Reducible Second Order Differential Equations, Missing X (Differential Equations 27)

hey welcome to another video we are ready to get going on some more reducible second-order differential equations now if you remember from the last video we did this and we talked about there's three cases when we can solve some second-order differential equations and that's when they're reducible right now that later on we'll learn some other techniques but from the last video we learned that there's really three cases where we can reduce a second-order differential equation into a first-order so that we can use some of the techniques that we know to solve them and a lot of times they end up being separable so I want to review the the cases with you and then we'll talk about our second case which we haven't haven't done so far we'll do about five examples and then we should call it good and that next video will move on to what exact equations exactly mean and then how to solve them so I know some of you are really looking forward to that video that's coming up next I'll do three videos on exact equations because I feel that they're so important so we'll talk about that a little bit later so for right now what a reducible second order difference equation means is that we can make this substitution so normally you'll find this under a substitution heading and like a textbook or something where it will change the second-order into a first-order and there's three times that we can do this if our second order differential equation is missing the Y variable so Y GaN or if it's missing the X variable so X gone or if it's meant the missing both x and y so both x and y are gone and we'll talk about how this is really a case which should be done like a case one so here's the idea if our Y variable is completely gone that means that we have Y prime or Y double prime for the first or second derivative and second derivatives but we don't have the variable Y in other words all the other variables besides the second derivative of Y and perhaps the first term Y are all X's we can make a substitution we say let's just call this the first derivative of Y P all right let make sense because if we take a second derivative then the second derivative of Y with respect to X on both sides would give us a derivative of P with respect to X so all right that's that's not so bad so we replace all the first derivatives of Y with P all the second derivatives of Y with DP DX and what that does is that reduces the the order so instead of a for a second order we have now a first order differential equation in terms of a dependent variable P where X is still our independent variable you see if Y is gone then everything's in terms of X and that actually makes sense okay I can actually take this derivative with respect to X because there's X's involved I can do integrals with respect DX because we have X's involved now let's look at the second case all right a little bit second case well if X is gondola it's let's let's try the same thing let's say um let's just make our first derivative of Y with respect to X equal to P okay that should reduce the order of ikana first derivative into a variable of some sort that's not we were looking at it as if it's not a derivative right now we're going to go backwards but if we treat this as just a variable and we take a derivative on both sides notice how the second derivative became like it looks like a first derivative of P and that's that's what we're doing here so let's let's try it again okay well the Y double prime would be the derivative of P with respect to X well that looks the same right but this is a problem because if exes are gone and we get down to here and say let's find let's treat this as DP DX or derivative of P with respect X and when we try to integrate well there's no X's so when we do it integral DX we're going to run into a lot of problems so if we had to do derivatives we'd run into a lot of problems because if X is gone that that doesn't really work for us so what we're going to do is a nice little use of the chain rule to switch this just just a bit know what's a chain rule do do you remember what the chain rule does the chain rule says man if I have if I have something like DP DX but this is hard to find I can write this as dp/dy or sometimes I'll see a DP D u times dy/dx and that's what the chain rule says it says I can do this derivative by considering this one if I have some intermediary variable and so that's we're going to treat the sets we're gonna say okay for Y gone this is fine because I got X's if Y's are gone I'll only have X's then dy/dx is P great Y double prime gives us the derivative of P with respect to X and I have X's in there no problem but if X is gone okay the river wives rate the XS p take the second derivative great y double prime is DP DX that's a problem I don't have any X's so we're going to use this chain rule and say instead of thinking of this as DP DX let's make this y double prime equals derivative of P with respect to Y times the derivative of Y with respect to X because I still don't have exes yeah you're right but here's the cool part battery this is exactly what you define P as so if you defined P as Y prime or the first derivative of Y and we get down here say okay let's take a second derivative then y double prime is DP DX no chain rule that derivative of P with respect to Y times do you my DX by the chain rule but that right there is P and so we get a nice substitution that looks like this since this is P we can have P times DP dy and and when we think about that I hope that makes sense that chain rules if this has only wisent no X's and we make it all the way down to here this is not a problem before we define that as the first derivative of Y from step y is in it it has X is implicitly in there but then we also have a derivative of P with respect to Y and why as the variable here so all of our intervals are going to be dy until I get to the very very end and the integrate with both sides and then we'll well we still treat it as dy but then our Y is gonna our functions can be implicitly defined so our Y's do have X is somewhere and there we assume perhaps they're just not in our differential equation let me walk through this one more time to make sure you understand the concept we'll do one example and do like four more after that so the first thing I want you to notice is there are three cases the first case if it Y's are gone it's a very very straightforward substitution we just call the first derivative peeing then the second derivative is DP DX and it works out nicely that's the last video watch it if you haven't let's jump down here it X is gone we do the same exact substitution the first derivative is P but we've run into a problem with DP DX because there are no X's so we use that chain rule alright DP dy dy DX oh hey that's what I defined PS so we get something that's totally defined in terms of why we treat Y as a variable works out great I'm going to show this one time on this example but after that I'm just gonna jump right into this substitution from here on out now the last case what if both are not what if you have no X's and no wives if you have no X's and you have no wise use case one because the substitution is nicer I did that in the last example the last video so if you don't remember that you might want to check that out I'm not gonna do this one in this video because it should just be done like that and that's our last video so let's start with this example let's make sure you can go through all I'll really kind of go slowly on this one and then we'll kind of fly through some of this form or why would I do four more when I explain this one so well is because there's some as always I like to give you some integration techniques as we go through because I know that we forgive them and that's okay it happens from time to time so I'm gonna review some integration technique some weird things that can't happen as we're going through here so while the technique will be the same for all five examples I want you to focus on solving them and how we should end our problem so let's start this if we have y double prime equals 2y y Prime we're looking at is that differential equation yeah is it the second order absolutely is it reducible well are we missing all our lies no we got a while right there we go I'm gonna Y double prime y prime we have a lie will be our missing our X's so let's see how this substitution actually works what we're going to look at doing the same you know it's reducible I'm gonna try to change the second derivative of Y into the first derivative of something else and we call it something else P so