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Particular solution given initial conditions for population Khan Academy.mp3	In the last video, we established that if we say the rate of change of a population with respect to time is going to be proportional to the population, we were able to solve that differential equation and find a general solution which involves an exponential, that the population is going to be equal to some constant times e to some other constant times time. In the last video, we assumed time was days. So let's just apply this just to feel good that we can truly model population in this way. So let's use it with some concrete numbers. And once again, you've probably done that before. You probably started with the assumption that you can model with an exponential function and then you use some information, some conditions, to figure out what the constants are. We probably did this earlier in pre-calculus or algebra class.
Particular solution given initial conditions for population Khan Academy.mp3	So let's use it with some concrete numbers. And once again, you've probably done that before. You probably started with the assumption that you can model with an exponential function and then you use some information, some conditions, to figure out what the constants are. We probably did this earlier in pre-calculus or algebra class. But let's just do it again just so we can feel that this thing right over here is useful. So let's give you some information. Let's say that at time equals zero, the population is equal to 100 insects or whatever we're measuring the population of.
Particular solution given initial conditions for population Khan Academy.mp3	We probably did this earlier in pre-calculus or algebra class. But let's just do it again just so we can feel that this thing right over here is useful. So let's give you some information. Let's say that at time equals zero, the population is equal to 100 insects or whatever we're measuring the population of. And let's say that at time equals, let's say at time equals 50, the population, so after 50 days, the population is 200. So notice it doubled after 50 days. So given this information, can we solve for C and K?
Particular solution given initial conditions for population Khan Academy.mp3	Let's say that at time equals zero, the population is equal to 100 insects or whatever we're measuring the population of. And let's say that at time equals, let's say at time equals 50, the population, so after 50 days, the population is 200. So notice it doubled after 50 days. So given this information, can we solve for C and K? And I encourage you to pause the video and try to work through it on your own. So this first initial condition is pretty straightforward to use because when T is equal to zero, P is 100. So we could say, let's just use, so based on this first piece of information, we could say that 100 must be equal to C, C times E, times E, to the K times zero.
Particular solution given initial conditions for population Khan Academy.mp3	So given this information, can we solve for C and K? And I encourage you to pause the video and try to work through it on your own. So this first initial condition is pretty straightforward to use because when T is equal to zero, P is 100. So we could say, let's just use, so based on this first piece of information, we could say that 100 must be equal to C, C times E, times E, to the K times zero. Well, that's just going to be E to the zero. Well, E to the zero is just one. So this is just the same thing as C times one.
Particular solution given initial conditions for population Khan Academy.mp3	So we could say, let's just use, so based on this first piece of information, we could say that 100 must be equal to C, C times E, times E, to the K times zero. Well, that's just going to be E to the zero. Well, E to the zero is just one. So this is just the same thing as C times one. And just like that, we have figured out what C is. So now we can write, we can now write that the population is going to be equal to 100, E to the KT. And so you can see, expressed this way, our C is always going to be our initial population.
Particular solution given initial conditions for population Khan Academy.mp3	So this is just the same thing as C times one. And just like that, we have figured out what C is. So now we can write, we can now write that the population is going to be equal to 100, E to the KT. And so you can see, expressed this way, our C is always going to be our initial population. So E to the KT, E to the KT. And now we can use the second piece of information. So our population is 200.
Particular solution given initial conditions for population Khan Academy.mp3	And so you can see, expressed this way, our C is always going to be our initial population. So E to the KT, E to the KT. And now we can use the second piece of information. So our population is 200. Let's write that down. So our population is 200 when time is equal to 50, after 50 days. So 200 is equal to 100E, 100E to the K times 50, right?
Particular solution given initial conditions for population Khan Academy.mp3	So our population is 200. Let's write that down. So our population is 200 when time is equal to 50, after 50 days. So 200 is equal to 100E, 100E to the K times 50, right? T is now 50. So we can, let me just write that, times K times 50. Now we can divide both sides by 100, and we will get two is equal to E to the 50K, E to the 50K.
Particular solution given initial conditions for population Khan Academy.mp3	So 200 is equal to 100E, 100E to the K times 50, right? T is now 50. So we can, let me just write that, times K times 50. Now we can divide both sides by 100, and we will get two is equal to E to the 50K, E to the 50K. Then we can take the natural log of both sides, natural log on the left-hand side, we get the natural log of two. And on the right-hand side, the natural log of E to the 50K, well, that's just going to be, that's the power that you need to raise E to to get E to the 50K. Well, that's just going to be 50K.
Particular solution given initial conditions for population Khan Academy.mp3	Now we can divide both sides by 100, and we will get two is equal to E to the 50K, E to the 50K. Then we can take the natural log of both sides, natural log on the left-hand side, we get the natural log of two. And on the right-hand side, the natural log of E to the 50K, well, that's just going to be, that's the power that you need to raise E to to get E to the 50K. Well, that's just going to be 50K. That's just going to be 50K. All I did is took the natural log of both sides. Notice that this equation that I've just written expresses the same thing.
Particular solution given initial conditions for population Khan Academy.mp3	Well, that's just going to be 50K. That's just going to be 50K. All I did is took the natural log of both sides. Notice that this equation that I've just written expresses the same thing. Natural log of two is equal to 50K, that means E to the 50K is equal to two, which is exactly what we had written there. And now we can solve for K, divide both sides by 50. And we are left with K is equal to the natural log of two, natural log of two over 50, over 50.
Particular solution given initial conditions for population Khan Academy.mp3	Notice that this equation that I've just written expresses the same thing. Natural log of two is equal to 50K, that means E to the 50K is equal to two, which is exactly what we had written there. And now we can solve for K, divide both sides by 50. And we are left with K is equal to the natural log of two, natural log of two over 50, over 50. And we're done. We can now write the particular solution that meets these conditions. So we can now write that our population, and I can even write our population as a function of time, is going to be equal to, is going to be equal to 100, 100 times E to the, now K is natural log of two over 50, so I'll write that.
