Exact equations example 2

Let's do some more examples with exact differential equations. And I'm getting these problems from page 80 of my old college differential equations books. This is the fifth edition of Elementary Differential Equations by William Boyce and Richard DiPrima. I want to make sure they get credit, that I'm not making up these problems. I'm getting it from their book. Anyway, so I'm just going to give a bunch of equations. We have to figure out if they're exact, and if they are exact, we'll use what we know about exact differential equations to figure out their solutions. So the first one they have is, 2x plus 3, plus 2y minus 2, times y prime is equal to 0. So this is our M of x and y-- although, this is only a function of x-- and then this is our N, right? You could say that's M, or that's N. You could also say that, if this is exact-- well, first let's [? test ?] this exact, before we start talking about psi. So what's the partial of this, with respect to y? The partial of M, with respect to y. Well, there's no y here, so it's 0. The rate of change that this changes with respect to y is 0. And what's the rate of change this changes, with respect to x? The partial of N, with respect to x is equal to-- well, there's no x here, right? So these are just constants from an x point of view, so this is all going to be 0. But we do see that they're both 0. So M sub y, or the partial with respect to y, is equal to the partial with respect to x. So this is exact. And actually, we don't have to use exact equations here. We'll do it, just so that we get used to it. But if you look here, you actually could have figured out that this is actually a separable equation. But anyway, this is exact. So knowing that it's exact, it tells us that there's some function psi, where psi is a function of x and y. Where psi sub x is equal to this function, is equal to 2x plus 3, and psi-- I shouldn't say sub x. I say the partial of psi, with respect to x. And the partial of psi, with respect to y, is equal to this, 2y minus 2. And if we can find our psi, we know that this is just the derivative of psi. Because we know that the derivative, with respect to x of psi, is equal to the partial of psi, with respect to x, plus the partial of psi, with respect to y, times y prime. So this is this just the same form as that. So if we can figure out y, then we can rewrite this equation as dx, the derivative of psi, with respect to x, is equal to 0. Let me switch colors, or it's going to get monotonous. This right here, if we can find a psi, where the partial with respect to x, is this, the partial with respect to y, is this, then this can be rewritten as this. And how do we know that? Because the derivative of psi, with respect to x, using the partial derivative chain rules, is this. This partial with respect to x, that's this. This partial with respect to y, is this, times y prime. So this is the whole point of exact equations. But anyway, so let's figure out what our psi is. Actually, before we figure out, if the derivative of psi, with respect to x, is 0, then if you integrate both sides, you just-- the solution of this equation is psi is equal to c. So using this information, if we can solve for psi, then we know that the solution of this differential equation is psi is equal to c. And then if we have some initial conditions, we could solve for c. So let's solve for psi. So let's integrate both sides of this equation, with respect to x. And then we get psi is equal to x squared plus 3x, plus some function of y. Let's call it h of y. And remember, normally when you take an antiderivative, you have just a plus c here, right? But you can kind of say we took an anti-partial derivative. So when you took a partial derivative, with respect to x, not only do you lose constants-- that's why we have a plus c, normally-- but you also lose anything that's a function of just y, and not x. So for example, take the partial derivative of this with respect to x, you're going to get this, right? Because the partial derivative of a function, purely of y, with respect to x, is going to be 0. So it will disappear. So anyway, we take the antiderivative of this, we get this. Now, we use this information. Well, we use this information. We take the partial of this expression, and we say, well, the partial of this expression, with respect to y, has to equal this, and then we can solve for h of y, then we'll be done. So let's do that. So the partial of psi, with respect to y, is equal to-- well, that's going to be 0, 0, 0. This part is a function of x. If you take the partial with respect to y, it's 0, because these are constants, from a y point of view. So you're left with h prime of y. So we know that h prime of y, which is the partial of psi, with respect to y, is equal to this. So h prime of y is equal to 2y minus 2. And then if we wanted to figure out what h of y is, we get h of y-- just integrate both sides, with respect to y-- is equal to y squared plus-- sorry-- y squared minus 2y. Now, you could have a plus c there, but if you watched the previous example, you'll see that that c kind of merges with the other c, so you don't have to worry about it right now. So what is our psi function, as we know it now, not worrying about the plus c? It is psi of x and y is equal to x squared plus 3x, plus h of y-- which we figured out is this-- plus y squared, minus 2y. And we know a solution of our original differential equation is psi is equal to c. So the solution of our differential equation is this is equal to c. x squared plus 3x, plus y squared, minus 2y is equal to c. If you had some additional conditions, you could test it. And I encourage you to test this out on this original equation, or I encourage you to take the derivative of psi, and prove to yourself that if you took the derivative of psi, with respect to x, here, implicitly, that you would get this differential equation. Anyway, let's do another one. Let's clear image. So the more examples you see, the better. So let's see, this one says 2x plus 4y, plus 2x minus 2y, y prime is equal to 0. So what's the partial of this with respect to y? So M, the partial of M, with respect to y-- this is 0-- so it's equal to 4. What's the partial of this, with respect to x, just this part right here? The partial of N, with respect to x, is 2. This is 0. So the partial of this, with respect to y, is different than the partial of N, with respect to x. So this is not exact. So we can't solve this using our exact methodology. So that was a fairly straightforward problem. Let's do another one. Let's see. I'm running out of time, so I want to do one that's not too complicated. Let's see, 3x squared minus 2xy-- actually, let me do this in the next problem. I don't want to rush these things. I will continue this in the next video. See you soon.