let's take P and say our first derivative of Y with respect to X or Y prime it's going to equal P well that means that the second derivative of Y is if I take the derivative of both sides with respect to X I get the second derivative of Y with respect to X equals the first derivative of P with respect to X and what that does is that changes our second derivative of Y into a first derivative of some other variable P that's why that's reducible we just reduced org now I hope you've seen the problem our problem here is that that's a DX I have no X's I run into a big issue and this is where that change will I mentioned earlier comes into play to see let's instead of doing this let's use that chain rule to our advantage let's call this derivative of Y with respect to Y times the derivative of Y with respect to X by the chain rule and then as I mentioned earlier are you seeing it or I seem like this first derivative is P that's great take a driven both sides it reduces the order calls a second derivative a first or some other variable do you see the problem if I have no Exodus that DX screws us up so we use the chain rule and say the second derivative Y is DP DX yeah but let's chain rule that thing and then recognize that this piece is already defined that piece is exactly what we called P that's pretty cool so we're gonna say y double Prime is not just derivative of P with respect to X like we did in case one what it is is P times D anyone you know a lot of students get confused right there we go where in the world does that P come why do I have a line right there you have a Y dy right there because you need it because you're in terms of Y only you don't have any X's so you have to have that the second thing I want to understand is that that P comes from a chain rule so you are making the exact same substitution I hope you see it in this case in this case start up exactly the same the only reason this P comes into play is because you are using the chain rule so if your if your teacher asks you your man how it works okay come from just a chain rule because that's exactly what you're doing here you sayin let's use the chain rule this is already defined as P and so we gain extra P so the substitution we're about to make is going to look just a little bit different from before we're still going to call this Y prime P so this is still P that Y is still gonna be there but that's gonna look a little bit different than case one let's let's go kind of slowly to make sure we're understand exactly what's going on here on the right hand side you're still going to have a two you're still going to have a Y look that Y prime that first derivative of the Y with respect to X we're gonna call that P on the left hand side we can't just make this because I have no X's look at what would happen if we did that and maybe this will make it sink in for you so really pay attention here if we call this DPD bags did you see how I have three variables floating around you see that we have a P we have a DP we have a P we have Y we have an X that's problem because there's no way to to use this as a separable equative you several equations here whether there's no way for us to deal with those three variables at one time with the techniques that we understand and so that's why this is a problem that and so that's what we did that thing and said well in order for this to be separable that's got to be really ID live that way through my T's on one side DP why isn't one side dy then I can integrate but in order to get that it involves a chain rule in order to get from here to here you have to tack on a dy/dx you have to tack on a P and that's exactly what this substitution says it says let's just let's go ahead and let's make liable prime chain rule of DP DX or P DP dy does that make sense to you if that doesn't make sense you I need you to relaunch the last couple minutes make sure that you can go through here understand that we're just using the chain rule understand the problem with having DX if your X's are all gone because you'd have wise in their X's I'll be all gone that'd be a problem last thing I want to mention before we go on the reason why we use this one for both cases is not because it's the only one that's possible you could actually use both imaginers imagine that that why wasn't there that what why wasn't there your DX would be perfectly valid your DUI would be perfectly bad be fine but we choose the Y's are gone because it's an easier substitution I just want to mention that to you that both techniques are valid if both your variables are gone that one's just easier so let's get back to it we uh man we've noticed reducible second-order we've made a great substitution to say Y prime is now key we're taking a second room and said DP DX is a problem because I have no exits I have wise though that's a problem so I'm gonna take a second derivative but I'm gonna use a chain rule and now we look at and go this is this is pretty nice we got something that's separable let's get our P's DP let's get our constants with y dy if we divide both sides well we're just going to get and you would make some some domain restrictions where appropriate so be careful with that most of time a lot of text we're gonna say just assume that these things are positive or not zero where you need them to be and that's that's very common so just take a look at your textbook there probably is a little blurb about that somewhere if you're working out of the textbook so we're gonna divide both sides by P and we're also going to move our dy so dividing by P we're just gonna get DP on the right hand side we'll get 2y dy and the nicest integral we've added a really long time if we integrate both sides with respect to y notice that we're integrating both sides but we are with respect to why we're treating Y is the variable here why because it's the only variable that we have there's no X's we can't really do that we're going to assume that Y is implicitly defining X on the left hand side we just get P on the right hand side and we get Y squared but notice you're going to get a constant we're going to call that c1 because we're definitely going to change that at the end so we're integrating oh yeah that's P this one's really nice y squared plus c1 and then we realize that what P is P is really a first derivative with respect to X so first derivative of Y with respect to X so just like case 1 we're going to have to go backwards and say because P is that guy up there because P is the derivative of Y with respect to X we're gonna change this that's what P is that's what we defined it as and then we have another integral to do when they have another differential equation to solve so the whole technique of reducible second-order differential equations boils down to this fact you're trying to take the second derivative reduce it to a first derivative of some dummy variable like like P then what we do once we could solve for P well that is a first derivative of Y with respect to X so we're basically doing two differential equations at one problem now we have this thing so well how do I solve that do I just integrate both sides no not really because right now notice this this real important right now because we have that DX X now come becomes our independent variable we're going to integrate with respect to X now and so this becomes separable just like we did before just like actually up here this was separable where we sort of treated Y is our dependent variable but now this is separable or between X as our dependent variable it's always that that looks like the denominator of a fraction that bottom part right there that's telling you what you're treating as you're integrating variable there that as well you're let's say independent variable there so here we're treating X like that that means that we need to get our wise dy on one side and our DX and other so if we divide by Y squared plus c1 and we get our constants DX do you see what I mean about that do you see what I mean like we just switched variables so with a case - it's a little bit weird because you're using Y is your variable until you get all the way down to solve for P and then you go alright what's what's P well it's the first rivet of life with respect X right there you're changing back into that independent variable being X so that's that's the idea hope that I hope that makes sense I'm gonna explain that well know if we understand that so right here we're saying okay now now we got it we can do one more step and go treat this as the first root of it is let's make sure that we understand that separable in this case Y is on one side with dy X and other sides of constants DX and now we can