Differential equation from slope field First order differential equations Khan Academy.mp3	What I want to do in this video is do an exercise that takes us the other way. Start with a slope field and figure out which differential equation is the slope field describing the solutions for. And so I encourage you to look at each of these options and think about which of these differential equations is being described by this slope field. I encourage you to pause the video right now and try it on your own. So I'm assuming you have had a go at it. So let's work through each of them. And the way I'm going to do it is I'm just gonna find some points that seem to be easy to do arithmetic with and we'll see if the slope described by the differential equation at that point is consistent with the slope depicted in the slope field.
Differential equation from slope field First order differential equations Khan Academy.mp3	I encourage you to pause the video right now and try it on your own. So I'm assuming you have had a go at it. So let's work through each of them. And the way I'm going to do it is I'm just gonna find some points that seem to be easy to do arithmetic with and we'll see if the slope described by the differential equation at that point is consistent with the slope depicted in the slope field. And I don't know, just for simplicity, maybe I'll do x equals one, y equals one for all of these. So I'm gonna see what, so when x equals one and y is equal to one. So this first differential equation right over here, if x is one and y is one, then dy dx would be negative one over one, or negative one.
Differential equation from slope field First order differential equations Khan Academy.mp3	And the way I'm going to do it is I'm just gonna find some points that seem to be easy to do arithmetic with and we'll see if the slope described by the differential equation at that point is consistent with the slope depicted in the slope field. And I don't know, just for simplicity, maybe I'll do x equals one, y equals one for all of these. So I'm gonna see what, so when x equals one and y is equal to one. So this first differential equation right over here, if x is one and y is one, then dy dx would be negative one over one, or negative one. dy dx would be negative one. Now is that depicted here? When x is equal to one and y is equal to one, our slope isn't negative one, our slope here looks positive.
Differential equation from slope field First order differential equations Khan Academy.mp3	So this first differential equation right over here, if x is one and y is one, then dy dx would be negative one over one, or negative one. dy dx would be negative one. Now is that depicted here? When x is equal to one and y is equal to one, our slope isn't negative one, our slope here looks positive. So we can rule this one out. Now let's try the next one. So if x is equal to one and y is equal to one, well then dy dx would be equal to one minus one, or zero.
Differential equation from slope field First order differential equations Khan Academy.mp3	When x is equal to one and y is equal to one, our slope isn't negative one, our slope here looks positive. So we can rule this one out. Now let's try the next one. So if x is equal to one and y is equal to one, well then dy dx would be equal to one minus one, or zero. And once again, I just picked x equals one and y equals one for convenience. I could have picked any other. I could have picked negative five and negative seven.
Differential equation from slope field First order differential equations Khan Academy.mp3	So if x is equal to one and y is equal to one, well then dy dx would be equal to one minus one, or zero. And once again, I just picked x equals one and y equals one for convenience. I could have picked any other. I could have picked negative five and negative seven. This just makes the arithmetic a little easier. Once again, when you look at that point that we've already looked at, our slope is clearly not zero. We have a positive slope here.
Differential equation from slope field First order differential equations Khan Academy.mp3	I could have picked negative five and negative seven. This just makes the arithmetic a little easier. Once again, when you look at that point that we've already looked at, our slope is clearly not zero. We have a positive slope here. So we could rule that out. Once again, for this magenta differential equation, if x and y are both equal to one, then one minus one is once again going to be equal to zero. And we've already seen the slope is not zero here.
Differential equation from slope field First order differential equations Khan Academy.mp3	We have a positive slope here. So we could rule that out. Once again, for this magenta differential equation, if x and y are both equal to one, then one minus one is once again going to be equal to zero. And we've already seen the slope is not zero here. So rule that one out. And now here we have x plus y. So when x is one, y is one, the derivative of y with respect to x is going to be one plus one, which is equal to two.
Differential equation from slope field First order differential equations Khan Academy.mp3	And we've already seen the slope is not zero here. So rule that one out. And now here we have x plus y. So when x is one, y is one, the derivative of y with respect to x is going to be one plus one, which is equal to two. Now this looks interesting. It looks like this slope right over here could be two. This looks like one, this looks like two.
Differential equation from slope field First order differential equations Khan Academy.mp3	So when x is one, y is one, the derivative of y with respect to x is going to be one plus one, which is equal to two. Now this looks interesting. It looks like this slope right over here could be two. This looks like one, this looks like two. I'd want to validate some other points, but this looks like a really, really good candidate. And you could also see what's happening here. When dy dx is equal to x plus y, you would expect that as x increases for a given y, your slope would increase.
Differential equation from slope field First order differential equations Khan Academy.mp3	This looks like one, this looks like two. I'd want to validate some other points, but this looks like a really, really good candidate. And you could also see what's happening here. When dy dx is equal to x plus y, you would expect that as x increases for a given y, your slope would increase. And as y increases for a given x, your slope increases. And we see that. If we were to just follow, if we were to hold y constant one, but increase x along this line, we see that the slope is increasing.
Differential equation from slope field First order differential equations Khan Academy.mp3	When dy dx is equal to x plus y, you would expect that as x increases for a given y, your slope would increase. And as y increases for a given x, your slope increases. And we see that. If we were to just follow, if we were to hold y constant one, but increase x along this line, we see that the slope is increasing. It's getting steeper. And if we were to keep x constant and increase y across this line, we see that the slope increases. And in general, we see that the slope increases as we go to the top right.
Differential equation from slope field First order differential equations Khan Academy.mp3	If we were to just follow, if we were to hold y constant one, but increase x along this line, we see that the slope is increasing. It's getting steeper. And if we were to keep x constant and increase y across this line, we see that the slope increases. And in general, we see that the slope increases as we go to the top right. And we see that it decreases as we go to the bottom left and both x and y become much, much more negative. So I'm feeling pretty good about this, especially if we can knock this one out here, if we can knock that one out. So dy dx is equal to x over y.
Differential equation from slope field First order differential equations Khan Academy.mp3	And in general, we see that the slope increases as we go to the top right. And we see that it decreases as we go to the bottom left and both x and y become much, much more negative. So I'm feeling pretty good about this, especially if we can knock this one out here, if we can knock that one out. So dy dx is equal to x over y. Well then when x equals one, y equals one, dy dx would be equal to one. And this slope looks larger than one. It looks like two, but since we're really just eyeballing it, let's see if we can find something that more clearly, where this more clearly falls apart.
