First Order Linear Differential Equation & Integrating Factor (idea/strategy/example)

everything will work out nicely in this video I'm going to talk about the idea and a strategy for how to solve a linear differential equation but before I do this with you guys let's talk about an integral first let me ask you guys this what's the integral of sine x over X and I know I did not put down with the X by was for good purpose anyways what if the answer to this and you should know that this right here has no answer right this is actually a non elementary integral so no matter what you do use substitution integration by parts Y or whatever the answer idea is just not going to be nice right but what if I tell you that maybe I can add something right here let me ask us this following two instead what if I'm going to ask you guys what's the integral sine x over X plus Ln X times cosine X and now suppose this is my actual question I want to ask you guys I'll put parentheses around this and a DX so I can be legitimate what's the answer to this integral now okay maybe you see it maybe not but let me tell you that this right here he actually has a much better answer and the answer to this integral just going to be sy x times Ln X and of course we are done I'll put on a plus C at the end well how come I can just get this for the answer well does look at this backwards okay if you just kind of pay attention to the form first of all you see that you have a addition in between of two things right and then here we have sex over X and then here we have ln x times cos x and guess what if you put this down you see signs x ln x that's the function part and if you just go ahead and differentiate this instead this is how to entire derivative right you put on the answer and you want to know if this is the right answer well now you just have to take that the root of this if you can get back that would be the antiderivative I will be the integral heart you have to use the product rule because this is a product of two functions and the product one to show you this well I will keep the first function which is sine X and then we multiply by the derivative second which is going to be one over X and then we add you see this is why I know that I will use the product to function because earlier this is a adding situation right we add the second function which is Ln X and that's a peahen DC and we multiply by the derivative of the first function the relative sex is positive cosine X you see we get back to the original integrand and of course this will be the antiderivative for that and that would be the answer earlier so that will be a way to do integral right you just kind of look at the form and then going to just notice that this right here he actually came from a product rule situation for the derivative okay now why do I bring this up let's talk about the linear differential equation if you look at this right here we have this is the phone now you want to begin with dy/dx which is the derivative plus another function times y and y it's a function of X and then you put everything else on the right-hand side and you see this is linear because the derivative okay it's just to a first power it's nine the cosine it's not being square whatsoever likewise for the Y it's only to the first power if you have Y squared if you have school of Y this wouldn't work okay so I'm going to show you guys the strategy of how to deal with this kind of situation look at this again we have the derivative and the original hmm and I'm going to use this idea again let me write it down for you guys here is the product rule I just want to put this down and I would like to write it down as if you want to take the derivative of two functions and I'll just use F times G F is the first and G is the second of course and the floor I want to write it down for you guys is this we will keep the first function and we multiply by the derivative second and we add the second function times the derivative of the first this is the product rule right and now I'm going to differentiate this for you guys you see we have the derivative and its original we have the derivative or in this case with the G Prime in front of match like right here and the original which is right here very similar right however over there we have an issue because there was nothing in front and of course it's not all the case that you have a one in front I guess a function wise anyways I want to just kind of match the phone for you guys today I will to differentiate something times y and I'm not sure what this is yet because I didn't have anything right here so that's right just leave it blank okay all right I want to differentiate something times y you can imagine this equity function okay all right using the product rule we'll have something times the derivative the second function which I will just put down dy/dx so I can match what's that from over there and then we add well this right here I don't know what it is yet and I'll just kind of like no deeper blank but then the second part is we will keep the second function which is y times y over this was and now you see this is why I put down for the linear differential equation I'm looking at the product rule once again I didn't have anything the front so that's why I leave a space right here and then for this right here is the one I have to differentiate all right now the issue is seriously we didn't have anything in front and this may not be a good phone that one to use so this is the idea and the strategy we're going to use something that we can force the left hand side here to be the derivative that came from the product of two functions and what we are going to do right here is that we'll just say we are going to use and