integrate so when we do well on the right hand side that's pretty nice we're just going to get X plus C something on the left hand side though that looks like a tan inverse now I'm gonna do something real interesting it's interesting to me when you do a tan inverse you can look at your table or something when you do a tan inverse of what it would that be alert variables Y then you have a constant C sub one do you remember what you really do you treat this as Y squared and you treat this like some other number squared so this is in Y squared plus the thing for you would get Y over 2 or the square root of 4 now I want to tell you that I'm not going to put the square root of C sub 1 Y if I put the square root of C sub 1 a square root of a constant is still a constant so here's what we're going to do instead because that number's arbitrary we don't care what it is this is still going to be why sure you're always doing that one over Y squared plus constant gives you tan inverse of Y over that square root of that constant well if that constant C sub one the square root of that is just going to be a different constant let's just call it C sub two and not have to worry about that square root let's say that one more time whatever this number is you'd be taking the square root but it'd still be a constant it would just be a different constant so who cares let's call it C sub two on the right hand side we're gonna get X but since we just used up a C sub two let's call this season three all right now let's solve for y as much as we can so we're gonna be taking tangent on both sides so Y over C sub two equals tangent of X plus C sub three ma'am we're almost done when we multiply by C sub two we're gonna be done we're gonna call this constant something slightly different but I want you to notice this also with these second-order differential equations we should be getting two different arbitrary constants well this should make sense we're doing the integral one time arbitrary constant we're seeing that's a first derivative do an integral again a second arbitrary constant you should have two of them in there so when we solve for y and get this C sub two tangent X plus C sub three most of the time you're going to see these written is like a and B because the the subscript oh yeah those subscripts right there they're not super important it doesn't really matter what they are and so a lot of times not all the time but a lot of times we'll see this as y equals a tangent X plus B so very similar idea just a little bit of a different substitution I'm always the type of teacher who likes to explain why we're doing this so I'm Anna never stuck with me unless it was fully explained like where's my even coming from where that's coming from as a chain rule that's why we came that extra pee while we have to have a dpdy instead of a deep PDX because we don't have any experience we have a y-variable that we're treating as an independent until we get down the pee and we're saying hey that's the first derivative the spec 2x then we go back to our independent variable thanks I wish I could ask you if it makes sense to you know yeah makes sense but it should be making sense right here if you're ok with the chain rule and you understand or substitution its kind of the same technique so what we're gonna do we're gonna come back we're gonna work through four examples I'm gonna go fairly quickly this is the only time I will show you the actual chain rule here from here on out I'm going to say here's a substitution and we can change into that by the chain rule and then we're gonna go through and I want to focus more on some of the integration techniques so I'll be back in just a second let's do a couple more so check out our next one it's clearly a differential equation it's clearly a second-order got that second derivative of Y with respect to both times the hell reviewing that adds so let's go ahead and see if it's reducible now think about what case you have here so our second order differential equations are reducible if we can make a substitution that changes the first rhythm into a variable and the second derivative into a first derivative of that variable and that is happening if we have missing an X variable or missing a Y variable here we clearly have a Y but we're missing an X so we're gonna do the same exact substitution to start that's how what you think of it there's only one substitution the only thing we're really doing so we're saying let's make the first derivative of Y with respect to X equal to P go alright well suppose it works so the second derivative of Y with respect to X is equal to the first derivative of P with respect to X that's the only substitution we ever make but because we're missing the exit here's how I want you to think about I do not want you to think of this as two completely different substitutions I want you to think about it as one substitution where this one doesn't particularly work exactly like it is for this type of a case when you're missing X's so same substitution same substitution but then in your head you go oh I don't have any X's so that DX is a real problem for me so we're going to just chain rule this and by the chain rule we say this would be derivative of P with respect to Y kind of derivative of Y with respect to X which is P that's where that P comes from now let's do the substitution and what that's gonna do is see it immediately it's going to reduce the second derivative into a first derivative of P so we're still going to have let's see what we got we're still gonna have a while that second derivative we're gonna call it this piece by the chain rule we have that P derivative of P with respect to Y plus also that Y Prime's got a change you see when we called that derivative of Y with respect to X equal to P that Y prime has changing your P so this is P squared equals zero let's make sure we've got this second-order yes missing exes okay same substitution same substitution oh that's substitution to work I don't have X's so we're going to use a chain rule we keep our variable Y we change all the derivatives that's why it's reducible is because you can change this into a variable first derivative into a variable the second derivative is the derivative of that variable P with respect to Y in this case X implicitly now well it's about you a little bit God you know first glance you might think that can be linear if you divide everything by P kind of it looks that way but could you make it easier and if you have a zero over here that answers yeah you can if we subtract P from both sides then we're gonna have something that is separable so let's do P squared any less let's do that so Y times P derivative of P with respect to wild if we subtract that P squared and get negative P squared now we're going to group our PCP or wise dy so let's divide both sides by that P squared you'll remember like with separable I always try to keep my my sides and my constants or even my command constant coefficients on the right hand side a little bit easier for us to deal with because we're trying to solve for P once we do that first integral so here let's let's divide by P squared but let's leave the negative let's divide by Y so divided by P squared we're giving 1 over P DP on the right hand side we're dividing by Y that's negative 1 or Y dy let's make sure that you guys are ok with that one so we've divided this right so P divided by P squared that's one of repeat you'd make some some domain restrictions if you have to we have DP great let's divide both sides by Y I know I'm doing a lot of algebra in my head right now I hope that you stick it with me on that one should be about that level that we can see divided by P squared you need to wonder what can you divide by Y you're gonna do one of the why you keep your negative here group our DP dy now we get something that's separable why don't you try right now to see if you can solve that let's see if you can get that all the way down to solving for P if you want to possibly do that now I mean we're going to continue so clean do an integral on both sides left hand side hey man that's nice that's gonna be Ln absolute value of the people on the right hand side we get negative Ln absolute value of 1 over y plus C sub 1 now do you remember what we're gonna do with those L ends so when we have this Ln Ln we're going to do that the evaluate both of those as an exponent on e so we're going to do an exponential on both sides so left hand side will just get the absolute value of the P on the right hand side let's move this negative and we're going to get e to the Ln yeah I'm sorry I kind of did this in my head a little bit too fast Ln Y and then