Differential Equations - AP Calculus Unit 7 Review

hey everybody today we are going to be doing unit 7 of our ap calculus review this is going to be a short one it's on differential equations um so yeah let's just get started so first off what is a differential equation so pretty much it is some equation with a derivative and the you're not trying to find a numerical answer i mean sometimes you might but typically you're trying to find what the function is so for for our purposes typically a differential equation is going to look something like d y over dx equals some function right you might have x's and y's just some function of that right but for our purposes let's first begin with a very easy one so so let me actually get rid of this for now and let's say we have f prime of x equals 2x now how do we find our original function well this is the derivative of some function right so if we want to undo the derivative we'll integrate and typically just might be like oh okay so then f of x so you can say this implies that f of x is just equal to x squared this is easy you don't even need to do any work it's just okay well this guy's derivative is that so you're done but let's do this more concretely instead of just kind of looking at it easily because some of these will not be that easy so let me just put that guy okay so yeah um how how do we think about this well let's assume that y equals f of x so i'm going to put that just so that we can write d y over dx because you could write df over dx but it's it's more common to write d y so d y over dx equals 2x these two mean the exact same thing because they're both the derivative but the difference is kind of in the connotations that one has so this guy really makes it makes it explicit that the derivative is a small change in y over a small change in x whereas this this notation kind of loses that meaning so that's an advantage of using this this one is an advantage of using it might be just that it's quicker it's quicker to write you might even just write y prime something like that um but this guy kind of makes it look like the derivative is a fraction now it kind of is because you find it by taking a fraction but it's not totally a fraction so you can't totally treat it as a fraction but we're going to do that anyway now you might be like well why if it's not totally legit then why are we doing it well the thing is it's not 100 legitimate but it is 100 acceptable so i don't know uh you can look more into that um but this is a very interesting way what we can do is multiply both sides by dx right so we're kind of getting rid of this fraction so if i if i multiply both sides by dx over here so i'm going to do dx and then i'm going to do dx right so then this and this cancel and so from here what we have left is a d y is equal to 2x dx so now we have the d y and the dx over here they're no longer in the fraction so what do we do well we can actually integrate both sides if we integrate both sides so we do this and we do that the integral of d y is equal to y plus some constant right now over here we're going to get another constant so i'm going to call this c 1 right and then on this side the integral of 2x is just x squared right this is kind of what we did over here so we have x squared plus c2 now you might be like oh my god this is there's so much right um like why why are we doing this whole thing if we already knew that it was x squared well that's because we have to kind of learn this process for in a second um and i'll just show you why because it'll it'll get a little harder in just a few minutes so okay so we did we have this we have two constants here well a constant minus another constant we can move this guy to the other side it's just gonna give us yet another constant so i can say y equals x squared plus c three and i'll just call this c3 just c because this is all we need so you can totally skip this middle step okay you can just kind of go straight to uh the fact that you're going to get one constant on one side so y equals x squared plus c right that's it right or if they ask for f of x um you can just say f of x equals x squared plus c either way works as long as this guy is true so now we have our original function and now let's say we want to find what c is because you typically don't really um you don't know c whenever you're integrating it could be any number but let's say they gave us that f of 0 is equal to 2 right so that means that 2 equals 0 squared plus c so then from there it's easy to see that 2 equals c so then your your function at the end is f of x equals x squared plus two so this is your final answer but that's only if they gave you this you know uh if they gave you another f of one equals ten you know then you then you'd have a different constant here but that's pretty much the premise so now what if we have a 2x y this changes things right so let me let me erase this okay so now what we have is this and again i'm just going to write that this is d y over dx is equal to 2xy okay so what i want to do now is i want to move everything that deals with y to the side with the with the d y and i want to move everything with x on this side everything with x is already here so we don't have to do anything um but pretty much this process is called separation of variables because we're separating the variables all the y's on this side all the x's on the other so let's do the whole thing we're going to multiply both sides by dx and divide both sides by y so this is what we have and we're going to again integrate both sides and the integral of 1 over y is just ln of absolute value of y and then we're going to get a constant there but i'm going to hold off on the constant i'm going to move it i'm going to take the constant on this side right because again we're trying to just isolate y so you know so this is again going to be x squared plus c 1 okay so now what i want to do is i just want to find what y is so i'm going to take e on both sides and these are going to cancel so we're going to get that the absolute value of y is equal to e to the x squared plus c1 now one interesting thing is that we can use a an exponent property to basically just say you know e to the x squared plus c one is equal to e to the x squared times e to the c1 okay so basically i'm just going to do that okay so i hope you can see that based on one of the like a major exponent property um and this e to the c1 e is just a number right so it's 2.7 to some constant that's going to be yet another constant so i'm going to call this c2 and it's pretty common to write write it in this order so we'll have and i'm just going to call this c2 just c so we'll have y equals c e to the x squared okay this is cool um okay but now we need you know it's either y equals the positive version of this or the negative because it's absolute value so let's say the question had an initial condition of let's say y of zero equals three right so y equals 3 so it's going to be absolute value 3 equals c times e to the 0 squared right so e to the 0 squared this is e to the 0 which is 1 so then we know that c is equal to 3. so what we have is y equals 3e to the x squared and that's our final answer but you might be like wait how do we know