Differential equation from slope field First order differential equations Khan Academy.mp3	So dy dx is equal to x over y. Well then when x equals one, y equals one, dy dx would be equal to one. And this slope looks larger than one. It looks like two, but since we're really just eyeballing it, let's see if we can find something that more clearly, where this more clearly falls apart. So let's look at the situation when, let's look at the situation when they both equal negative one. So x equals negative one and y is equal to negative one. Well in that case, dy dx should still be equal to one because you have negative one over one.
Differential equation from slope field First order differential equations Khan Academy.mp3	It looks like two, but since we're really just eyeballing it, let's see if we can find something that more clearly, where this more clearly falls apart. So let's look at the situation when, let's look at the situation when they both equal negative one. So x equals negative one and y is equal to negative one. Well in that case, dy dx should still be equal to one because you have negative one over one. Do we see that over here? So when x is equal to negative one, y is equal to negative one, our derivative here looks negative. It looks like negative two, which is consistent with this yellow differential equation.
Differential equation from slope field First order differential equations Khan Academy.mp3	Well in that case, dy dx should still be equal to one because you have negative one over one. Do we see that over here? So when x is equal to negative one, y is equal to negative one, our derivative here looks negative. It looks like negative two, which is consistent with this yellow differential equation. The slope here is definitely not a positive one. So we could rule this one out as well. And so we should feel pretty confident that this is the differential equation being described.
Differential equation from slope field First order differential equations Khan Academy.mp3	It looks like negative two, which is consistent with this yellow differential equation. The slope here is definitely not a positive one. So we could rule this one out as well. And so we should feel pretty confident that this is the differential equation being described. And now that we've done it, we can actually think about, well okay, what are the solutions for this differential equation going to look like? Well it depends where they start, if you have, or what points they contain. If you have a solution that contains that point, it looks like it might go, looks like it might do something, looks like it might do something like, something like this.
Differential equation from slope field First order differential equations Khan Academy.mp3	And so we should feel pretty confident that this is the differential equation being described. And now that we've done it, we can actually think about, well okay, what are the solutions for this differential equation going to look like? Well it depends where they start, if you have, or what points they contain. If you have a solution that contains that point, it looks like it might go, looks like it might do something, looks like it might do something like, something like this. If you had a solution that contained, I don't know, if you had a solution that contained this point, it might do something, it might do something like that. And of course it keeps going, it looks like it would asymptote towards y is equal to negative x, this downward sloping, this essentially is the line y is equal to negative x. Actually no, that's not the line y equals negative x, this is the line y is equal to negative x minus one.
Differential equation from slope field First order differential equations Khan Academy.mp3	If you have a solution that contains that point, it looks like it might go, looks like it might do something, looks like it might do something like, something like this. If you had a solution that contained, I don't know, if you had a solution that contained this point, it might do something, it might do something like that. And of course it keeps going, it looks like it would asymptote towards y is equal to negative x, this downward sloping, this essentially is the line y is equal to negative x. Actually no, that's not the line y equals negative x, this is the line y is equal to negative x minus one. So that's this line right over here. And it looks like if you had, if you had your, in a solution, if you had a, if the solution contained, say this point right over here, you would actually, that would actually be a solution to the differential equation. Y is equal to, y is equal to negative x, whoops, y is equal to negative x minus one.
Differential equation from slope field First order differential equations Khan Academy.mp3	Actually no, that's not the line y equals negative x, this is the line y is equal to negative x minus one. So that's this line right over here. And it looks like if you had, if you had your, in a solution, if you had a, if the solution contained, say this point right over here, you would actually, that would actually be a solution to the differential equation. Y is equal to, y is equal to negative x, whoops, y is equal to negative x minus one. And you can verify that. If y is equal to negative x minus one, then the x and negative x cancel out, you're just left with dy dx is equal to negative one, which is exactly what's being described by this slope field. Anyway, hopefully you found that interesting.
Laplace transform solves an equation 2 Laplace transform Differential Equations Khan Academy.mp3	In the first part of this problem, we just had this fairly straightforward differential equation. And I know it's a little bit frustrating right now, because this is such an easy one to solve using the characteristic equation. Why are we doing Laplace transforms? Well, I just want to show you that they can solve even these problems. But later on, there are going to be classes of problems that frankly, our traditional methods aren't as good as the Laplace transform. But anyway, how did we solve this? We just took the Laplace transform of both sides of this equation.
Laplace transform solves an equation 2 Laplace transform Differential Equations Khan Academy.mp3	Well, I just want to show you that they can solve even these problems. But later on, there are going to be classes of problems that frankly, our traditional methods aren't as good as the Laplace transform. But anyway, how did we solve this? We just took the Laplace transform of both sides of this equation. We got all this hairy mess. We used the property of the derivatives of functions when you take the Laplace transform. And we ended up, after doing a lot of algebra, essentially, we got this.
Laplace transform solves an equation 2 Laplace transform Differential Equations Khan Academy.mp3	We just took the Laplace transform of both sides of this equation. We got all this hairy mess. We used the property of the derivatives of functions when you take the Laplace transform. And we ended up, after doing a lot of algebra, essentially, we got this. We got the Laplace transform of y is equal to this thing. We just took the Laplace transform of both sides and manipulated algebraically. So now our task in this video is to figure out what y's Laplace transform is this thing.
Laplace transform solves an equation 2 Laplace transform Differential Equations Khan Academy.mp3	And we ended up, after doing a lot of algebra, essentially, we got this. We got the Laplace transform of y is equal to this thing. We just took the Laplace transform of both sides and manipulated algebraically. So now our task in this video is to figure out what y's Laplace transform is this thing. And essentially, what we're trying to do is we're trying to take the inverse Laplace transform of both sides of this equation. So another way to say it, we could say that y, we could take the inverse Laplace transform of both sides. We could say that y is equal to the inverse Laplace transform of this thing.