we're gonna get that negative 1 it's going to be 1 over Y my dad absolute value 1 over y time e to the C sub 1 you're very very careful here when you're doing this e to the Ln yes those are gone those are composition inverse functions your absolute value P same thing happens here but do you remember when we take and we treat this as an exponent on e we can separate exponents on on on a base by saying those are multiplied together with the common base so if with a common base you add exponents added expose means you have a common base so we can separate this and get either the C sub 1 so let's make sure we're here this right here would give us a composition inverse functions in just a minute we have Ln of absolute value why I mess up the first time I put 1 over Y cuz I was taking out of here we move that negative right there that's going to give us 1 over Y now we can simplify this e to the Ln on the left hand side let's get absolute value of P on the right hand side we're going to have this e to the C sub 1 we're going to have this 1 over Y and do you remember what we can do with those absolute values because we can treat them as a plus and minus because we can wrap that up and our ISA bond we can say P equals plus or minus e to the C sub 1 times 1 over Y and then we say this whole entire thing this thing is C sub 2 I know that I've gone a little bit quickly on that one but this is all some old technique that we've seen before so P would equal C sub 2 over Y or times 1 over 1 that's fine as well now what now are we done how we solved for y whoa no actually I'm going to solve the differential equation dy/dx yet so all we've done here is we've said that's a reducible second-order we made a great substitution we got all the way down to here we said that's separable fantastic do separable equations there do certainly separation of variables let's integrate both sides let's let's understand that in order to get rid of Ln we do that be on both sides to make sure that we understand that that e gets raised to this exponent and this exponent multiplicatively that can be a constant and we can draw our absolute values because of that plus or minus that's fantastic so we say hey that's P yes that's one of our Y great but let's wrap up those absolute value that I did that positive negative inside of this constant it's arbitrary anyway and we get down to P equals C sub 2 over Y but we're not done because we have to remember what exactly P is when we did that substitution P is dy DX let's go back and let's make that dividing X so P is C sub 2 over Y then we know that P really came from dy DX you know some of you might be wondering letter D huh then you're calling that Y prime Y do you have that is Eli the X there instead of Y prime and why do you have that it's just y double prime Y the notation switch I like that as dy DX because when you do your substitution right here it's much easier to see that you have a separable equation it's a lot easier to see that if you just call that Y prime you down here Y prime you got what's that mean you're gonna have to really change into dy DX anyway so you don't have to but it's nicer to see it that way so that's why I write Y Prime like this within Y double Prime it's not really necessary to do that it's a direct substitution you never have to go back to it I hope that makes sense to you so we've done all of our work we use our exponential appropriately we've simplified some things we've got all the way down here we've said that's great I've solved my differential equation in terms of that dummy very liquid in there but now realizing that what that P means is dy DX I've got another usually basic differential equation to solve it is separable that's great it's now in terms of our independent variable X I'm going to group our Y's be Y our constants and X is the S and so when we do that let's multiply both sides by Y on the right hand side we're going to leave that constant DX if you have something that's really nice to integrate so when we do let's integrate both sides well that's that's pretty good on the left-hand side we get this y squared over two on the right hand side we get C sub two x plus we're going to happen have another arbitrary constant just like every time we have that second order differential equation we're going to get two of them so if we multiply both sides by two check this out check this out you're going to multiply both sides by two right well that would be two C sub two x plus two C sub three is that it's still an arbitrary constant is that still an arbitrary constant yeah so you can wrap all those up in like an A and B and that's how we're gonna end this so we'll say that Y squared is a X plus two B and that's about as good as that we can as we can get we really don't want to go solving for y we don't want to deal with that plus or minus square root idea so we leave that implicitly defined are you with me are you getting the idea do you understand the idea that this substitution is the same substitution as for when Y is missing but when X is missing you need to just change that and that's the chain rule do you understand the same exact idea separable equations that we've done for a long time I hope that's really just from here just kind of easy for you I hope that's easy for you and then getting down to our P just remember what P is P is dy/dx which means you're going back to it in their independent variable X they're typically separable always but then we can do an integral of both sides and solve for y how we want to let's do another one and then what I'd like to do is give you give you two of them this next one is kind of awkward I want to show you something about it so let's do this one together and then I want to give you two of them and give you a chance to do that so I'll write two on the board I really want you to pause the video after we do this example together so let's take a look at it well know one thing we notice is that yeah it's a differential equation and hopefully you're seeing a second-order and hopefully you've seen its reducible how do we know it's reducible if we're missing a variable its reducible so if we're missing X or we're missing Y we have a do that just same technique it's just when we're missing exes it gets a little more complicated so let's look at how to do it do it together so a second order differential equation we're missing exes let's go ahead let's make our substitution understanding that we're going to call the first derivative of Y with respect to X equal to P all right but wait there's no first derivative of of life so is that a problem it's not a problem you're just not gonna have the variable P by itself that's totally okay all right so if the first derivative of Y with respect to X is P then the second derivative of Y with respect to X is the derivative of P with respect to X but that's not gonna work why is it not gonna work the x's so we're gonna change it just slightly we're gonna use the chain rule and say if this is DP over D Y times dy over DX that dy/dx is P that leaves us with a derivative of P with respect to Y which we're gonna treat as our variable for right now now let's go ahead let's plug everything back in so we don't have any y Prime's any first derivatives of Y to call people we do get 1p up there but there's nothing else that we're gonna call feed that that's okay that's another problem so our second derivative of Y with respect to X made a substitution we took a derivative on both sides we used a chain rule since this piece right here is P times DP dy plus 4y don't ever change the variables alright so if you have X's you want to leave those if you have Y is to the first power y is just saying whatever leave those now what you don't leave are Y primes and Y telephones you have to change those on the right hand side we have 0 what is it get a 0 it's not linear that Y doesn't count that Y is treating this our independent variable right now so this is not linear what this is a separable so we're going to subtract that 4y we're going to group our peas on one side y lies on the other side we're gonna integrate and it looks pretty innocuous right now so you'll that's that's really easy that doesn't look too bad so I'm gonna group my PvP I'm going to subtract that for Y on both sides and then we're gonna move our DUI so we're subtracting the for y we're moving our dy on the right hand side is a nice convention that we use to integrate these for these different equations and we'll take an interval on both sides on the left hand side starts off looking kind of nice we get this P squared over two on the right hand side let's see we add add 1 so it's Y squared over new exponent of 2 so