Calculus: 1st-Order Linear Differential Equations

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Differential Equations
Solving First Order Linear Differential Equations
First Order Linear Differential Equations Page [1 of 2]
Another type of differential equation that we may see has the following general shape: plus some function we'll call P(x) times y = Q(x). So what we're looking at here is a derivative - - plus a function of x times one equals another function of x. If you see an equation like this, this actually is called a first-order linear differential equation - linear because, you see, in terms of y, just one y occurring to the first power, and first order because we just see one derivative, the first derivative.
Now how can you actually tackle this? Well, there's actually a powerful technique that I want to sort of illustrate the method, and then that technique often allows us to crack open this particular type of differential equation - the first-order linear variety. The idea is the following: Let's take this equation and let's just pretend for a moment that we can find a function which I'm going to call I(x), and I(x) has an amazing property, and I'm going to tell you the property right now. Let's suppose that we can find this function, which I'll call I(x), and it has the following spectacular property - that if I multiply I(x) by the left-hand side here - - so this is like a huge supposition, then what that actually will equal is nothing more than the derivative with respect to x of I(x) times y.
I'm not looking for this to be intuitive, I'm trying to show you a method that's going to allow us to crack these first-order linear differential equations. So what I'm trying to set up for us is a way of thinking about this. Let's pretend for a moment that we can find an I such that I times this side is just the derivative of I(x)y. If we assume that - if we pretend that we can do that - then what would I see? Well, then what I would see is that if I multiplied this thing through by I for real, then what I'd see would be is I times all of that, and I times all of that we know is just tis term now. So I would see . Let me just say again what I did: I just took this entire equation and multiplied both sides by I(x) - you can see it right there. But when I multiplied this side by I(x), we're assuming that that actually equals this thing. This is a huge assumption - a huge assumption. But if we assume that, then I have this equation. Then if I wanted to actually solve this, all I have to do is integrate both sides. Since the derivative or this equals that, that means that this must be the anti-derivative of this. So if I integrate both sides...let me show you a little more detail here. I could write it this way: I could say the differential of this equals I(x)Q(x), and now I can integrate. And if you integrate a derivative, you just get the function back, and I'll just keep it like that because who knows what that is.
Now you can solve this for y by just dividing through by this integrating factor. So what I see is y would equal one over this integrating factor - y = - times this integral - - plus a constant...if you want to throw that in right here, you can - - and that will give the answer.
So, in fact, if we could evaluate that integral, which is just an integral in terms of x's and maybe it's not as bad as its original-looking thing, we would have actually uncovered this. The question is: How do you find the integrating factor? So what does I equal? That's the question mark. Well, let's see if we can now figure out what I would have to be. Remember that we just assumed that I was some function that actually satisfies that. Let's come back and see if we can actually make progress on this.
Well, what would I have to be? Well, let's actually just distribute and see. What happens if I distribute here, what I would see is I(x) + I(x)P(x)y - I just distribute it - that would have to equal the derivative of this, which would require the product rule. It's a product, so the derivative is the first times the derivative of the second - so that's - plus the second, which is y - times the derivative of the first - so that would be y times the derivative of I. Well, now notice there's a cancellation because I have I(x) and I(x)on this side. So, in fact, they cancel. So now what am I left with? Well, now what I seem to be left with is just I(x)P(x)y = yI^1(x). Now the y's can actually be divided through on both sides, and they go away. So whatdo I see? What I see now is that I(x)P(x) = I^1. If I divide both sides by I, I'd see P(x) = . And why do I like that? Because, actually I can now integrate. If I were to integrate both sides with respect to x, if their things are equal, their integrals will be equal. And the integral of that, well I don't know what that is, so that's just the integral of P(x)dx. But that integral I actually know because if you make a substitution, if you let u equal the denominator, then the derivative is sitting right on top. So that integral just becomes du, which is the natural log of the absolute value of u. So in this case, that would become the natural log of the absolute value of I(x).
So if I were to solve this, I just would e both sides. So if I e both sides, on this side I'd see a cancellation, so I just would see I(x) - assuming I(x) positive - and on this side I would see e. So that's what the integrating factor must be. We actually found it. We said, "Oh, let's assume it exists," but now we actually see what it is. So the question is: What is the integrating factor?" Now we see the complete resolution. It equals e.
So let me recap what we do whenever we see a first-order linear differential equation. It's of this form, so it always looks like this: Derivative plus some function of x times just the y equals another function of x. When you see that, you first go off and create what's called an integrating factor. This integrating factor I is just equal to e raised to the power the integral of P(x)dx. So take this thing, integrate that little key piece right there, and then put that answer as an exponent on e. Once you do that, then take I, and now it's called an integrating factor, which means you just multiply it through on both sides. And once you multiply it through on both sides, the integral should be correct. I'll show this in action with an example at the next lecture. This is now really thinking about some more exotic types of differential equations, in particular we're not thinking about first-order linear differential equations. Create an integrating factor, multiply through the integral, and then the differential equation will be crackable. We'll see for ourselves in the next lecture.