A lot of what you have learned I differential equations is really just different bags and tricks and in this video, I'll show you one of these tricks and it is useful beyond this because it is always good. Maybe, one day you become a mathematician or a physicist and you have an unsolved problem some of these tricks that solve simpler problems back in you education might be a useful trick that solves some unsolved problems. So it is good to see it. If you’re taking differential equations, it might be on the exams. So it’s good to learn. We’ll learn about integrating factors.
Let’s say we have an equation that has this form. Let’s say this s my differential equation, 3xy, I’m trying to write this neatly as possible, +y2 + x2 +xy (y’) = 0. So the first - especially since we’ve covered these aggressive videos, whenever you see an equation of this from where you have some function of xy and then you have another function of x and y times y prime equal 0, you said “Oh this looks like this could be an exact differential equation.” How do we test that? Well, we can take the partial derivative of this with respect to y and what you call these is these function of x is ym. So the partial of that with respect to y, so m partially different to y so it would be 3x + 2y. If this function right here, that expression right here, that’s our function n which is the function of x and y. We take the partial with respect to x then we get, that is equal to 2x + y. And in order for this to have been an exact differential equation, the partial of this with respect to y would have to equal the partial of this with respect to x but we see here just by looking at these two. They don’t equal each other. They are not equal.
At least superficially is the way we look at just now. This is not an exact differential equation but what if there were some factor or I guess some function that we could multiply both sides of this equation by that would make it an exact differential equation. Let’s call that Mu. I want to do is I want to multiply both sides of this equation by some function Mu and then see if I can solve for that function Mu that would make it exact. Let’s try to do that.
Let’s multiply both sides by Mu. Just as a simplification, Mu could be a function of x and y, it could be a function of x, it could be a function of just x, it could be a function of just y. I’ll assume it’s just the function of x, you could assume as just a function of y and try to solve it or you could just assume it’s a function of x and y. If you assume it’s a function of x and y, it becomes a lot harder to solve for but that doesn’t mean that there isn't one. Let’s say that Mu is a function of x, and I'm going to multiply it by both sides of this equation so I get Mu of x times 3xy + y2 + Mu of x times x2 + xy times y’ and then what’s zero times any function? That’s just going to be zero. Zero times mu of x is just going to be zero but I did multiply the right hand side times Mu of x.
Remember what we are doing. Where this Mu of x is when we multiply it; the goal is after multiplying both sides of equation by it we should have an exact equation. Now, if we consider these whole things our new M, the partial derivative of this with respect to y should be equal to the partial derivative of this with respect to x. What’s the partial derivative of this with respect to y? Well, if we are taking the partial with respect to y here, Mu of x which is only a function of x is not a function of y, it’s just a constant term. If we take partial with respect to y, x is just a constant or function of x can be viewed just as the constant. So the partal of this with respect to y is going to be equal to Mu of x times 3x + 2y. That’s the partial of this with respect to y, and then what’s the partial of this with respect to x?
Well, here we will use a product rule. So we will take the derivative of the first expression with respect to x. Mu of x is no longer constant anymore since we are taking the partal with respect to x. So the derivative of Mu of x with respect to x, well that’s just Mu’ of x or Mu’ not u. Mu prime of x, Mu is a Greek letter it’s for the ma sound but it looks a lot like a u. So Mu’ of x times the second expression, x2 + xy + just the first expression. This is just a product rule, Mu of x times the derivative of the second expression with respect to x. So times, run out of space on that line, 2x + y.
Now, for this new equation where I multiplied both sides by Mu, in order for this to be exact, these two things have to be equal to each other. Let’s just remember the big picture. We are saying this is going to be exact. Now, we are going to try to solve for Mu. Let’s see if we can do that. On this side, we have Mu of x times 3x + 2y and let’s subtract this expression on both sides, so it’s (-Mu) of x times 2x + y. You see a lot of these differential equation problems; it can get kind of hairy. They are really just lot algebra. That equals what do we have left? That equals just this term right here that equals Mu prime of x times x2 + xy. Let’s see if we factor out the Mu of x here, we get Mu of x times 3x + 2y - 2x - y = Mu’ of x, the derivative of Mu with respect to x times x2 + xy.
Now, we could simplify this so we get Mu of x times. what is this? 3x - 2x = x, 2y - y so x + y is equal to, now, I'm just going to simplify this side a little bit; is equal to Mu prime of x. Let’s try an x here. The reason why I am doing that is because it seems like if I factor out an x here, I'll get an x plus y. So this is Mu prime of x times x times x + y, x times x + y is x2 + xy. That’s why I did it and I have this x + y on both sides of the equation which I will now divide both sides by since you divide both sides by x + y or we could maybe assume that is not zero. We get, that simplifies things pretty dramatically, we get Mu of x is equal to Mu’ of x times x.
Now, just the way my brain works, I like to rewrite this expression just kind of in our operator from where we write it, instead of writing it new prime of x, we could write that as D Mu D x so let’s do that. So we could write Mu of x is equal to D, the derivative of Mu with respect to x times x. This is actually a separable differential equation in of itself. This is kind of a sub differential equation to solve or broader one. We are just trying to figure out the integrated factor right here. So let’s divide both sides by x so we get Mu over x. I'm just doing this as separable equation now, is equal to D Mu, Dx and then let’s divide both sides by Mu of x then we get 1/x is equal to 1/Mu. That’s Mu of x, I'll just write one of the Mu right now for simplicity, times D Mu, Dx. I'm actually going to a horizontal right here. Multiply both sides by Dx, you get 1/x Dx is equal to 1 over Mu of x D Mu.
Now, you could integrate both sides of this and you will get the natural log of the absolute value of x is equal to the natural log of the absolute value Mu etcetera, etcetera. But it should be clear from this that x is equal to Mu or Mu is equal to x. They are identical, if you look at both sides of this equation there, you can just change x from u and it becomes the other side. So this is obviously telling us that Mu of x is equal to x or Mu is equal to x.
So we have our integrating factor and if you want, you can take the antiderivative of both sides with the natural logs and all that and you’ll get the same answer but this is just by looking at it. By inspection, you know that, Mu is equal to x because both sides of this equation are completely the same. Anyway, we now have our integrating factor but I am running out of time. So in the next video, we are now going to use this integrating factor, multiply it times our original differential equation. Make it exact and then solve it as an exact equation. I'll see you in the next video.
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