How to Solve Bernoulli Differential Equations (Differential Equations 23)

since we're studying substitution techniques to solve differential equations we may as well talk about Bernoulli equations because this is another substitute substitution technique and a marvelous one of that that's going to take care of solving some of these first-order differential equations that we can't otherwise solve and it's a really really cool idea so before what we've been talking about our substitution techniques that change homogeneous or like an obvious substitution that's a composition into something that we can use separable equations technique for now Bernoulli what this technique does what Bernoulli equation czar is a very specific form of a differential equation that when we do a substitution will change it not into something that's separable but something that is linear and here's how it works it's a really cool idea very fascinating that that this equation exists force and the substitution works but here's the idea if you have any differential equation where you have here's the key or you have I don't care what this is but you have Y to the first power and I don't care what this is but you have Y to some other power now why can't it be the same now if this isn't power 0 then this is 1 and you have something that's linear already if this is a power 1 then these two things can be factored the Y and the Y to the first then you have something that's separable so if this is 0 or 1 you either have linear or you have separable and this doesn't them this isn't batter but if n is any other power besides 0 or 1 you have something is called a Bernoulli equation so differential equation plus something that's Y to the first power and something that's why to a power greater than that those sorry power different than 0 or 1 don't have to be greater but a power that's not 0 or 1 so we're looking for two terms here we're looking for one term that has Y in it and another term that has Y to a power that's not 0 or 1 now here's the idea this is really cool if that's the case I think it's cool four times are you counting my cools that means that this is really cool if that's the case and we make this substitution B equals y to the 1 minus n here's what happens with these Bernoulli equation so we're looking at it we're going not separable not linear not a easy basic substitution oh wait I have one term with the Y the first one term of the Y naught to the 0 1 power let's make the substitution here's what happens with it when we make this substitution when we solve this for y we're gonna get some some factor that when we when we have this substitute of back into our padule-- equation it's going to make something that is linear it's going to eliminate these and a lot of places it's going to leave one B raised to the first power right there and everything else would be X's an awesome idea very very important for us to understand how to do these things if this is not something that we can do directly maybe renew the equation would work and what we're looking for is y is the first power Y to any other power and we're going to make this substitution v equals y to the 1 minus N and like every other substitution technique we've had we're going to solve this for y we're going to go ahead and take the derivative dy/dx we're gonna make a substitution for dy/dx we're going to make a substitution for y we're going to make a substitution for y to the nth power and then what we're gonna do is we're going to multiply to get rid of whatever pieces over here because you're gonna have one it's gonna be something DV DX and when we multiply by that piece that gets rid of it this is going to change to B to the first power and then it's gonna be linear and we use these in linear differential equations techniques to solve the rest of it let's look at one example let's go through it I'm gonna explain all of this again as we're going through because this was this is a little bit vague because you haven't seen it yet but this is this is how it works so let's firstly look at our differential equation we look at it we go hey does that anything that we can do with is that something can take a basic integral with no could I view it as separable yes you could you could divide both sides by y plus y to the third and get 1 over Y cubed plus y and you could factor out the Y and get Y times y squared plus 1 and you could use partial fractions and see where that leads you if it's possible cool maybe there's an easier way can you use a substitution technique well not a non-homogeneous there's I don't really wouldn't want to build any X's in there is it separable or linear it's closed is it Bernoulli can we try it Bernoulli and make this into a Bernoulli equation here's what Bernoulli looks for but really says look at your differential equation if you have one term with Y to the first power and one term with Y to not the first power that one that satisfies the Bernoulli equation here's what we want to do we're gonna try to structure this as a linear what linear meant is you want the dy/dx with the function that has Y to the first on one side you want everything else on the other side you know well that that's not linear exactly that's why we need a substitution so the substitution we're gonna make is going to create for us something that when we multiply both sides to get rid of a piece right over here it's going to create some the first power it's going to get rid of all of this all of these views on the right-hand side going to be gone and