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# Exact equations example 2

## Video transcript

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 is the fifth edition of elementary differential equations by William voice and Richard DePrima 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 equation 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 the first lesson is exact before we start talking about Zhai 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 derivative the rate of change that this changes with respect Y is 0 and what's the rate of change of district 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 and 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 there's some functions I where's I as a function of x and y where's i sub x is equal to this function is equal to 2x plus 3 and Zhai I shouldn't say sub X I say the partial of Z with respect to X and the partial of Z with respect to Y is equal to this 2 y minus 2 and if we can find the partial of if we can find ours I we oh that this is just the derivative of Zhai right because we know that the derivative with respect to X of Z is equal to the partial of Z with respect to X plus the partial of Z with respect to Y times y Prime so this is just the same form as that so if we can figure out why then we can subject we could rewrite this equation as DX the derivative of Z with respect to X is equal to zero right this let me switch colors just going to gets monotonous this right here if we can find his I where the the partial with respect to X is this the partial with respect to Y is this then this can be rewritten as this right and how do we know that because the derivative of Z with respect to X using the partial derivative chain rules is this and this is the same this partial with respect to X that's this partial with respect to Y is this times y Prime so this is a whole point of exact equations but anyway so let's figure out what hours is actually before we figure out if if the derivative of Z with respect to X is zero then if you integrate both sides you just the solution of this equation is AI is equal to C so using this information if we can solve for Z if we can solve for Z then we know that the solution of this differential equation is AI is equal to C and then if we have some initial conditions we could solve for C so let's solve for is AI so let's integrate both sides of this equation with respect to X and then we get Z 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 we have 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 purely of Y with respect to X is going to be zero 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 disrespectin this expression with respect to Y has to equal this and then we can solve for H of Y and we'll be done so let's do that so the partial of Z with respect to Y is equal to well that's going to be 0 0 0 it's going to be right this is where this part as a function of X 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 Z 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 and but if you watch 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 what is ours I function as we know it now not worrying about the plus C it is zai of x and y is equal to x squared plus 3 X 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 zai is equal to C so the solution of our differential equation is this is equal to C is x squared plus 3 X plus y squared minus 2y is equal to C and if you had some initial conditions you could test it and I encourage you to to test this out on the on this original equation or I encourage you to take the derivative of zai and prove to yourself that if you took the derivative of Z with respect to X here implicitly that you would get this 2 initial equation anyway let's do another one clear image so more examples you see the better so let's see this one says 2x plus 4y 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 was effective Y is different than the partial of n with respect to X so this is not exact 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 do another one see I'm running out of time so I want to do one that's not too complicated see we have 3x squared minus 2x Y 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