In the last video, we had this differential equation that it at least look like it could be exact, where we took the partial derivative of this expression which will be called m with respect to y. It was different to the partial derivative of this expression which is kind of n in the exact differential equations world. It was different than n with respect to x. We said oh boy it’s not exact or we said what if we could multiply both sides of this equation by some function that would make it exact? We call that Mu and in the last video, we actually solved for Mu. We said well if we multiply both sides of this equation by Mu of x is equal to x, it should make this into an exact differential equation. It’s important to note, there might have been a function of y that if I multiplied by both sides, it would also make into exact. There might have been a function of x and y that would have done the trick. But really our whole goal is just to solve, is to make this exact. So it doesn’t mater which one we pick. Which integrating factor and this is called the integrating factor which integrating factor we pick.
Anyway, let’s do it now. Let’s solve the problem. So let’s multiply both sides of this equation by Mu which is a Mu of x is just x and we multiply both sides by x. So if you multiply this term by x, you get 3x2y + xy2 +, we’re multiplying these terms by x now, + x3 + x2y; y’ is equal to 0. Now, first of all just as a reality check, let’s make sure that this is now an exact equation. So what’s the partial of this expression or this kind of sub function with respect to y. Well, that’s 3x2 that’s a kind of constant coefficient on y, + 2xy, that’s the partial of with respect to y of that expression. Now, let’s take the partial of this with respect to x. So we get 3x2 + 2xy and there we have it. The partial of this with respect to y is equal to the partial of this with respect to n, so we now have an exact equation whose solution should be the same as this. All we did is we multiplied both sides of this equation by x. So really it shouldn’t change the solution of that equation or that differential equation. So it’s exact? Let’s solve it. How do we do that?
Well what we say is and since we’ve shown this exact, we know that there are some functions I where the partial derivative of zhi with respect to x is equal to this expression right here. So it is equal to 3x2y + xy2. Let’s take the entire derivative of both sides with respect to x and we will get zhi is equal to what. So x to the third y plus x and you can write ½ x squared y squared. Of course, this zhi is a function of x and y so when you take the partial with respect to x when you go that way, you might have lost some function that’s only a function of y. So instead of a + C here, there could have been a whole function of any that we lost so we’ll add that back when we take the antiderivative. So this is our zhi but we are not completely done yet because we have to somehow figure out what this function of y is and the way we figured that out as we are used the information that the partial of this with respect to y should be equal to this. So let’s set that up.
What’s the partial of this expression with respect to y? So I could write zhi. The partial of zhi with respect to y is equal to let’s see; x to the third plus say two times ½ so it’s just x squared y plus h prime of y, that’s the partial of a function purely of y with respect to y. and then that has to equal our new n or the new expression we got after multiply it by the integrating factor. So that’s going to be equal to this right here. This is hopefully be making sense to you at this point so that should be equal to x to the third plus x squared y. And interesting enough, both of these terms are on this side so let’s subtract both of those terms from both sides so x to the third, x to the third, x squared y, x squared y and we are left with h prime of y is equal to zero. Or you could say that h of y is equal to some constant. There’s really no y I guess, the extra function of y, there’s just some constant left over.
For our purposes, we can just say that zhi is equal to this because this is just a constant. We are going to take the antiderivative anyway and get a constant on the right hand side. In the previous videos, in the constants all merged together so we’ll just assume that, that is our zhi. And we know that this differential equation up here can be rewritten as the derivative of zhi with respect to x and that just falls out of the partial derivative chain rule. The derivative zhi with respect to x is equal to zero. If you took the derivative zhi with respect to x, it should be equal to this whole thing just using the partial derivative chain rule. Well, we know that zhi is so we can write or actually, we don’t even have to. We could use this fact to say well if you integrate both sides, that a solution of this differential equation if that zhi is equal to C, I just took the antiderivative on both sides.
A solution to the differential equation is zhi is equal to C so zhi is equal to x3y + 1/2x2y2. We could have said plus C here but we know the solution is at zhi is equal to C so we’ll just write that there. I could write a plus C here but then you have a plus C here and you have another constant there and you could just subtract them from both sides and then just emerge in to another arbitrary constant.
Anyway, there we have it. We have a differential equation that at least superficially looked exact. It looks exact but then when we tested the exact and then solve it was not exact but we multiply it by an integrating factor. In the previous video, we figured out that a possible integrating factor is that we could just multiply both sides by x and when we did that, we tested t and true enough it was exact and so given that was exact, we knew that this zhi would exist where the derivative of zhi with respect to x would be equal to this entire expression. So we could rewrite our differential equation like this so we’d know that the solution of zhi is equal to C. And to solve for zhi, we just say okay the partial derivative of zhi with respect to x is going to be this thing. Antiderivative of both sides, there are some constant h of y like constant. There’s some function y, h of y that we might have lost when we took the partial with respect to x so they figured that out. We take this expression.
We take the partial with respect to y and set that equal to our n expression. By doing that, we figured out that that function of y is just really some constant and we could have written that here. We could have written that plus C. But we know, we can call that C1 or something but we know that the solution of our original differential equation is zhi is equal to C. So the solution of our differential equation is zhi x to the third y plus one half x squared y squared is equal to C. We could have this plus C1 here and then subtract with both sides but I think I've said it so many times if you understand why. If h of y is just a C, you can kind of ignore it.
Anyway, that’s all for now and I will see you in the next video as you now know a little bit about integrating factors. See you soon.
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