Laplace transform solves an equation 2 Laplace transform Differential Equations Khan Academy.mp3	So now our task in this video is to figure out what y's Laplace transform is this thing. And essentially, what we're trying to do is we're trying to take the inverse Laplace transform of both sides of this equation. So another way to say it, we could say that y, we could take the inverse Laplace transform of both sides. We could say that y is equal to the inverse Laplace transform of this thing. 2s plus 13 over s squared plus 5s plus 6. Now we'll eventually actually learn the formal definition of the inverse Laplace transform. How do you go from the s domain to the t domain?
Laplace transform solves an equation 2 Laplace transform Differential Equations Khan Academy.mp3	We could say that y is equal to the inverse Laplace transform of this thing. 2s plus 13 over s squared plus 5s plus 6. Now we'll eventually actually learn the formal definition of the inverse Laplace transform. How do you go from the s domain to the t domain? Or how do you go from the frequency domain to the time domain? We're not going to worry about that right now. What we're going to do is we're going to get this into a form that we recognize and say, oh, I know those functions.
Laplace transform solves an equation 2 Laplace transform Differential Equations Khan Academy.mp3	How do you go from the s domain to the t domain? Or how do you go from the frequency domain to the time domain? We're not going to worry about that right now. What we're going to do is we're going to get this into a form that we recognize and say, oh, I know those functions. That's the Laplace transform of whatever and whatever. And then we will know what y is. So let's try to do that.
Laplace transform solves an equation 2 Laplace transform Differential Equations Khan Academy.mp3	What we're going to do is we're going to get this into a form that we recognize and say, oh, I know those functions. That's the Laplace transform of whatever and whatever. And then we will know what y is. So let's try to do that. So what we're going to use is something that you probably haven't used since algebra 2, which is, I think, when it's taught in eighth, ninth, or tenth grade, depending. And you finally see it now in differential equations that it actually has some use. We're going to use partial fraction expansion.
Laplace transform solves an equation 2 Laplace transform Differential Equations Khan Academy.mp3	So let's try to do that. So what we're going to use is something that you probably haven't used since algebra 2, which is, I think, when it's taught in eighth, ninth, or tenth grade, depending. And you finally see it now in differential equations that it actually has some use. We're going to use partial fraction expansion. And I'll do a little primer on that in case you don't remember it. So anyway, let's just factor the bottom part right here. And you'll see where I'm going with this.
Laplace transform solves an equation 2 Laplace transform Differential Equations Khan Academy.mp3	We're going to use partial fraction expansion. And I'll do a little primer on that in case you don't remember it. So anyway, let's just factor the bottom part right here. And you'll see where I'm going with this. So if I factor the bottom, I get s plus 2 times s plus 3. And what we want to do is we want to rewrite this fraction as the sum of 2 partial fractions. I think that's why it's called partial fraction expansion.
Laplace transform solves an equation 2 Laplace transform Differential Equations Khan Academy.mp3	And you'll see where I'm going with this. So if I factor the bottom, I get s plus 2 times s plus 3. And what we want to do is we want to rewrite this fraction as the sum of 2 partial fractions. I think that's why it's called partial fraction expansion. So we want to write this as a sum of a over s plus 2 plus b over s plus 3. And if we can do this, then you might already, bells might already be ringing in your head. We know that these things that look like this are the Laplace transform of functions that we've already solved for.
Laplace transform solves an equation 2 Laplace transform Differential Equations Khan Academy.mp3	I think that's why it's called partial fraction expansion. So we want to write this as a sum of a over s plus 2 plus b over s plus 3. And if we can do this, then you might already, bells might already be ringing in your head. We know that these things that look like this are the Laplace transform of functions that we've already solved for. And I'll do a little review on that in a second. But anyway, how do we figure out a and b? Well, if we were to actually add a and b, if we were to, let's do a little side right here.
Laplace transform solves an equation 2 Laplace transform Differential Equations Khan Academy.mp3	We know that these things that look like this are the Laplace transform of functions that we've already solved for. And I'll do a little review on that in a second. But anyway, how do we figure out a and b? Well, if we were to actually add a and b, if we were to, let's do a little side right here. So if we said that a, so if we were to give them a common denominator, which is this, s plus 2 times s plus 3, then what would a become? We'd have to multiply a times s plus 3, right? So we'd get a s plus 3a.
Laplace transform solves an equation 2 Laplace transform Differential Equations Khan Academy.mp3	Well, if we were to actually add a and b, if we were to, let's do a little side right here. So if we said that a, so if we were to give them a common denominator, which is this, s plus 2 times s plus 3, then what would a become? We'd have to multiply a times s plus 3, right? So we'd get a s plus 3a. Right? This, as I've written it right now, is the same thing as a over s plus 2. You could cancel out an s plus 3 on the top and the bottom.
Laplace transform solves an equation 2 Laplace transform Differential Equations Khan Academy.mp3	So we'd get a s plus 3a. Right? This, as I've written it right now, is the same thing as a over s plus 2. You could cancel out an s plus 3 on the top and the bottom. And now we're going to add the b to it. So plus, I'll do that in a different color. Plus, well, if we have this as the denominator, we could multiply the numerator and the denominator by s plus 2, right?
Laplace transform solves an equation 2 Laplace transform Differential Equations Khan Academy.mp3	You could cancel out an s plus 3 on the top and the bottom. And now we're going to add the b to it. So plus, I'll do that in a different color. Plus, well, if we have this as the denominator, we could multiply the numerator and the denominator by s plus 2, right? To get b times s plus 2b. And that's going to equal this thing. That's going to equal, all I did is I added these two fractions.
Laplace transform solves an equation 2 Laplace transform Differential Equations Khan Academy.mp3	Plus, well, if we have this as the denominator, we could multiply the numerator and the denominator by s plus 2, right? To get b times s plus 2b. And that's going to equal this thing. That's going to equal, all I did is I added these two fractions. Nothing fancier than there. That was algebra 2. Actually, I think I should do an actual video on that as well.
Laplace transform solves an equation 2 Laplace transform Differential Equations Khan Academy.mp3	That's going to equal, all I did is I added these two fractions. Nothing fancier than there. That was algebra 2. Actually, I think I should do an actual video on that as well. But that's going to equal this thing. 2s plus 13, all of that over s plus 2 times s plus 3. Notice, in all differential equations, the hairiest part's always the algebra.