what that substitution is is always be to the notice our V equals y to the 1 minus whatever the other power happens to be in this case is 3 y to the 1 minus 3 you know okay well that's V equals y to the negative 2 power I'm going to pause right there I want to show you exactly what we're doing so we have something that is a Bernoulli equation if we have a derivative respectable X if we have Y to the first power notice how the on one site just like every linear ever this looks exactly like a linear right now but we have another function of X with otherwise this is what creates for us the Bernoulli equation rather than just a linear if I erase that that's a linear differential equation we can solve it we know techniques for that but with this you go well what am I supposed to do now get rid of that thing how do I get rid of that thing if we make a substitution V equals y to the 1 minus n if we solve that for y every substitution technique ever if we solve that for y and well that we've done before solving for y and take a derivative dy/dx to replace T with D by DX then what we're what we're gonna get is we're gonna get some function of V times DV DX we're gonna have this as a function of V we're gonna have this as a function of V to a power and when we go ahead and multiply it to get rid of the piece that's in front of DV DX it's going to cancel all this out it's going to create this as a difference of 1 what difference of 1 why do I want a difference of 1 so that I get V to the first power we're taking this making a simple substitution that's going to create something that's linear let's see how that works so Y the first power write it as linear Y to the third power leave it over here the N part the end part is this end so you okay what's my substitution let's do V equals y to the 1 minus 3y we're choosing this guy is because this is the piece that's causing the problems if that wasn't there that'd be linear to be easy but now we go I want to get rid of that how's it gonna work well since derivatives subtract 1 and we're multiplying both sides we're going to get rid of that difference of 1 for this piece and over here but it's gonna build a 2 e to the first power we're gonna see that right now so if we have V equals y to the 1 minus 3 that's V equals Y to the negative 2 if we solve that for y and probably easier way to do that now so y equals B to the negative 1/2 power so solving that for wife we need to remember how these substitutions work in general we need to remember that right here this multiply both sides by 1/2 power we need to remember that when we do a substitution we're wrapping up in this case it's not all of the Y's so that's different with the the substitution techniques that change in the separable equations we want to wrap up all of the Y's inner substitution here we don't because what we want is we want to be able to substitute the V in both places but then have this one change it to V to the first not just disappear completely but change them to be the first row in this one disappear this substitution does it so all of our substitutions to wrap up some lies in these solves for y and then takes a derivative so we can replace it basically two ideas all the stuff involving Y's and the dy/dx let's try that right now so when we take dy/dx we bring down the power subtract one and we get DV DX based on the chain rules this is implicit V was a function of X you're always going to have that hopefully this is familiar to you because you've done lots of substitutions right now so we look for our terms you said I wanted to be linear right in this linear oh I got another piece with Y to some power cool make your substitution that's now Bernoulli equation if the equals y2 that it's always this it's always 1 minus whatever that power it is and I want you to see what's gonna happen if we do 1 months what if that power is and we subtract and we solve for y think in your head right now why do we need to solve for y it should have bounced up out of that that we need to replace this along with our wives so we need to replace this as well so once you solve for y and take a derivative we can now replace dy/dx we can also replace every instance of Y so let's do that and be careful with it there's lots of things go so from here I'm gonna start with this guy so dy/dx I'm gonna replace it with this I'm not going to have dy/dx anymore I'm gonna have negative 1/2 of V to the negative three halves DV DX so dy no that garbage - - why no I'm not going to have Y anymore I have solved for Y right here I'm gonna replace this Y with the V to the one half well negative what happened equals and then I want to replace this Y with V to the negative one half as well let's pause for a moment kisses is typically where students get lost go wow that's a lot of stuff this looks different from the other substitution techniques you're right this is a very specific form where we're trying to build a linear and in doing so we're going to make us a very special substitution we're gonna make a substitution that it is 1 if it's 1 minus of that power and we solve for y instead of going directly and say hey this one part is V notice how V equals something is not even up there but what it does do is if I solve for y it gives me V such that when I'm when I take a derivative of it this is separated by a 1 but that that negative 1 if I multiply both sides times whatever it does whatever it happens to undo this thing it's going to create V to the first power here that's why we do that 1 minus whatever the power is when