B that's negative 4 over 2 that's our negative 2 y squared and we will have an arbitrary constant now every other time we've solved for P now we're gonna solve for P here but do you see that we're gonna run into a problem well let's multiply both sides by 2 first let's get this P squared equals negative 4 y squared plus 2c someone yes we're gonna change that into a cease of two in just a minute but this is an issue this is going to cause a problem for us and we're gonna have to define what we want here so when you solve for P you know this you're gonna take a square root of both sides and you shouldn't be giving either positive or negative we oftentimes choose just the positive could you choose the negative yes but it's gonna lead you to a different solution set all right so I'm not gonna show that but you could do that so normally normally we would just choose the positive square root also depends our problems we can make that that determination but but I do want to understand that when we're about to do this when I want to take a square root on both sides and when I omit the plus and minus what I'm choosing is to just look at the positive square root I'm not choosing the negative in front of that square root okay could you do it what the negative yes it's just going to lead you to a different solution set with a different interval let's call this thing C sub 2 right now since that's just a constant I want to pause right there and make sure that we are a hundred percent before we go any further so are you okay that it's a second-order differential equation it happens to be reducible and we're in the technique of having a missing X so we use the same substitution but you say that doesn't work I need a chain rule because I have to treat Y as my variables this finally variable we do our substitution it's separable we do a nice integral get although okay we gotta solve down for P we take a square root or understanding that I'm taking the positive because my choice here and then we say all right well what is what is if you don't stop here some students that don't get this are going to stop here you get that right because they've got lots of their no but but you sold it for a dummy variable it's like doing a substitution like doing a u sub and ending your integral and used and not going back to X's it's essentially what we've done we've just not using you so to use a P sub so fix it we want P that we want to understand that he is really you my DX so dy/dx what we have our substitution equals and I'm gonna start changing things around I'm gonna start putting this as C sub two minus 4y squared is it separable yes because now our independent variable is X so are we want dy with Y's on the left hand side we want DX with constants and X's on the right hand side so we're going to divide both sides by this we'll have our DX on the right side oh this is why we're doing the problem is for the integration technique right here when we're when we're doing this when we run into this sort of a problem some of your tables might actually give this to you they will hold that that for out and call it a two in front of your sine inverse which we're about to get I want to show you a slightly different technique where if you don't remember that that you can still get the sine inverse and still see where that two comes from so we're going to integrate yes we're gonna do them both sides right hand side hopefully I don't have an issue with that left hand side though if you want to use your table right here do it that's totally fine I do not have a problem with that if you don't or if you don't have a table or if you forgotten it or you just wants to said cool something cool look you can force this to factor out of four notice how if I distributed that I get C sub two minus 4y squared because because square roots are exponents because exponents distribute across multiplication and division you can simplify that four so we have two square root of C sub two over four minus y squared that's where that number comes from on the integration table white has a number in front sine inverse there's because of that because you can simplify that out of the times usually the square root of whatever that number is the next thing I want you to notice is that because this is a constant let's just call that C so we're gonna burn up a lot of constants right here let's call that C two three or leave it I don't care whatever whatever it is it's just an arbitrary constant so what I'm going to do I'm gonna come down and put this right here we'll talk about the integral and then that will be good we're almost done so we'll factor out the 4 we'll take the square root of four which would give us two so this would be 1 over 2 square root of C sub 2 over 4 minus y squared dy equals integral the X on the left-hand side yeah you're gonna get it you're gonna get that one half so that one over - yeah I guess I said it wrong the sine inverse you're gonna have a 1 over something I just meant that that that's gonna be 2 so you can have a 1/2 you can have a sine inverse from the integration table but then what it looks like it says here's your usual 1/2 here's your sine inverse sine inverse works a lot like the tan inverse does so in that you're going to get Y over the square root of this number now listen we're gonna say to see some three-tiered Y because that's an arbitrary constant divided by 4 you're still going to get an arbitrary constant take the square root of you're still gonna get an arbitrary constant it doesn't matter what it is so when we do all this and you're seeing sine inverse but that's screwing me up who cares just call it C sub 3 it's just some number that really doesn't matter what it is in order to make this differential equation work so with differential equations remember we're talking about families of curves those arbitrary constants are exactly that they're arbitrary unless we have an initial value a point where we need one of those curves to go through we leave them and we say this is representing a family of curves and we can't really solve it the first particular solution without an initial condition therefore it doesn't matter what that is that could be 500 million I don't know what it is so when we were doing this we're simplifying or we're going straight from the table and you're having trouble with that guy and you're saying I don't know what to do with that arbitrary constant I know that it should be like the square root or something this should be something squared so we get the square root who cares the square root of C sub two is still like arbitrary contour the sphere sees two or four still arbitrary constant just put on the right-hand side we get x + we'll have to call it C sub 4 it's already burned through 1 2 & 3 we're pretty close to being done we're almost almost there so let's go ahead and let's do this yeah I get it a little bit different on my notes but this'll work just as well mine I did this before and I move the two to the right hand side and integrated to DX but you'll see that that's the same thing right what what I mean is I before I even did an integral I pulled the two out I multiplied both sides by two instead of having a one-half I had to over here be the same because I just get 2x plus constant now I have one half I'm going to multiply by two anyway so where that is that's fine so sine inverse Y over C sub 3 equals 2x plus 2 C sub 2 C sub 4 and we're gonna do the same thing we did with tan inverse we're going to take sign on both sides and we're going to multiply it by C sub 3 let's multiply both sides by C sub 3 and that's perfectly appropriate but a lot of times what you're gonna see is you're gonna see an arbitrary constant called a and an arbitrary constant this - C sub 4 and we're gonna call that B so all right back because I ran out of the room I'll write that right up here so here's what we should we should be getting we should be giving y equals this a some arbitrary constant sign it looks pretty good we got this 2x that's great and then plus B guys that's about as good as we can get we've solved for y explicitly which is pretty fantastic considering where we started I hope this this this techniques making sense to you I wanted to go through this integration technique and show you what to do with the constants because I've seen students get really all jacked up a mountain do about that stuff why do I do with that don't worry about it use your integration table and when you have an arbitrary constant you're trying to figure out where that go it doesn't it doesn't matter it's still arbitrary so you don't worry about the square root or whatever they have to do there so anyhow I hope