Laplace transform solves an equation 2 Laplace transform Differential Equations Khan Academy.mp3	Actually, I think I should do an actual video on that as well. But that's going to equal this thing. 2s plus 13, all of that over s plus 2 times s plus 3. Notice, in all differential equations, the hairiest part's always the algebra. So now what we do is we match up. We say, well, let's add the s terms here. And we could just say that the numerators have to equal each other because the denominators are equal.
Laplace transform solves an equation 2 Laplace transform Differential Equations Khan Academy.mp3	Notice, in all differential equations, the hairiest part's always the algebra. So now what we do is we match up. We say, well, let's add the s terms here. And we could just say that the numerators have to equal each other because the denominators are equal. So we have a plus bs plus 3a plus 2b is equal to 2s plus b. So the coefficient on s on the right-hand side is 2. The coefficient on the left-hand side is a plus b.
Laplace transform solves an equation 2 Laplace transform Differential Equations Khan Academy.mp3	And we could just say that the numerators have to equal each other because the denominators are equal. So we have a plus bs plus 3a plus 2b is equal to 2s plus b. So the coefficient on s on the right-hand side is 2. The coefficient on the left-hand side is a plus b. So we know that a plus b is equal to 2. And then on the right-hand side, we see 3a plus 2b must be equal to, oh, this is a 13. Did I say b?
Laplace transform solves an equation 2 Laplace transform Differential Equations Khan Academy.mp3	The coefficient on the left-hand side is a plus b. So we know that a plus b is equal to 2. And then on the right-hand side, we see 3a plus 2b must be equal to, oh, this is a 13. Did I say b? This is a 13. It looks just like a b, right? That was 2s plus 13.
Laplace transform solves an equation 2 Laplace transform Differential Equations Khan Academy.mp3	Did I say b? This is a 13. It looks just like a b, right? That was 2s plus 13. Anyway, so on the right-hand side, I get 3a plus 2b is equal to 13. Now we have two equations with two unknowns. And what do we get?
Laplace transform solves an equation 2 Laplace transform Differential Equations Khan Academy.mp3	That was 2s plus 13. Anyway, so on the right-hand side, I get 3a plus 2b is equal to 13. Now we have two equations with two unknowns. And what do we get? I know this is very tiresome, but it'll be satisfying in the end because you'll actually solve something with the Laplace transform. Let's multiply the top equation by 2. Let's just say minus 2.
Laplace transform solves an equation 2 Laplace transform Differential Equations Khan Academy.mp3	And what do we get? I know this is very tiresome, but it'll be satisfying in the end because you'll actually solve something with the Laplace transform. Let's multiply the top equation by 2. Let's just say minus 2. So we get minus 2a minus 2b equals minus 4. And then we add the two equations. You get a is equal to, these cancel out, a is equal to 9.
Laplace transform solves an equation 2 Laplace transform Differential Equations Khan Academy.mp3	Let's just say minus 2. So we get minus 2a minus 2b equals minus 4. And then we add the two equations. You get a is equal to, these cancel out, a is equal to 9. Great. If a is equal to 9, what is b equal to? b is equal to 9 plus what is equal to 2?
Laplace transform solves an equation 2 Laplace transform Differential Equations Khan Academy.mp3	You get a is equal to, these cancel out, a is equal to 9. Great. If a is equal to 9, what is b equal to? b is equal to 9 plus what is equal to 2? Or 2 minus 9 is minus 7. And we have done some serious simplification because now we can rewrite this whole expression as the Laplace transform of y is equal to a over s plus 2 is equal to 9 over s plus 2 minus 7 over s plus 3. Or another way of writing it, we could write it as equal to 9 times 1 over s plus 2 minus 7 times 1 over s plus 3.
Laplace transform solves an equation 2 Laplace transform Differential Equations Khan Academy.mp3	b is equal to 9 plus what is equal to 2? Or 2 minus 9 is minus 7. And we have done some serious simplification because now we can rewrite this whole expression as the Laplace transform of y is equal to a over s plus 2 is equal to 9 over s plus 2 minus 7 over s plus 3. Or another way of writing it, we could write it as equal to 9 times 1 over s plus 2 minus 7 times 1 over s plus 3. Why did I take the trouble to do this? Well, hopefully you'll recognize this was actually the second Laplace transform we figured out. This was the second Laplace transform we figured out.
Laplace transform solves an equation 2 Laplace transform Differential Equations Khan Academy.mp3	Or another way of writing it, we could write it as equal to 9 times 1 over s plus 2 minus 7 times 1 over s plus 3. Why did I take the trouble to do this? Well, hopefully you'll recognize this was actually the second Laplace transform we figured out. This was the second Laplace transform we figured out. What was that? I'll write it down here just so you remember it. It was the Laplace transform of e to the at was equal to 1 over s minus a.
Laplace transform solves an equation 2 Laplace transform Differential Equations Khan Academy.mp3	This was the second Laplace transform we figured out. What was that? I'll write it down here just so you remember it. It was the Laplace transform of e to the at was equal to 1 over s minus a. That was the second Laplace transform we figured out. So this is interesting. So if we were to take, this is the Laplace transform of what?
Laplace transform solves an equation 2 Laplace transform Differential Equations Khan Academy.mp3	It was the Laplace transform of e to the at was equal to 1 over s minus a. That was the second Laplace transform we figured out. So this is interesting. So if we were to take, this is the Laplace transform of what? So if we were to take the inverse Laplace transform, actually let me just stay consistent. So that means that this is the Laplace transform of y is equal to 9 times the Laplace transform of what? If we just do pattern matching, if this is s minus a, then a is minus 2.
Laplace transform solves an equation 2 Laplace transform Differential Equations Khan Academy.mp3	So if we were to take, this is the Laplace transform of what? So if we were to take the inverse Laplace transform, actually let me just stay consistent. So that means that this is the Laplace transform of y is equal to 9 times the Laplace transform of what? If we just do pattern matching, if this is s minus a, then a is minus 2. So 9 times the Laplace transform of e to the minus 2t. Does that make sense? Take this, put it in this, which we figured out, and you get 1 over s plus 2.