the thing is that we will multiply what we mean by the inter grading factor okay so we can force it the left hand side to be the product of two functions and we have to use the product rule to match that well we are going to use another function and that's called the integrating factor and our label uses mu of X okay so I'm going to kind of just draw a line between this and this is what we're going to do we will multiply everything by mu of X so from here we will end up mu of x times dy/dx plus mu x times P of x times y and then on the right hand side we have to just multiply this in that and I'll just have mu of x times Q of X just like that all right now you see I'm just going to put on mu of X right that's a Newseum to have otherwise 0 see I just have blanks that's no fun right so the blank will be the mu of X and right here let me just write down mu instead of MU effects for simplicity purposes hopefully gets don't mind today if we differentiate mu which is the function of x times y which is also the function of X we use the product rule we keep the first function which is mu and then for the second part right here see we will have to differentiate the mu and then put it here and now you see we can mix and match the form mu of x times dy/dx which is this right here and then we add the words we have the Y right here that's matched right we are in order for us to come over to formula from U of X what must happen is you'll see me up x times P of X which P of X was finally original you must match moon Prime so the key is we are going to set this to be Moon Prime and they mean ready stand right here for you guys mu of X times P of X we want this to be so let's set this to be mu prime of X because this way you will see the left-hand side will be the derivative of a product of two functions right and found this right here will be able to come up with a formula to get mu of X and we will multiply everything by the mu of X the actual function and you'll see everything will work out nicely of course I'll show you guys an example with actual functions so be sure to continue to watch this video and also I would suggest you guys to watch another video it's called fake prada rule you can click the link in the description it's a lot of fun and a strategy right here it's very similar okay now let's continue what I want to do next is I would like to divide both sides by mu of X okay so that they can so and let me write this down first right here for you guys we will have mu prime of X over mu of X and this is equal to P of X and now let's look at the left hand side we have the original function in the denominator and the derivative of the numerator this it's a form that came from the derivative of a logarithm right in another word if I integrate this and the reason that were integrating is toda at a somehow get rid of the derivative right anyways let me integrate this of course we'll have to integrate both sides this right here it's going to be Ln X the value of MU of X well double check today if we differentiate this the derivative of Ln of X it's going to be 1 over the inside 1 over mu of X and then because of the chain rule we multiply by the derivative the inside that's how we get the new prime of X ok so the integral of this is equal to Ln of this if you differentiate that you get that don't worry about the plus D and on the right hand side I will just write D dance how this integral of P of X yet and once again don't worry about the plus C because we don't care about the constant we just care about the function part we just care about that function part so we can multiply the original equation by that function part every single will come next at the end we see that we have milk eggs inside the Ln what can we do to clear off the iron we do e to that e to the right so that E and now can so finally this is the formula and let me write this down right here for you guys to get milk bags and you don't even need to worry about the absolute value because once again you can have positive and one negative function both of them will work you just need to have one function that you need to make things work so let me just write down mu of X this is equal to e okay and the power is an integral you have to integrate P of X DX this is how you are going to find out what that integrating factor is and now let me show you guys an example so every single make much more sense so let me show you guys how can solve this linear differential equation with the strategy that we just talked about it and if you're doing this on your own you should pay attention to the following we see that we have the Y DX the first derivative right and here we have the Y to the first power there's no otherwise there's no other derivative this is indeed the first water linear differential equation so you can use the integrating factor the one that just showed you guys earlier right now this right here it's not in the correct folder because we have this cosine of X in the front but it's okay because we can just go ahead and divide everything by cosine of X right this way we'll be able to get into the correct form you want we will have dy/dx plus sine of X over cosine of X we can just write that down tangent of X and you see right here I put a parentheses around the X because we are to emphasize X is the only thing that insert the tangent the Y it's being multiplied it with tangent of X and this is equal to 1 over cosine of X which is just secant of X right let's put it down from here we are in business because we can see that the P of X this right here is positive tangent of X from here we will be able to work out a formula to find ourselves an integrating factor and yes you have to know that formula so right here let me write it down