this making sense I hope that you can do it I hope that it I'm getting this to to be internalized with you so that you will remember that these techniques for missing X and missing life identical but the X screws us up so you realize that that's gonna lead you to the chain rule that's gonna lead you to having that PEP dy and then our integration is exactly the same to solve that differential equation terms P and y when you get down to P you need to understand what that is that's the derivative of Y with respect to X X now becomes an independent variable and we do in another technique so it's like two problems rolled up into one with a very nice substitution I'll be back with a couple more in just a minute which I'm gonna have you try on your own all right let's get to it so what I would like you to do if you can I would like everyone to do this actually whether you're in a class or not man it's really good for you to practice on your own if you screwed up now who cares but you will learn from those mistakes so if you can pause the video and try these two examples man that we rock solid alright so be thinking about is it a second-order differential equation is it reducible which means it's missing a variable what variable is it missing make the same substitution in either case both these are missing X's but then realize that a DX would be a problem if you're missing X's that's the thought process of one I'd just modeled my thinking on how I go through it I'm gonna start right now but you really should pause the video and try so I'm looking at that saying second order differential equation man absolutely is it reducible yes it's missing a variable what variables missing it's missing X so in my head right now it's triggering DX is gonna be a problem I'm gonna make the same system substitution but I'm gonna have to use a chain rule here so we're gonna try it and what's funny is that problem looks a lot harder than that problem this one's gonna be real easy this one's gonna take a little bit of time so let's let's look at this so when we do our substitution we say hey let's make the first derivative of Y with respect X equal to P and I'm doing this I'm doing the dy/dx instead of a Y prime because when we have a substitute back for P that dy/dx is very nice it shows us that we have a differential equation where X is are independent variable and y prime does not do that that's what we're doing that so we say alright on then the second derivative of Y is the derivative of P with respect to x sure that's what that means we're taking a second derivative with respect to X on both sides so we have a derivative of P with respect to X but since we're missing X's we need this chain rule that we've talked about Lots right now so that we can treat Y as our variable until we get down to people that's been put into practice so every single inner instance of Y prime is gonna change to P every instance of y double prime is going to change to P DP divided by the chain rule we leave all the other wise alone so we're gonna leave this Y but this is gonna be our P DP dy Y great y double prime got it y prime oh that's a P but it's squared on the right hand side why we leave this why this is gonna be P so Y stays Y double prime changes Y prime color P we have a square Y stays Y prime P if you did on your own I hope you made it that far now taking a look at it that's a whole lot of stuff floating around so what in the world are we gonna do well what I notice and we take and talk about some domain restrictions here we talk about think about at least usually said in a footnote assume positive or not zero where you have to we're gonna divide everything by P because every single term is a peanut so we're going to do that so dividing everything by P we get Y times DP dy plus just P to the first power equals one go alright well the world's that wait a minute hey that's great so that right there that's that's P that's P the first power here's our will retreat in its independent variable Y we get Y on one side we get a Y right here that looks so many earth because that's P to the first power that is linear in fact don't go any further don't find any green in this bag actually don't try if you want to find an integrating factor right here you're gonna do this you're gonna get back exactly this because take a look at it think outside the box a little bit this left hand side when we deal with linear remember remember linear we want to get this integrated factor to make the left hand side a product rule and have just a function of our independent variable on the right you remember that we're on the left hand side to become a product rule we have the right hand side to be whatever this variable is well if you found an integrating factor you would be integrating e to the flash you'd be divided by Y wouldn't you so B divided by Y everywhere you get 1 over Y here you get one here I'm Kevin shouldn't be doing this but I'm gonna show you you'd be divided by Y so you get 1 over Y that's important that's a white you get one right here you're integrating factor B 1 over Y dy you get e to the Ln Y you get Y so you will live waited you tell me you divide everything by Y you get your integrating factor Y and they multiply everything back my life wait a minute if that's gonna happen then what that means is that on the left hand side you've already got the result of some sort of product rule look at that if you take the derivative of P leave the first during the why leave the P that right there is already once you would get if you took Y times P and took a derivative with respect to Y so this is the result of some derivative with respect to Y of Y times P leave the first one alone the river the second there it is derivative the first one that's one leave the second alone that's already there so that's kind of cool so when we're saying this I almost did that any rating factor and did that on my own whoa hey that's that's already a the result of some sort of profit rule and that's what linear wants anyway on the right hand side we still just have one so since that the result of some sort of product rule we just showed the product if we take an integral on both sides just like we've seen a nonlinear a lot of times the left hand side is just going to deal this Y times P remember we're taking this with respect to Y that's where you have this little decent Y here derivative with respect to Y on the left hand side we get this one-half Y squared or Y square root 2 plus C sub 1 now what hopefully no now what we're always trying to solve for P because that means that we can substitute dy/dx and get something that's really nice so let's multiply both sides by Y I'm sorry I meant divide so if we divide both sides by Y we get Y the first power divided by Y over 2 and then C sub 1 over Y you know what before we go any further and you're gonna see this a lot whenever you get these two terms and you're gonna have a dy/dx you're probably gonna make something inseparable not all the time but 99% of the time you're going to get that it would benefit us to make a common denominator and get one fraction before that happens because if this is all in terms of Y which it is and you're gonna put a dy/dx here which you are and there's no X's over here then this is all gonna be in terms of Y and you're gonna have to move it to the other side and find a common denominator anyway it's easiest to do that here so let's take a moment let's multiply and I'm gonna say that anything that was that was really fast look you're about to put dy/dx here right and you have all-wise many low ones you missing the X whole reason for this technique so this is gonna be separable but all your Y's are here with no X's you're going to have to move them to the left hand side and make one fraction anyhow so that you can integrate the separable equation that we're about to create let's get one fraction out so we're gonna multiply by Y on the left hand traction and two on the right hand fraction so Y squared plus two C sub 1 over 2y and that's our common - number that we're gonna have that that looks a little bit nicer as far as making a separable equation later a month I don't look any nicer it's gonna be in just a moment so and so now let's let's go through it again reducible yes made a good substitution great we already had the result of some sort of a product rule for that linear showed it we took an integral both sides with respect to Y because that's what we're treating this or variable right now we solve for P we understand that this is going to be separable I'm gonna want one fraction and now we see it P is dy/dx let's change this on the right-hand side we've got a y squared + 2 