Laplace transform solves an equation 2 Laplace transform Differential Equations Khan Academy.mp3	If we just do pattern matching, if this is s minus a, then a is minus 2. So 9 times the Laplace transform of e to the minus 2t. Does that make sense? Take this, put it in this, which we figured out, and you get 1 over s plus 2. And let me clean this up a little bit because I'm going to need that real estate. I'll write this. I'll leave that there because we'll still use that.
Laplace transform solves an equation 2 Laplace transform Differential Equations Khan Academy.mp3	Take this, put it in this, which we figured out, and you get 1 over s plus 2. And let me clean this up a little bit because I'm going to need that real estate. I'll write this. I'll leave that there because we'll still use that. And then we have minus 7 times, this is the Laplace transform of what? This is the Laplace transform of e to the minus 3t. This is just pattern matching.
Laplace transform solves an equation 2 Laplace transform Differential Equations Khan Academy.mp3	I'll leave that there because we'll still use that. And then we have minus 7 times, this is the Laplace transform of what? This is the Laplace transform of e to the minus 3t. This is just pattern matching. You're like, wow, if you saw this, you would go to your Laplace transform table if you didn't remember it. You'd see this. You're like, wow, that looks a lot like that.
Laplace transform solves an equation 2 Laplace transform Differential Equations Khan Academy.mp3	This is just pattern matching. You're like, wow, if you saw this, you would go to your Laplace transform table if you didn't remember it. You'd see this. You're like, wow, that looks a lot like that. It's just I just have to figure out what a is. I have s plus 3, I have s minus a. So in this case, a is equal to minus 3.
Laplace transform solves an equation 2 Laplace transform Differential Equations Khan Academy.mp3	You're like, wow, that looks a lot like that. It's just I just have to figure out what a is. I have s plus 3, I have s minus a. So in this case, a is equal to minus 3. So if a is equal to minus 3, this is the Laplace transform of e to the minus 3t. So now we can take the inverse Laplace transform. Actually, before we do that, we know that this, because the Laplace transform is a linear operator, and actually now I can delete this down here.
Laplace transform solves an equation 2 Laplace transform Differential Equations Khan Academy.mp3	So in this case, a is equal to minus 3. So if a is equal to minus 3, this is the Laplace transform of e to the minus 3t. So now we can take the inverse Laplace transform. Actually, before we do that, we know that this, because the Laplace transform is a linear operator, and actually now I can delete this down here. We know that the Laplace transform is a linear operator, so we can write this. And you normally wouldn't go through all of these steps. I just really want to make you understand what we're doing.
Laplace transform solves an equation 2 Laplace transform Differential Equations Khan Academy.mp3	Actually, before we do that, we know that this, because the Laplace transform is a linear operator, and actually now I can delete this down here. We know that the Laplace transform is a linear operator, so we can write this. And you normally wouldn't go through all of these steps. I just really want to make you understand what we're doing. So we could say that this is the same thing as the Laplace transform of 9e to the minus 2t minus 7e to the minus 3t. Now we have something interesting. The Laplace transform of y is equal to the Laplace transform of this.
Laplace transform solves an equation 2 Laplace transform Differential Equations Khan Academy.mp3	I just really want to make you understand what we're doing. So we could say that this is the same thing as the Laplace transform of 9e to the minus 2t minus 7e to the minus 3t. Now we have something interesting. The Laplace transform of y is equal to the Laplace transform of this. Well, if that's the case, then y must be equal to 9e to the minus 2t minus 7e to the minus 3t. And I never proved to you, but the Laplace transform is actually a one-to-one transformation. That if a function's Laplace transform, if I take a function and get its Laplace transform, and then if I were to take the inverse Laplace transform, the only function whose Laplace transform that is is that original function.
Laplace transform solves an equation 2 Laplace transform Differential Equations Khan Academy.mp3	The Laplace transform of y is equal to the Laplace transform of this. Well, if that's the case, then y must be equal to 9e to the minus 2t minus 7e to the minus 3t. And I never proved to you, but the Laplace transform is actually a one-to-one transformation. That if a function's Laplace transform, if I take a function and get its Laplace transform, and then if I were to take the inverse Laplace transform, the only function whose Laplace transform that is is that original function. It's not like two different functions can have the same Laplace transform. Anyway, a couple of things to think about here. Notice we had that thing that kind of looked like a characteristic equation popped up here and there, and we still have to solve a system of two equations with two unknowns.
Laplace transform solves an equation 2 Laplace transform Differential Equations Khan Academy.mp3	That if a function's Laplace transform, if I take a function and get its Laplace transform, and then if I were to take the inverse Laplace transform, the only function whose Laplace transform that is is that original function. It's not like two different functions can have the same Laplace transform. Anyway, a couple of things to think about here. Notice we had that thing that kind of looked like a characteristic equation popped up here and there, and we still have to solve a system of two equations with two unknowns. Those are both things that we had to do when we solved an initial value problem when we used just traditional, the characteristic equation. But here, it happened all at once. And frankly, it was a little bit hairier because we had to do all this partial fraction expansion.
Laplace transform solves an equation 2 Laplace transform Differential Equations Khan Academy.mp3	Notice we had that thing that kind of looked like a characteristic equation popped up here and there, and we still have to solve a system of two equations with two unknowns. Those are both things that we had to do when we solved an initial value problem when we used just traditional, the characteristic equation. But here, it happened all at once. And frankly, it was a little bit hairier because we had to do all this partial fraction expansion. But it's pretty neat. The Laplace transform got us something useful. In the next video, I'll actually do a non-homogeneous equation and show you that the Laplace transform applies equally well there.
Laplace transform of t L{t} Laplace transform Differential Equations Khan Academy.mp3	Let's try to fill in our Laplace transform table a little bit more. And a good place to start is just to write our definition of the Laplace transform. The Laplace transform of some function f of t is equal to the integral from 0 to infinity of e to the minus st times our function f of t dt. That's our definition. And the very first one we solved for, we can even do it on the side right here, was the Laplace transform of 1. We can almost do that as t to the 0. And that was equal to the integral from 0 to infinity.
