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Current time:0:00Total duration:8:26

in the last video we had this differential equation and it at least looked like it could be exact but we took the partial derivative of this expression which we could call M with respect to why it was different than the partial derivative of this expression which is kind of an inexact differential equations world with respect to it was different than n with respect to X we said oh boy it's not exact but we said what if we could multiply both sides of this equation by some function that would make it exact and we call that mu and in the last video we actually solved from you 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 that would also make it exact there might have been a function of x and y that would have done the trick but you really our whole goal is just to solve is to make this exact so it doesn't matter which one we pick which integrating factor and this is called the integrating factor which integrating factor we pick so anyway let's do it now let's solve the problem so let's multiply both sides of this equation by mu which is n mu of X is just X so we multiply both sides by X so we get C if you multiply this term by X you get 3x squared y plus X Y squared plus we're multiplying this term by these terms by X now plus X to the third plus x squared Y y prime is equal to 0 well 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 it's three x squared that's just kind of constant coefficient on y plus two X Y 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 three x squared right plus 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 it really shouldn't change the solution of that equation or that differential equation so it's exact let's solve it so how do we do that well we say is if since we've shown this exact we know that there's some functions I where the partial derivative of Z with respect to X is equal to this expression right here so it's equal to 3x squared y plus X Y squared let's take the antiderivative of both sides with respect to X and we'll get Z is equal to what see it's X to the third Y plus C X we'll get right 1/2 x squared Y squared and of course this is I is a function of x and y so when you 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 plus C here there could have been a whole function of Y that we lost so we'll add that back when we take the antiderivative so this is hours I but we're not completely done yet because we have to somehow figure out what this function of Y is and the way we figure that out is we use the information that the partial of this with respect to Y should be equal to this so let's set that up so what's the partial of this expression with respect to Y so it would be so I could write Z the partial of Z with respect to Y is equal to let's see X to the third plus c2 times one half so it's just x squared y plus h prime of Y right 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 we multiplied by the integrating 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 these 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're left with H prime of y is equal to 0 or you could say that you could say that H of Y is equal to some constant and in in in the so there's really no there's really no y I guess the extra function of Y there's just some constant left over so for the for our purposes we can we can just say that zai is equal to this because this is just a constant so we're going to take the antiderivative anyway and get a constant on the right hand side and in the previous videos the constants all merge together so we'll just assume that that is RSI and we know that this this differential equation up here can be re-written as the derivative of Z with respect to X and that just falls out of the partial derivative chain rule the derivative of Z with respect to X is equal to 0 right if you took the derivative of Z with respect to X it should be equal to this whole thing just using the partial derivative chain rule well we know what Z is so we can write or actually we don't even have to we could use this fact to say well if we integrate both sides that the solution of this differential equation is that Z is equal to C alright I just took the antiderivative of both sides so a solution to the differential equation is AI is equal to C so zai is equal to X to the third y plus one-half x squared Y squared and we could have said plus C here but we know the solution is that Z is equal to C so we'll just write that there right I could have write in a plus C here but then you have a plus C here and you have another constant there and they kind you could just subtract them from both sides and they just merge into another arbitrary constant but anyway there we have it we add a a differential equation that at least superficially looked exact it looked exactly the exactness of it it was not exact but we multiplied it by an integrating factor and the previous video we figured out that in a possible integrating factor is that we could just multiply both sides by X and when we did that we tested it and true enough it was exact and so given that it was exact we knew that as I would exist where the derivative of Z with respect to X would be equal to this entire expression so we could rewrite our differential equation like this and we know that a solution is AI is equal to C and to solve for Z we just say okay the partial derivative of Z with respect to X is going to be this thing antiderivative of both sides there's some constant H of Y that constant there's some function of Y H of Y that we might have lost when we took the partial with respect to X so to figure that out we take this expression take the partial with respect to Y and set that equal to our n expression and by doing that we figured out that that function of Y is really just it's 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 C 1 or something but we know that the solution of our original differential equation is y is equal to C so the solution of our differential equation is sy X third y plus one-half x squared Y squared is equal to C we could have had a just you know this plus c1 here and then subtracted both sides but I think I've said it so many times and you understand why I if H of Y is just to see it you can kind of ignore it anyway that's all for now and I will see you in the next video you now know a little bit about integrating factors see you soon