it's not negative that well you can really easily check because if it was negative 3 e to the x squared right based on this you know since we have the absolute value it could be the positive or the negative version if we had the negative version y equals negative three e to the x squared then it would be um then the initial condition wouldn't work because here we'd have negative three equals three that that doesn't work so i'll just you know check that way so yeah um okay now i want to do something very similar to this we have we have a good expression here but i want to do something pretty similar so let me just erase this okay so let's say we have d y over d x i'm just going to write it d y over d x equals actually let's use d y over d t uh this is going to be equal to uh k y alright where k is just some constant okay so now let's solve this again so we're going to have 1 over y d y equals k dt i'm going to integrate both sides and here we get the ln of absolute value of y i'm just going through this quickly because you guys should know what to do okay the integral of k with respect to t where k has no relation to t is just going to be kt then plus some constant c1 and then we have the absolute value of y is going to be equal to again the same thing we did last time it's going to be c2 times e to the kt and again i'll just call it c2 equal to c right so we have c e to the kt um and you should you should probably recognize this from your algebra class and i'm just going to get rid of these you should get rid uh like pretty easily recognize this maybe it had different letters maybe you had like a like p equals a e to the rt or something of that flavor this is a very important equation and it models a lot of different things compound interest population growth any sort of exponential growth anything like that can be modeled in some form like this this is a very important formula um that's used pretty much everywhere in in anything physics finance uh any sort of science you know in general um and it comes from this differential equation so this guy being derived from this means that this is also super important right so so this is where that comes from i hope i hope that that made made some sense like that some of you guys were like oh really well that's cool um and this is where that comes from so this is a very important differential equation and this basically highlights why differential equations are so important they model population growth um any sort of growth like compound interest like i said uh any any pretty much anything that changes can be modeled by a differential equation um i'll show you guys some other examples of how you can use differential equations in the next section okay so i realized i put unit one for this part i meant section one so here's section two we're gonna talk about slope fields so you might see this like weird like thing i have here um and that's what a slope field will look like in a second that's kind of a grid that i'm going to fill in so pretty much i have here the differential equation negative x over y um and i basically showed all the work and pretty much you end up with x squared plus y squared equals c2 so it's a circle the radius of c2 all right so this is interesting and that's basically kind of what you get at the end just a circle so we're going to take this guy and we're going to figure out its slope field so basically we're going to take a graph and then we're going to see the slopes at any possible point so let's start at the middle zero over zero well that's undefined right because you can't you know divide by zero um so so we can't have a slope at zero well actually i'm sorry i'm sorry oh yeah no there is no slope of zero okay so no slope there um at any place where x is equal to zero so your x value is zero so it's going to be on this line your slope your your derivative your slope is zero right so let me actually use a red marker here so you're going to have stuff like this and again we're not going to have anything in the middle because 0 0 is undefined now you you will also have like other slopes going that way but it's good to just have this you'll see why in a second and then any place where your y is zero your slope is going to be something divided by zero which is undefined but whenever we have a slope like that that means it's a vertical slope so any place where your y zero you have a vertical slope so those are those are really good um now what about the rest of the points we still have 4 8 12 16 points left so let's let's just figure it out so this is at one comma one right so if we plug in one and one here we're going to get negative one and then over here we're going to have this is this point is negative 1 comma no this is 1 comma negative 1. so plug that in here we're going to get a positive 1. here we're going to get a negative 1 and then here we're going to get a positive one if you just plug those in right so we're getting this interesting kind of spiral pattern so let's just keep going so if we plug in this point this is two comma one right so we're going to get negative two over one that's a slope of negative two that's steeper than this guy um and then here we're gonna get you know if you plug in these points you're gonna get slopes that look like this um and eventually you know you start to see what this looks like and then i'm just gonna actually leave out these points because then it looks kind of weird um but you can see that it's it's looking like a spiral like a circle and you can basically if you dropped um let's see let me get another color so let's let's say this is some water here and if you if you if you're thinking of this as like a field of just water basically let's say we dropped droplet water there it's going to rotate in this sort of pattern and it's going to keep going like that in a spiral and that's exactly what our slope field tells us so another application of differential equations is that they can model fluid flow wind patterns anything that really flows around and whenever you um take because there's a whole course on differential equations in college if you take that class you're gonna you're gonna be very familiar with these if you take calculus three you're also going to be super familiar with these that this like these are a really big portion of that at a higher level um and it's just it's just like it's it's really cool um now i didn't get too much into the details but again this should not be like teaching you this should be a review and i just want to show you the methods on on how to solve them there are a lot of other differential equations um that you should solve so don't go into the ap exam thinking oh yeah this is all i'm going to see make sure that you're really good with this i just wanted to show you this kind of stuff just at a kind of a basic level but like really so that you understand the intuition so yeah that's it for this video thank you guys so much and i will see you in the next video bye guys

Channel: Math Café

Views: 578

Keywords: math, calculus, ap, ap calculus, integral, integration, derivative, differential equation, diff eq, ODEs, ODE, DE, DEs

Id: f8D-c3pRBpA

Length: 17min 1sec (1021 seconds)

Published: Tue Feb 15 2022