C sub 1 over 2 1 now do you see whatever copy I might not have sunk in at first and that's okay right now I hope you see what we're talking about we have all wise we have dy over DX so we have our independent variable X we want that on the right hand side you are wise dy is here so we're going to have to multiply by the reciprocal here we would think about some domain restrictions when we're doing this but again a lot of books get a little lazy and some math these are you'll amazing sometimes I do too and we just say we're because they happen so often where appropriate you're gonna have to assume that some things are positive or not negative it happens it just happens a lot so we're going to multiply by the reciprocal get our wise dy our X is DX and now we can integrate them that looks really good as far as integrating goes because what we're gonna do is integrate both sides on the right hand side would be something real nice for you keep X on the left hand side you have a little bit of a substitution to do so this is not like a tan inverse or anything we don't have that we have this making use of 4y squared then D u equals 2y dy that's great that's exactly what we have that's so nice and little when it happens no need to divide anything what we get is this oh sorry plus 2c something I made a mistake I should have put this whole thing I was thinking of the derivative only so take your whole denominator make that Y squared plus 2 C sub 1 the derivative is 2y dy never that's just a constant it goes away when you take a derivative and then we have this real nice integral 1 over u this is are you one of you 2y dy well that's now D u so let's do the integral right now the integral of 1 over u D U is Ln u we'll call that C sub two we've already burned it the C sub one so now let's see well we're gonna have this this outline of what was you well you was this y squared plus two C sub 1 e equals x plus C sub 2 hey I mean even exponential on both sides let's do it if we do an exponential on both sides we're gonna get Y squared plus 2 C sub 1 equals I'm going to I'm going to do this this e to the nd to remember that what we're doing we're going to have an e to the X but we're also going to have this e to the C sub 2 you remember that if we can separate that egg the addition of excellence by multiplying the common bases so we have an e to the X we have a times e to the C sub 2 and I'm wrapping up the absolute value in that plus and minus so we're just doing exponential on both sides its composition inverse functions we're getting this y squared plus 2 C sub 1 but now we're gonna start doing a couple other things let's call this because this is sort of look real nasty let's call this piece right here let's call that C sub 3 back is a constant no matter whether it's positive or negative that's a constant e to the X and if we subtract that to C sub 1 well now we have two arbitrary constants we have assumed a let's call that C sub 3 2 it doesn't really matter we're about to change it to an A but now we have this solution to a differential equation implicitly defined where y squared equals e to the X plus being the women that's a minus who cares so call this plus negative 2 C sub 1 and then you can call that negative 2 C sub 1 equal to B or something so that's the final form of this I wanted to do this to show you our substitution to show you that some things can look really hard within the integral for the person was really easy we have just a basic u sub and then just make sure that when going back we're remembering that if you're using a you you're going back to the original whatever you substituted for we're doing Exponential's on both sides remember that we can wrap up absolute values with pluses and minuses remember that we can separate these addition of exponents by multiplication of common bases with those exponents and then we're solving for y as much as we can and we should be ending with two different arbitrary constants like a and B are you with me is it making sense to you are you are you getting it because if you are and that is cool so very valuable stuff that we're learning what we will eventually go on to solve some second-order differential equations that aren't reducible and I'll give you techniques for that we're out there yet so last one we're gonna we're going to power through it and take a look at it it is a difference equation it is definitely set the second order is it reducible it looks a little funny because I don't have them of Y Prime we've dealt with that before so that's not such a huge issue but it's not a it's not a it's not one that's missing X I'm sorry it's not one that's missing why does ever widen to their power it's missing X and it's missing that Y Prime that's okay we've dealt with it before we don't have a handle it let's make that substitution and see what happens so the first thing we're gonna do is realize that if I'm missing a variable then I can reduce the order of my difference equation by one by calling the first derivative equal to a variable and in the second derivative would equal the first derivative of that variable if we're missing X's that's a problem so we use that chain rule we use that chain rule correctly and let's make our substitution and it looks really easy at the start but we're gonna see some things in the second integral that I want to just make you aware not super hard stuff just just to remember okay so uh Wow what do we do do we ever change Y's did we ever change once no doing while crimes yes into peas and why don't rhymes yes into either dpdx if we have X's or PD PD widens we don't so that Y to the third that has to stay that Y double prime becomes P times DP dy and that one is right it is what it is alright well hey that's separable in fact I've already get my piece on one side DP but I need to group my y's dy is on the other side so we're going to divide by Y cubed and that's so nice we can do this by simple equation but again what we're seeing out of this is that on the left hand side we're going to get that P squared over 2 or that one-half of P squared on the right hand side well we think of this as an integral of Y to the negative third power so one-half P squared looks real good on the right hand side or adding 1 that's negative 2 we're dividing by the new exponent and we always get that arbitrary constant you should get another one by the time we're done let's simplify this just a bit because it looks kind of nasty you know what I'm going to do it is a couple things the first thing I'm gonna do is I'm gonna leave that 1/2 for just a moment so we don't be confused we'll have that 1/2 T squared on the right hand side wait surely right this is negative 1 over 2 y squared let's see someone so we're keeping that that negative but moving to the numerator we have a 2 and a bi that over Y squared just wants your negative exponents in your sign so we're keeping this 1/2 P squared great we've got that negative okay over 2 y squared let's multiply everything by 2 when we do that we'll have negative 1 over Y squared Plus to see someone now do you remember me telling you that when you're about to go to dy/dx you're going to want to have one fraction over there because if you're missing X's all of this is in terms of Y this is not the same as if you're missing Y's all that would be excess that would be fine because you just do dy/dx moving the X and you're already good to go but when you're missing X's and not missing Y's all this stuff is going to be in terms of Y you're gonna have to move it to the left hand side you'll get dy over DX so Y is d Y's have to be on the left with X's again you're good to go you're fine there's move your DX and integrate that's what we saw a last video but here we're gonna have to move this left what that means is that we're gonna want to spend some time right now to just find a common denominator so when we do to make that one fraction let's go ahead that's x y squared over Y squared and we'll get P squared equals well let's see here's a minus one this is gonna have a common denominator for y squared so it's real good well this gives us some sort of a constant y squared minus one are you with me on that one you see it if I get my my common denominator I can move these around there they're not commutative but I can move my terms at long distance you know I can move my terms of their sign says to C sub 1 y squared minus 1 all over Y squared why are we doing that for two reasons number one and you see the square we have to get rid of that and we're gonna choose to keep the positive square root here so if we do let's take a square root on both sides because we have that one fraction already it makes it much