well, in the real world, sometimes you do have something that essentially jolts something, that moves it from this positi... |
Laplace transform of the unit step function Laplace transform Khan Academy.mp3 | Why is this type of a function useful? Well, in the real world, sometimes you do have something that essentially jolts something, that moves it from this position to that position. And obviously, nothing can move it immediately like this, but you might have some system. It could be an electrical system or a mechanical ... |
Laplace transform of the unit step function Laplace transform Khan Academy.mp3 | It could be an electrical system or a mechanical system, where maybe the behavior looks something like this, where maybe it moves it like that or something. And this function is a pretty good analytic approximation for some type of real world behavior like this, when something just gets moved. Whenever we solve these d... |
Laplace transform of the unit step function Laplace transform Khan Academy.mp3 | Eventually, we'll see that it doesn't perfectly describe things, but it helps describe it enough for us to get a sense of what's going to happen. Sometimes it will completely describe things, but anyway, we can ignore that for now. So let me get rid of these things right there. So the first question is, well, what if s... |
Laplace transform of the unit step function Laplace transform Khan Academy.mp3 | So the first question is, well, what if something doesn't jar just like that? What if I want to construct more fancy unit functions or more fancy step functions? Let's say I wanted to construct something that looked like this. Let me say this is my y-axis, this is my x-axis. And let's say I wanted to construct somethin... |
Laplace transform of the unit step function Laplace transform Khan Academy.mp3 | Let me say this is my y-axis, this is my x-axis. And let's say I wanted to construct something that is at let me do it a different color, let's say it's at 2 until I get to pi. And then from pi until forever, it just stays at 0. So how could I construct this function right here using my unit step function? So what if I... |
Laplace transform of the unit step function Laplace transform Khan Academy.mp3 | So how could I construct this function right here using my unit step function? So what if I had written it as, so my unit step function is 0 initially. So what if I make it 2 minus a unit step function that starts at pi? So if I define my function here, will this work? Well, this unit step function, when we pass pi, is... |
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