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Current time:0:00Total duration:10:16

a lot of what you'll learn in differential equations is really just different of different bags of tricks and in this video I'll show you one of those tricks and it's it's useful beyond this just because it's always good when if maybe one day you become a mathematician or a physicist and you have an unsolved problem some of these tricks that solve simpler problems back in your education might be a useful trick that solves some unsolved problems so it's good to see it and if you're taking differential equations it might be on an exam so it's good to learn so we'll learn about integrating factors so let's say we have an equation that has this form let's say this is my differential equation 3 X Y I'm trying to write as neatly as possible plus y squared plus x squared plus XY times y prime is equal to 0 so the first especially since we've covered this in recent videos whenever you see an equation of this form where you have some function of X Y and then you have another function of x and y times y prime equals 0 you said oh this looks like this could be an exact differential equation and how do we test that well we can take the partial derivative of this with respect to Y and we could call this our this function of x and y m so the partial of that with respect to Y so M the partial with respect to Y C would be 3x plus 2y and if this function right here that expression right there that's our that's our function n which is a function of x and y we take the partial with respect to X and we get that is equal to 2x plus y and in order for this to have been an exact differential equation the partial of this with respect to Y would have to equal the partial of this with respect to X but we see here just by looking at these two they don't equal each other they're not equal so at least superficially the way we looked at just now this is not an exact differential equation but what if there were some factor or I guess some function that we could multiply both sides of this equation by that would make it an exact differential equation so let's call that mu mu so what I want to do is I want to multiply both sides of this equation by some function mu and then see if I can solve for that function mu that would make it exact so let's try to do that so let's multiply both sides by mu and just as a simplification mu could be a function of x and y it could be a function of X it could be a function of just just X it could be a function of just Y I'll assume it's just a function of X you could assume it's just a function of Y and try to solve it or you could just assume it's a function of x and y if you assume it's a function of x and y it becomes a lot harder to solve for but that doesn't mean that there isn't one so let's say that mu is a function of X and I'm going to multiply it by both on this equation so I get mu of x times 3x y plus y squared plus mu of x times x squared plus X Y times y Prime and then what's 0 times any function well it's just going to be 0 right 0 times mu of X is just going to be 0 but I did multiply the right hand side times mu of X and remember what we're doing we are this this mu of X is when we multiply the goal is after multiplying both sides equation by it we should have an exact equation so now if we consider this whole thing our new M the partial derivative of this with respect to Y should be equal to the partial derivative of this with respect to X so what's the partial derivative of this with respect to Y well if we're taking the partial with respect to Y here mu of X which is only a function of X it's not a function of Y it's just a constant term right when you take a partial with respect to Y X is just a constant or a function of X can be viewed just as a constant so the partial of this with respect to Y is going to be equal to MU of X times we could just say x 3x plus 2y that's the partial of this with respect to Y and then what's the partial of this with respect to X well here we'll use the product rule so we'll take the derivative of the first expression with respect to X mu of X is no longer constant anymore since we're taking the partial with respect to X so the derivative of MU of X with respect to X well that's just just u prime of X mu prime not u mu prime of X mu is the Greek letter it's for the sound but it looks a lot like a u so mu prime of x times the second expression x squared plus XY plus just the first expression this is just the product rule mu of X times the derivative of the second expression with respect to X so x ran out of space on that line 2x plus y and now for this this new equation that I multiply I multiply both sides by mu in order for this to be exact these two things have to be equal to each other so let's just remember the big picture we're kind of we're saying this is going to be exact and now we're going to try to solve for MU so let's see if we can do that so let's see we have on this side we have mu of x times 3x plus 2y and let's subtract this expression from both sides so it's minus mu of x times 2x plus y you'll see a lot of these differential equation problems that get kind of hairy they're really just a lot of algebra and that equals what do we have left I'll write it in yellow that equals I think I'm going to run out of space so I'm going to do it a little bit lower that equals just this term right here that equals mu prime of x times x squared plus X Y and let's see if we factor out a mu of X here we get mu of x times 3x plus 2y minus 2x minus y is equal to MU prime of X the derivative of U with respect to x times x squared plus XY now we can simplify this so we get mu of X mu of x times what is this 3x minus 2x is X 2y minus y so X plus y is equal to and I'm just going to simplify this side a little bit is equal to MU prime of X let's factor out an X here and I'm the reason why I'm doing that is because it seems like if i factor out an X here I'll get an X plus y so so this is mu prime of x times X times X plus y right x times X plus y is x squared plus XY so that's why I did it and I have this X plus y on both sides this equation which I will now divide both sides by so if you divide both sides by X plus y we could maybe assume that it's not 0 we get that simplifies things pretty dramatically we get mu of x is equal to MU prime of x times X and now for just the way my brain works I like to rewrite this expression just kind of in our operator form where we write instead of writing it u prime of X we could write that as D mu DX so let's do that so we could write mu of X is equal to D the derivative of U with respect to X times X and this is actually a separable differential equation in of itself it's kind of a sub differential equation to solve our broader one we're just trying to figure out the integrating factor right here so let's divide both sides by X so we get we get mu over so over X I'm just doing this is just a separable equation now is equal to D mu DX and then let's divide both sides by mu of X and we get 1 over X is equal to is equal to 1 over mu that's mu of X I'll just write 1 over mu right now for simplicity times D mu DX I'm actually going to go a horizontal right here multiply both sides by DX you get 1 over X DX is equal to 1 over mu of X D mu now you could integrate both sides of this and you'll get the natural log of the absolute value of X is equal to the natural log of the absolute value etc etc but it should be pretty clear from this that X is equal to MU or mu is equal to X right they're identical if you look at both sides of this equation there you can just you know change X from you and it becomes the other side so this is obviously telling us that mu of X is equal to X right or mu is equal to X so we have our integrating factor and if you want you can take the antiderivative of both sides with the natural logs and all of that and and you'll get the same answer but this is just a just by looking at it by inspection you know that mu is equal to X because both sides of this equation are completely the same anyway we now have our integrating factor but I am running out of time so in the next video we're now going to use this integrating factor multiplied it times our original differential equation make it exact and then solve it as an exact equation I'll see you in the next video