Laplace transform of t L{t} Laplace transform Differential Equations Khan Academy.mp3	That's our definition. And the very first one we solved for, we can even do it on the side right here, was the Laplace transform of 1. We can almost do that as t to the 0. And that was equal to the integral from 0 to infinity. f of t was just 1, so it's e to the minus st dt, which is equal to the antiderivative of e to the minus st is minus 1 over s, minus 1 over s, e to the minus st. And then you have to evaluate that from 0 to infinity. When you take the limit as this term approaches infinity, this e to the minus, this becomes e to the minus infinity. If we assume s is greater than 0, this whole term goes to 0.
Laplace transform of t L{t} Laplace transform Differential Equations Khan Academy.mp3	And that was equal to the integral from 0 to infinity. f of t was just 1, so it's e to the minus st dt, which is equal to the antiderivative of e to the minus st is minus 1 over s, minus 1 over s, e to the minus st. And then you have to evaluate that from 0 to infinity. When you take the limit as this term approaches infinity, this e to the minus, this becomes e to the minus infinity. If we assume s is greater than 0, this whole term goes to 0. So you end up with 0 minus this thing evaluated at 0. When you evaluate t is equal to 0, this term right here becomes 1, e to the 0 becomes 1. So it's minus minus 1 over s, which is the same thing as plus 1 over s. So the Laplace transform of 1, of just the constant function 1, is 1 over s. We already solved that.
Laplace transform of t L{t} Laplace transform Differential Equations Khan Academy.mp3	If we assume s is greater than 0, this whole term goes to 0. So you end up with 0 minus this thing evaluated at 0. When you evaluate t is equal to 0, this term right here becomes 1, e to the 0 becomes 1. So it's minus minus 1 over s, which is the same thing as plus 1 over s. So the Laplace transform of 1, of just the constant function 1, is 1 over s. We already solved that. Now let's increment it a little bit. Let's see if we can figure out the Laplace transform of t. So we could view this as t to the 0, now this is t to the 1. This is going to be equal to the integral from 0 to infinity of e to the minus st times t dt.
Laplace transform of t L{t} Laplace transform Differential Equations Khan Academy.mp3	So it's minus minus 1 over s, which is the same thing as plus 1 over s. So the Laplace transform of 1, of just the constant function 1, is 1 over s. We already solved that. Now let's increment it a little bit. Let's see if we can figure out the Laplace transform of t. So we could view this as t to the 0, now this is t to the 1. This is going to be equal to the integral from 0 to infinity of e to the minus st times t dt. Now, I can tell you right now that this isn't, I don't have the antiderivative of this memorized, I don't know what it is, but there's a sense that the integration by parts could be useful, because integration by parts kind of decomposes this into a simpler problem. I always forget integration by parts, so I'll re-derive it here in this purple color. So if we have u times v, if we take the derivative with respect to, say, t of that, that's equal to the derivative of the first times the second function, plus the first function times the derivative of the second function.
Laplace transform of t L{t} Laplace transform Differential Equations Khan Academy.mp3	This is going to be equal to the integral from 0 to infinity of e to the minus st times t dt. Now, I can tell you right now that this isn't, I don't have the antiderivative of this memorized, I don't know what it is, but there's a sense that the integration by parts could be useful, because integration by parts kind of decomposes this into a simpler problem. I always forget integration by parts, so I'll re-derive it here in this purple color. So if we have u times v, if we take the derivative with respect to, say, t of that, that's equal to the derivative of the first times the second function, plus the first function times the derivative of the second function. It's just the product rule. Now if we take the integral of both sides of this equation, we get uv is equal to the antiderivative of u prime v plus the antiderivative of u v prime. Now, since we want to apply this to an integral, maybe let's make this what we want to solve for.
Laplace transform of t L{t} Laplace transform Differential Equations Khan Academy.mp3	So if we have u times v, if we take the derivative with respect to, say, t of that, that's equal to the derivative of the first times the second function, plus the first function times the derivative of the second function. It's just the product rule. Now if we take the integral of both sides of this equation, we get uv is equal to the antiderivative of u prime v plus the antiderivative of u v prime. Now, since we want to apply this to an integral, maybe let's make this what we want to solve for. So we can get the integral of u v prime is equal to, we can just subtract this from that side of the equation, so it's equal, I'm just switching, swapping the sides, so I'm just solving for this, I just subtract this from that, so it's equal to u v minus the integral of u prime v. So there you go. Even though I have trouble memorizing this formula, it's not too hard to re-derive as long as you remember the product rule right there. So if we're going to do integration by parts, it's good to define our v prime to be something it's easy to take the antiderivative of, because we're going to have to figure out v later on, and it's good to take u to be something that's easy to take the derivative of.
Laplace transform of t L{t} Laplace transform Differential Equations Khan Academy.mp3	Now, since we want to apply this to an integral, maybe let's make this what we want to solve for. So we can get the integral of u v prime is equal to, we can just subtract this from that side of the equation, so it's equal, I'm just switching, swapping the sides, so I'm just solving for this, I just subtract this from that, so it's equal to u v minus the integral of u prime v. So there you go. Even though I have trouble memorizing this formula, it's not too hard to re-derive as long as you remember the product rule right there. So if we're going to do integration by parts, it's good to define our v prime to be something it's easy to take the antiderivative of, because we're going to have to figure out v later on, and it's good to take u to be something that's easy to take the derivative of. So let's make t is equal to our u, and let's make e to the minus st as being our v prime. If that's the case, then what is v? Well, v is just the antiderivative of that.
Laplace transform of t L{t} Laplace transform Differential Equations Khan Academy.mp3	So if we're going to do integration by parts, it's good to define our v prime to be something it's easy to take the antiderivative of, because we're going to have to figure out v later on, and it's good to take u to be something that's easy to take the derivative of. So let's make t is equal to our u, and let's make e to the minus st as being our v prime. If that's the case, then what is v? Well, v is just the antiderivative of that. In fact, we've done it before. It's minus 1 over s, e to the minus st. That's v. And then if we want to figure out u prime, because we're going to have to figure out that later anyway, u prime is just the derivative of t. That's just equal to 1. So let's apply this.
