Introduction to differential equations

Khan Academy Presents: 3 basic differential equations that can be solved by taking the antiderivatives of both sides.

Welcome to the video on the introduction to differential equations. What's the differential? Well we've been doing things like you know if I said that y is equal to I don’t know 3x if we took the derivative of both sides of this equation we said one of the notations we use is we said dy/dx = 3 dx. We took the derivative of both sides of that, but this you could almost do this differential notation, dy is a differential and dx is the differential. And what does the differential mean? A differential just means an infinitely small change in y in this case so that’s the differential in y or an infinitely small change in x. So a differential it's a like a difference. When we learn slope we said slope is change in y over change in x or you could say the difference in y over difference in x but once we started taking the slope of curves we had to make this difference and this different approach zero and that’s what a differential essentially is. A differential is a really, really, really infinitely small difference. So what is a differential equation? Well it's an equation that involves differentials so for example this is a differential equation and this will be the first example that we solve, dy/dx = x2 + 1 and so how do we solve this differential equation. Well we could, there's a couple of ways you could say. You say “Oh! Well you know this is the same thing as f1(x) if you know y was a function of x and you just take the antiderivative of both sides” but let's solve this properly as a differential equation. So you can actually manipulate these differentials the same way that you would manipulate numbers for the most part. So what can we do here? Well let's multiply both sides of this equation by dx so you get for a very small change in y, dy you get x2 plus one times a very small change in the x or another way to say it is for very small change in x if you want to figure out how much does y change. You multiply times x2 +1 wherever you are on the curve. So what does that do for us? Well, let's take the antiderivative of both sides. Let's take the integral of both sides and if you watched all of the videos on integration and the definite integral and area on the curve you realized that an integral is essentially a sum. It's kind of an infinite sum of a bunch of this infinitely small dy. So if you take the sum of all of the changes in dy’s you're left with the y on this side of this equation and you might want to re-watch all the videos and that we did on integration and then what is this? This is essentially just the indefinite integral of x2 +1 so it's the antiderivative of that and what's the antiderivative of x2 + 1? What's x3/3 and we’re just taking the derivative backwards and once again watch the videos on the antiderivative. There are I think eight or nine of them plus x and then plus c and where does the c comes from? Well we know that when we take the derivative of a constant it goes to zero. So when you take the antiderivative we’re like well there could have been a constant there and that’s where that plus c comes from. So this is the general solution. This is the general solution to this differential equation and that’s something interesting. So with traditional equations the solution tended to be a number. If I just told y = 2y – 1 then you would just get- y = -1, you get y + 1. So this is a traditional equation and your solution was just a value. You solve for the variable. The first equations are something different. The solution is actually a function. You're saying what function satisfies this differential equation? So that’s something to keep in mind and right now we’re doing very basic differential equations but that’s something to keep in mind the whole time you learn differential equations. I think I’ll eventually do a play list on essentially but introductory course on differential equations that you would take at college and that applies even you start doing partial differential equation etcetera, etcetera. It's the solution to a differential equation is not a number. It is a function. So anyway this was the general solution to this differential equation and then if you want the particular solution, people normally give you initial conditions or they give you points on the function and then you can substitute back. So in this problem they said that you know dy-dx + x2 + 1 and then they said that y = 1 at x =1 or y = 1 when x =1. So we can use this information now to solve for c, how do you do that? Well it says y =1, so one is equal to an X is equal to one is equal to one third, right one over three, one to the third power over three plus one plus c and let's see what I can do. Subtract one from both sides this would come zero. Subtract one third from both sides and you get c= -1/3. So using these conditions or point where this function crosses through we can now give you the particular solution to this differential equation and that is y = x3/3 x – 1/3 and we just solved for the c and if you don’t believe me, take this expression and substitute it here and you will see that it equals. If you were to take the divertive of y with the respect of x you would see that it equals x2 + 1. Let's do another one. So this is a little bit more interesting. It says dy-dy-dx = x/y and it has the same conditions. It says y = 1 at x = 1 or when x = 1, y = 1. So let's find the function that satisfies this equation. This was interesting because we have x and a y on the right hand at the side of this equation. So this kind of looks like something we got after we did some type of implicit differentiation but let's see, let's see where we go. So let's do the same thing. Well let's multiply both sides of this equation by dx until you get dy = x/y dx and let's get this y over on to the dy side because it will be easy to take the antiderivative then. So we get y dy = x dx. Now we could take the integral of both sides of this equation. We’ll take the antiderivative on this side with respect to y on this side with respect to x. So that’s the anti derivative here. Well it is y2/2 and I could per say plus c some constant and then let me do that. There could be a constant here. I call it c1. We don’t know what that constant is and that equals x2/2 + c2 or some other constant. Maybe it's the same number I don’t know but we don’t know what either of these is. I could rewrite. Well let me take the x on to that side so I could have y2/2 – x2 is equal and let me subtract this from that side so I get the constant on the right hand side of the equation, c2 - c1. I just took the c1 and put it on the right hand side, take the x2/2 put on the left hand side that’s why it's negative and we didn’t know the reason this could be any constant and this could be any constant, so the difference between two arbitrary constants could just be a third arbitrary constant so I just rewrite that as c. And so the general solution to this differential equation is y2/2 – x2/2 = c and actually let's do something else just too clean it up a little bit because once again this could be any constants. So let's multiply both sides of this equation by two and you get y2 – x2 = 2c. Well now this is still any constant number so we could still write this a c so we have y2 – x2 = c. Now let's use our initial conditions to c, what c is? So when y is one, so 12 minus when X is one, 12 =c. Well this is 1 - 1 so it’s zero right, c = 0. So what is the particular solution to this differential equation? I will do it in green. It is y2 – x2 = 0 or we could add X squared to both sides of that we could also write it as y2 = x2. Now you might be tempted to take the square root of both sides of this and say that y = x and the reason why this would not be accurate is because here x could be minus two and y could be plus two or vice versa, right. So this would satisfy this equation but it would not satisfy this equation so be careful, when you take that square root. You have to worry about the plus or minus, let's do one more. So this one says the derivative of y with the respect to x is equal to, this is the well okay, so this is even a little bit more interesting. This is equal to the square root of x/y and it says that y = 4 when x = 1. All right, yup, I think I'm reading it right but we could do this one very similarly. So at end well let's just do the same step so multiply both sides equation times dx so you get dy is equal to and I'm actually just to skip a step. I'm going to rewrite square root of x/y is the square root of x over the square root of y and I multiplied both sides of that times dx. And now let's multiply both sides of this equation by the square root of y and I'm just going to rewrite it as y1/2. That’s the same thing as square root of y, dy = x1/2 dx. I just multiplied that there and rewrote it as y½ and so what is the antiderivative of y1/2? Well it's just this plus one so it's y3/2 and then times the inverse of this, so times 2/3 or you're going to said divide it by this either way. If you're not sure about this because sometimes it is confusing with the fractions and whatever else, take the derivative 3/2 of 2/3 is one and then you subtract this one from the exponent and you get right at the ½ so that works. And of course, we have we kind of go through the same drill plus c1 is equal to, it's going to be the same thing on the side. Two thirds x to the three halves plus c2 and what can we do. Well, let's take the x to the left hand side of this equation so we get 2/3 y3/2 – 2/3 x3/2 = c2 – c1 which can we just rewrite as you know c and let's multiply both sides of this equation by three halves. So this two will become one, so you get y3/2 – x3/2 is equal to well what's three half’s times some constancy, well we didn’t know what it was. We haven’t solved for it so we can still write C, I hope that doesn’t confuse you. You know three half c. We didn’t know what it was so we could call this c3 and now this is c4. It's a different constant but we still have to solve for it. And now let's use our initial conditions. So the initial conditions should tell us that four, it don’t have to be initial conditions you can guys kind of say conditions or points where we know that the function, the general, the particular solution to this differential equation are satisfied. So 43/2 – 13/2 = c, what’s 43/2? So 41/2 = 2 and then to the three power, that’s eight and then one to any power but in especially in this case one to the square root of one is one and then to the third power is one. So eight minus one is equal to c and so c is equal to seven and so the particular solution of this differential equation I will do in a different color and it is y 3/2 – x 3/2 = 7.