mooom of X for that integrating factor it's going to be e where's the base and the power is an integral right we have to integrate P of X and let me just write this down for you guys we have to integrate tangent of X right here in this situation and let's just go ahead and work things out first work out the integral the integral of tangent of X is Ln absolute value of C and X right so we still have the e for the base but this is going to give us Ln absolute value of secant of X don't worry about putting down classy right here we're not because we just need one integrated factor with just in that function part the plus see later on yes you will be found the process of solving the differential equation but for the integrating factor don't worry about it and in fact what I'm going to do next is I will just try to cancel the e and ELN and I'm just going to say mute legs instead of saying absolute value of secant X I'm just gonna use the positive version secant of X this is it okay and the reason for that is we just want to get we just need to have one function you don't want to multiply with absolute value of secant X you don't multiply by negative nobody that's negative use the positive version pass the secant of X this right here is a function that we are going to multiply throughout with our equation everything will work out nicely and I know I didn't put on the parentheses right here right here but they shouldn't bother you too much right anyways let's just multiply by secant of eggs throughout this equation and name it ready stand right here for you guys you want to apply everything by secant of X and of course we want to be sure to distribute distribute distribute and let me put on the result right here first we'll have secant of x times dy/dx and let me write it down right here next we add your words secant of x times tangent X right so let me put this down secant of x times tangent of X and then we still have this Y right here and on the right hand side secant of x times secant of X we have secant square of X all right here is the punch line on the left hand side this you see we have secant of x times the derivative Y plus y times the derivative of the secant X right this right here is nothing but just the derivative of secant x times y you see if you go back to the phone earlier this was the move X and the y and once again you can always write down the derivative and you just differentiate it to double-check the relative secant x times y is you keep the second egg and times the derivative of Y and then you add you keep the white right here and you multiply by the derivative second X which is secant tangent X so nice and you see this right here is just that will force you to be the derivative of the product is a nap the right hand side is still the secant squared of X okay and the left hand side we have a derivative how can we get rid of this derivative well we can just integrate right and of course we will have to integrate both sides on the death inside this and that will cancel and we just have this right so that's right down secant of x times y I know this looks sexy but this is secant of x times y and then this is equal to the integral secant squared X this is just tangent of X and then right here I will put down the plus C and this is because you know we are solving that differential equation this is the constant that we need right now at the end of course we want to isolate the Y it's Napper to do right but let me just write this as 1 over cosine of X first and this is x y and this is equal to tangent of X let me just ready S sub X over cosine of X and we still have this plus C because if I do care the equation this for now we can just multiply everything by cosine of X this and that cancel out and we will just have the wire left on the left hand side and then be sure you have to distribute right multiply everything by cosine of X cosine of X times this will just have sine of X and then cos of X times plus C we will just put down plus C times cosine of X and this right here is the general solution to that differential equation and of course suppose you have initial value let's say if you have this equation here along with an initial value that say y of pi is equal to 2 well now we can just plug in these values into this phone and so for C this means X is equal to PI and Y will be 2 now that's just plug in 2 into this Y will have 2 this is equal to sine of pi right because X is PI so let's write that down sine of PI plus C we don't know yet and we multiply by cosine of X which is the same as constant pi and just work this out real quick sine of pi is 0 cosine of PI is negative 1 so this means we have the 2 right here and this is just 0 and we have just C times negative 1 that's negative C and of course we can justify what set by negative 1 C will give us maybe 2 at the end if you have this initial value the solution will be Y it's equal to sine of X and here we have the plus C but you know C is negative two so we have minus two times cosine of X this right here will be the answer with this initial condition so hopefully you guys like this

Channel: blackpenredpen

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Keywords: differential equations tutorials videos, blackpenredpen differential equations, steve chow pierce college, blackpenredpen, Nagle differential equations homework solutions, calculus homework solutions, first order linear differential equation, linear differential equation, patrictjmt differential equations, kristaking differential equations, solve dy/dx+P(x)y=Q(x), first order differential equations integrating factor, integrating factor, how to find integrating factor

Id: DJsjZ5aYK_g

Length: 20min 34sec (1234 seconds)

Published: Fri Dec 30 2016