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Current time:0:00Total duration:12:09

okay I filled your brain with a bunch of partial derivatives and XY's with respect to X's and Y's I think now it's time to actually do it with a real differential equation and make things a little bit more concrete so let's say I have the differential Y the differential equation y cosine of X plus 2x e to the y plus sine of X plus already running out of space x squared e to the Y minus 1 times y prime is equal to 0 well you could probably already your brain is already hopefully in the exact differential equations mode but if you were to just see this pattern in general where you see a function of x and y here right this is just some function of x and y and then you have another function of x and y times y prime or x dy d of X your brain should immediately say if this isn't separable and I'm not going to try to make it separable because that will take a lot of time but if it's not separable your brain said oh maybe this is an exact equation and you say how is let me let me test whether this is an exact equation so if this is an exact equation this is our function M which is a function of x and y and this is our function n which is a function of x and y now the test is to see if the partial of this with respect to Y is equal to the partial of this with respect to X so let's see the pet the partial of M with respect to Y is equal to CY is so this cosine of X tends to constant so it's just cosine of X cosine of X plus now what's the derivative well 2 X is just a constant what's the derivative of e to the Y with respect to Y what's just e to the Y right so we have the constant on the outside two x times the derivative with respect to Y so it's 2x e to the Y fair enough now what is the partial derivative of this with respect to X so n sub x or the partial of n with respect to X so what's the derivative of sine of X with respect to X well that's easy that's cosine of X plus two x times e to the Y right the Y is just a constant because Y is constant when we're taking the partial with respect to X so plus 2x e to the Y and then minus one the derivative of a constant with respect to anything is going to be zero so the derivative of n with the partial of n with respect to X is cosine of X plus 2x e to the Y which lo and behold is the same thing as the derivative the partial of M with respect to Y so there we have it we've shown that M of Y is equal to or the partial of M with respect to Y is equal to the partial of n with respect to X which tells us that this is an exact equation now given that this is an exact equation given that this is an exact equation oh yeah my wife snuck up behind me I was wondering whether I thought there was some critter in my house or so anyway so so we know that this is an exact equation so what does that tell us well that tells us that there's some zai where the partial derivative of Z with respect to X is equal to m and the partial derivative of Z with respect to Y is equal to N and if we know that zai then we can rewrite our differential equation as the derivative of Y with respect to X is equal to 0 so let's solve for zai so we know that the partial of Z with respect to X is equal to M so we could write that we could write we could write the partial of Z with respect to X is equal to M which is y cosine of X plus 2x e to the Y right that's just here that's my M of X we could have done it the other way we could set the partial of Y the partial of Z with respect to Y is this thing over here but let's just do it with X now to at least get kind of a first approximation of what size an approximation but start to get a sense of it let's take the derivative of both sides with respect to I'm sorry take the antiderivative take the integral of both sides with respect to X so if you take the derivative of this with respect to X if you if you integrate aside if you were to take the antiderivative of this with respect to X so this is so let me just write that down the partial with respect to X we're gonna take the integrate respect to X that is going to be equal to the integral of this whole thing with respect to X cosine of X plus 2 X e to the Y we're integrating with respect to X and normally when you integrate with respect to X you'd say ok plus C right but it actually could be a plus it could be since this was a partial with respect to X we could have had some function of Y here in general it couldn't it because Y we treated as a constant right and that makes sense because if you were to take the partial of both sides of this with respect to X if you were to take the partial of a function that is only a function of Y with respect to X we've gotten a 0 here so when you take the antiderivative we're like oh well there might have been some function of Y here that we lost when we took at the partial with respect to X so anyway this will simplify to zài right zài is going to be equal to the integral with respect to X or the antiderivative with respect to X here plus some function of Y that we might have lost when we took the when we took the partial with respect to X so let's do that let's figure out this integral I'll do it in blue so Y is just a constant so the antiderivative of Y cosine of X is just Y sine of X plus e to the Y is constant so 2 X the antiderivative 2 X with respect to X is x squared so it's x squared e to the Y and then plus plus some function some function of Y and if you want to verify this just you take the partial of this with respect to X if you take the partial of this with respect to X you're going to get this in here which is our function M up here and then we take the partial of this with respect X you'll get 0 and it'll get lost ok so we're almost there we've almost figured out hours.i but we still need to figure out this the function of this function of Y well we know that if we take the partial of this with respect to Y since this is an exact equation we should get this we should get our end function so let's do that so the partial I'll switch notation just to expose you to it the partial of Z with respect to Y is going to be equal to so here Y sine of X sine of X is just a constant Y is just Y so the derivative this respect to Y is just sine of X plus derivative of e to the Y is e to the Y X square is just a constant so it's just x squared e to the Y plus what's the partial of f of Y with respect to Y which is going to be F prime of Y and we also so we what what did we do we took em we integrated with respect to X and we said well we might have lost some function of Y so we added that to it and then we took the partial of that that Zhai that we've almost constructed and we've took it we took the partial of that with respect to Y now we know since this is exact that that is going to equal our n so our n is up there cosine of X plus so that's going to be equal to I want to make sure I can read it up there to our n right oh no sorry n is up here all right is up here sine of X let me write that sine of X plus x squared e to the Y minus 1 so sine of X sine of X plus x squared e to the Y minus 1 plus x squared e to the Y minus 1 that was just our n from our original differential equation and now we can solve for f prime of Y so let's see we get sine of X plus x squared e to the Y plus F prime of Y is equal to sine of X plus x squared e to the Y minus 1 so let's see we can delete sine of X from both sides we can delete x squared e to the Y from both sides and then what are we left with we're left with F prime of Y is equal to one and then we're left with we're left with F of Y is equal to well it equals y plus plus some constant C right so what is ours I now we wrote ours I up here and we had this F of Y here so we can rewrite it now so zai is a function of x and y we've actually pretty much almost done solving it zai as a function of x and y is equal to Y sine of X plus x squared e to the y plus y oh sorry this is F prime of y minus 1 so this is a minus 1 so this is a minus y plus C this is going to be a minus y plus C so we solved for zai and so what does that tell us well we said that original differential equation up here using the partial derivative chain rule that original differential equation can be re-written now as the derivative DX of Zhai is equal to Z as a function of x and y is equal to 0 or if you were to integrate both sides of this you would get that zai of XY is equal to C is a solution of that differential equation so if we were to set this as equal to C that's the differential equation so we could say Y sine of X plus x squared e to the Y minus y now we could say plus you know this C plus this C call that c1 is equal to c2 well you can subtract the C's from both sides and just be left with the C at the end but anyway we have solved this exact equation one first by recognizing it it was exact by taking the partial of this with respect to Y and say seeing if that was equal to the partial of n with respect to X once we saw that they were equal well I okay this is going to be exact so let's figure out zai since this is exact M is going to be the partial of Z with respect to X n is the partial of Z with respect to Y then to figure out why we integrated M with respect to X and we got this but since we said oh well you know instead of a plus C it could have been a function of Y there because we took the partial with respect to X and this might have been lost to figure out the function of Y we then took ours put ours I that we figured out took the partial of that with respect to Y got this and we said this was an exact equation so this is going to equal our n of X our n of X Y we set those equal to each other and then we solved for f of Y and then we had our finals I our finals I was this and then the differential equation because of the chain rule of partial derivatives we could rewrite the differential equation as this the solution is this and so this is the solution to our differential equation see you in the next video