much nicer to deal with so if you know you're gonna be taking the square root get that one fraction also if you're missing X's all of this right hand side of P equals whatever is always going to be wise you're gonna change that P into D by DX if the right hand side is all-wise and you have dy over DX you're gonna eventually have to move your wive's to left hand side so we want to create with these this white a this missing next technique we're going to want to create this one fraction pretty much every time especially we're taking square roots especially when we have this dydx we're about to get on the left hand side man we just get p on the right hand side why we got one fractious because you can distribute excellence across multiplication division because square roots are exponents they're one-half power we can simplify that denominator which is nice now that's something that's something that we can deal with because now they walk through the game maker substitution you go all the way down you're doing do our nice little and enroll say okay that's that's kind of nasty let's multiply by two great ok I'm going to take a square root it's real hard to take a square root of two terms because exponents don't distribute across addition and subtraction I can't distribute it across terms I can across multiplication division make a fraction so that you have division then we can take a square root of the numerator and denominator and then we realize we realize what P actually means he is dy/dx and now we're hopefully seeing it if you have no X's over here these are all wise they're gonna have to move to the left hand side so we're going to multiply by the reciprocal your X DX has to be on the right hand side I've made a little blurb made a blurb I'm a blurb here guys I've said a little blurb in a while ago that said if you have X's and are missing lies then if all these are X's this is a lot easier you just have to move your DX integrate that's the last bit of it so let's let's let's do this it's get our wise be wise constants DX the right hand side is always obviously going to be very easy because you have no X's you should have a DX and we integrate [Applause] I'm gonna move this up here to have a few more steps so now that we have moved up there um there's there's something that's starting to look a little awkward it's that to see someone so a lot of time this especially when we're about to make a substitution we want to make that part as concise as possible because we're going to be using it again what I mean is we're about to call that you this is not some sort of a trick subarray while this is just a you equals that inside that square root because the derivative is going to give us a Y that's going to cancel out that Y in the numerator so we're gonna make a substitution with that but before we do that it's really nice to make this look better because when you say you equal to that expression we're gonna have to resub stitute it so what I'm going to do is take a moment to say hey instead of back to C sub 1 since that's an arbitrary constant anyway 2 times C sub 1 gives us just a constant what's called C sub 2 y squared minus 1 then when we make our visa and say u equals C sub 2 y squared minus 1 the reason why I want this as nicely written as possible is because you're going to substitute this back in somewhere so you can be using that c2 again we don't want to have to write to C sub 1 everywhere think it's really awkward but what we're also not going to do we're not going to be changing this C sub 2 throughout this integral even if we get a concept or not going to change it because we have to plug in that C sub 2 in let me see if they're the same constant and I'll show you what I'm talking about so our D nu equals those two a C sub 2 y dy hey that's nice I've got a Y dy right here so we're gonna solve for a variable dy and that lets us make a very nice substitution so we do have an integral we've got a 1 over square root of U we call this whole thing you that's 1 over square to you we don't have a Y dy anymore we've got a d u over 2 C sub 2 so what can we do with that well we could call this one over to season two we have to all that constant mallet you have been negative one half you notice what I'm not doing right now what I'm not doing I'm not calling that cease of three why because I'm still in terms of you right and so when we substitute that back in I'm gonna have a cease of two here I have a season two here I'm gonna have to cease of twos and I want to show that those are the same constant that needs to make sense I'm not going to call this season three because I lose that so 1 over 2 C sub 2 let's see how we take integrals we add 1 we divide by noon exponents so you to the positive 1/2 over 1/2 hey that's pretty nice look at that we've got a 2 cancelling with a 1/2 that's great let's do two things number one we haven't even dealt the right side on the key right now let's do two things let's come back over here let's make that the square root of u 1 over C sub 2 square root of u well let's call you what it is seize the 2 y squared minus 1 this is what I was talking about now it's very clear that this constant and this constant are the same constant if we go ahead and call them like an A or something we're gonna show that so we've taken a nice use of but made it cleaner before we did it we did our use so just like normal we did our integral just like normal but when we're seeing constant that C 2 we're not changing that because we understand we've had the C sub 2 that we're going to substitute back in for that you so I'm leaving this so they can see when I do this when I step back for you these are the same two constants I hope that makes sense in your head the next thing we're gonna do is say hey you know what what this equals to is the integral of this X plus some constant and that is a different arbitrary constant so should you end with three arbitrary constants no this is the reason why we can't change in season two right they're this because you end with two different arbitrary constants not Celine so we have one over cease of two okay square root of u by nu with C sub two Y squared minus one and then a very easy in the bold let's multiply both sides by C sub two that looks like we can do that lets get kind of nasty I suppose we could square both sides maybe that one I don't think it's going to be much better if we do anything else besides that so let's look at what's going on we have an implicitly defined solution to our differential equation it is general it's got two arbitrary constants but this constant and this constant are the same this right here that C sub two times C sub three you can go ahead and call that one arbitrary constant now it does have the same concept here here here but that C sub three since we don't know what that is it doesn't really matter so we can call this like a y squared equals ax plus let's call that whole thing B and that is about as good as we can get and I hope it's making sense I hoped I know there's a longer video hope the technique works for you though I want to spend some time going through some integrals showing you some techniques showing you exactly how to deal with those constants because frankly when you're going through this on your own and they're not really showing you how they're doing it you're just gonna answer your own idea I have no idea where those concepts are coming from it can make you feel like you don't understand and guys you're very smart and you do understand this stuff if you're here you would definitely have the ability to do this so just just take your time learn from some of the stuff that I'm doing here that way it's the little things really then hold you back so let's let's make sure those things don't hold you back the technique itself I'm sure that you guys get so keep your heads up you're gonna do fine I hope that this has made sense to you and that you have a better feel about how to deal with these reducible second-order differential equations missing either X or Y now we've dealt with that X with both again you're going to want to use the missing why it's it's a little bit easier Missy next we have some more work to do so I'll come back with another video just a little while I'm gonna try to be doing one video a week from for the foreseeable future we'll talk about exact equations but I know a lot of you in the menasset pool have a wonderful day and I'll see you for that you

Channel: Professor Leonard

Views: 35,237

Keywords: Math, Leonard, Professor Leonard, Differential Equations, Reducible, Second Order, Substitution

Id: -gi4GfbhrnI

Length: 79min 49sec (4789 seconds)

Published: Wed Mar 06 2019