Laplace transform of t L{t} Laplace transform Differential Equations Khan Academy.mp3	Well, v is just the antiderivative of that. In fact, we've done it before. It's minus 1 over s, e to the minus st. That's v. And then if we want to figure out u prime, because we're going to have to figure out that later anyway, u prime is just the derivative of t. That's just equal to 1. So let's apply this. Let's see. So the Laplace transform of t is equal to u v. u is t. v is this right here. It's times minus 1 over s, e to the minus st. That's the u v term right there.
Laplace transform of t L{t} Laplace transform Differential Equations Khan Academy.mp3	So let's apply this. Let's see. So the Laplace transform of t is equal to u v. u is t. v is this right here. It's times minus 1 over s, e to the minus st. That's the u v term right there. And this is a definite integral, right? So we're going to evaluate this term right here from 0 to infinity. And then it's minus the integral from 0 to infinity of u prime, which is just 1, times v. v, we just figured out here, is minus 1 over s. This is my v right here.
Laplace transform of t L{t} Laplace transform Differential Equations Khan Academy.mp3	It's times minus 1 over s, e to the minus st. That's the u v term right there. And this is a definite integral, right? So we're going to evaluate this term right here from 0 to infinity. And then it's minus the integral from 0 to infinity of u prime, which is just 1, times v. v, we just figured out here, is minus 1 over s. This is my v right here. So minus 1 over s, e to the minus st dt. And this, let's see if we can simplify this. So this is equal to minus t over s, e to the minus st, evaluated from 0 to infinity.
Laplace transform of t L{t} Laplace transform Differential Equations Khan Academy.mp3	And then it's minus the integral from 0 to infinity of u prime, which is just 1, times v. v, we just figured out here, is minus 1 over s. This is my v right here. So minus 1 over s, e to the minus st dt. And this, let's see if we can simplify this. So this is equal to minus t over s, e to the minus st, evaluated from 0 to infinity. And let's see, we could take this. Well, this is just 1. This 1 times anything is 1.
Laplace transform of t L{t} Laplace transform Differential Equations Khan Academy.mp3	So this is equal to minus t over s, e to the minus st, evaluated from 0 to infinity. And let's see, we could take this. Well, this is just 1. This 1 times anything is 1. We can just not write that. And then bring the minus 1 over s out. So if we bring the minus 1 over s out, this becomes plus 1 over s times the integral from 0 to infinity of e to the minus st dt.
Laplace transform of t L{t} Laplace transform Differential Equations Khan Academy.mp3	This 1 times anything is 1. We can just not write that. And then bring the minus 1 over s out. So if we bring the minus 1 over s out, this becomes plus 1 over s times the integral from 0 to infinity of e to the minus st dt. And this should look familiar to you. This is exactly what we solved for right here. It was the Laplace transform of 1.
Laplace transform of t L{t} Laplace transform Differential Equations Khan Academy.mp3	So if we bring the minus 1 over s out, this becomes plus 1 over s times the integral from 0 to infinity of e to the minus st dt. And this should look familiar to you. This is exactly what we solved for right here. It was the Laplace transform of 1. So let's keep that in mind. So this right here is the Laplace transform of 1. And I want to write it that way, because we're going to see a pattern of this in the next video.
Laplace transform of t L{t} Laplace transform Differential Equations Khan Academy.mp3	It was the Laplace transform of 1. So let's keep that in mind. So this right here is the Laplace transform of 1. And I want to write it that way, because we're going to see a pattern of this in the next video. I'm going to write that as Laplace transform of 1. But what is this equal to? So we're going to evaluate this as it added infinity, and then subtract from that, evaluate it at 0.
Laplace transform of t L{t} Laplace transform Differential Equations Khan Academy.mp3	And I want to write it that way, because we're going to see a pattern of this in the next video. I'm going to write that as Laplace transform of 1. But what is this equal to? So we're going to evaluate this as it added infinity, and then subtract from that, evaluate it at 0. So I'm just going to make, you can kind of view it as a substitution. So this is equal to, well, let me write it this way. As a approaches infinity of minus a over se to the minus sa.
Laplace transform of t L{t} Laplace transform Differential Equations Khan Academy.mp3	So we're going to evaluate this as it added infinity, and then subtract from that, evaluate it at 0. So I'm just going to make, you can kind of view it as a substitution. So this is equal to, well, let me write it this way. As a approaches infinity of minus a over se to the minus sa. So that's this evaluated at infinity. And then from that, we're going to subtract this evaluated at 0. So minus all of this, but we already have a minus sign here, so we could write a plus.
Laplace transform of t L{t} Laplace transform Differential Equations Khan Academy.mp3	As a approaches infinity of minus a over se to the minus sa. So that's this evaluated at infinity. And then from that, we're going to subtract this evaluated at 0. So minus all of this, but we already have a minus sign here, so we could write a plus. Plus, we could write 0 over s times e to the minus s times 0. And then, of course, we have this term right here. So let me write that term.
Laplace transform of t L{t} Laplace transform Differential Equations Khan Academy.mp3	So minus all of this, but we already have a minus sign here, so we could write a plus. Plus, we could write 0 over s times e to the minus s times 0. And then, of course, we have this term right here. So let me write that term. I'll do it in yellow. Or let me do it in blue. Plus 1 over s, that's this right there, times the Laplace transform of 1.
Laplace transform of t L{t} Laplace transform Differential Equations Khan Academy.mp3	So let me write that term. I'll do it in yellow. Or let me do it in blue. Plus 1 over s, that's this right there, times the Laplace transform of 1. And what do we get? So what's the limit as this, as a approaches infinity? You might say, wow, as a approaches infinity right here, this becomes a really big number.
Laplace transform of t L{t} Laplace transform Differential Equations Khan Academy.mp3	Plus 1 over s, that's this right there, times the Laplace transform of 1. And what do we get? So what's the limit as this, as a approaches infinity? You might say, wow, as a approaches infinity right here, this becomes a really big number. There's a minus sign in there, so it's going to be a really big negative number. But this is an exponent. a is an exponent right here.

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