we solve for y it creates this difference of 1 so that when I multiply both sides it's going to leave me with nothing methylene in terms of V and then the first power game so the whole idea of a Bernoulli equation is make it look linear as much as you can do a substitution Y minus whatever that power is solve for y replace your Y with what you solve for replace your dy/dx with what the derivative is and then that's going to finish the job of creating this linear for us so I know this is this a lot of work already we're not even to like the integral yet but we've looked at and we said why the first y to not the not the first take v equals one minus that number all right we got that solve it for y take a derivative and then we're making that substitution so dy/dx becomes whatever dy/dx is after you solve for y taking a derivative y itself it's not a direct substitution for your V Y itself is you need to find this after you solve for V so Y is gonna be beaten theta 1/2 Y the third is V to the negative 1/2 sure yeah that's still why I say Y but now to the third power let's see how this works when we when we go a little bit further would simplify the right hand side I want you to notice something what I mentioned earlier was it what this technique does it creates this this idea that there's a difference of 1 between here and here and so what we're doing is we need to make this look linear remember the idea is to get a linear here if we need to make this look linear we need to get rid of that thing but in doing so we're gonna multiply by negative to get rid of the negative 1 now we're also going to multiply by V to the positive 3 halves look at the right hand side because we base this on this power right there when we multiply that it's going to get rid of both of these pieces at the same time not only that but because we have this one that difference of the difference of one that we have here when we multiply here this power is going to be one less or more depends on what we have positive than that power and what happens is we're going to get me to the first so when you're doing these Bernoulli equations and you have the correct substitution and you substitute this in here it's going to leave you with something that you need to get rid of multiply every term that's both sides with distribution to get rid of that and what's going to happen is that these two sides that these are going to be gone this right here will be a difference of one between these two powers so that when you when you get rid of those terms when you multiply those terms and you add your exponents together the difference of one says you're going to get B to the first it's going to leave you with a linear so let's do that let's multiply this whole thing by negative two V to the positive three has negative 2 V to the positive three-halves so if I distribute to every single one of these terms negative two times negative one half gives me positive one V to the negative three halves times B to the three-halves remember when we multiply common bases we add our exponents so this and this done we get DV DX minus it's gonna be plus negative two times negative one gives us plus two but look what happened you know what I'm gonna write this out a little bit differently so I'm going to show you this I'm showing doing too much weight in my hands a lot so I'm gonna show you what happens here's the negative to be to the three-halves that we have or that we have times what we had already our negative one-half V to the negative three-halves DV DX - we have the V to negative one half up here already but now we're multiplying by negative - we did three hats on the right hand side we have B didn't e of three-halves times negative to be the positive three-halves this is what's going on here we're gonna cancel this whole thing out negative to negative one half positive 1v2 three-halves times B to the negative three-halves positive one we get just dvdx we have - oh my gosh negative - that would be plus two we have V to the three-halves remember I told you there's a difference of one of those exponents that's what this substitution does being in the three-halves plus negative one half so that when I add these exponents together we're going to get one every single time V to the 1st that's what Bernoulli equations do they create a linear for you equals on the right hand side V to the negative three-halves times V to the positive house it's the same thing that happens here they have to be the same that's how the substitution is working the substitution is based on that power so that when I subtract one already take it to derivative and subtract one well that difference of 1 here makes it when I multiply to get rid of whatever this thing is it's gonna be this positive 1 but it's gonna create exactly the same exponent over here really neat we get just negative 2 this is exactly what the Bernoulli equation is supposed to do it says that's pretty close to linear let's write it as much as possible it says this guy's got to go let's make a substitution that has Y to the basically 1 minus 3 would give you a this difference of 1 in absolute value let's talk about it that way so that when I solve for y and I take my derivative this piece replaces all of my lives well what's it what's creating for us is this idea that when you multiply whatever it takes to get this piece these two sides are going to be identical I'm going to multiply by whatever it is to get rid of this it's going to also get rid of this but there's a difference of one here between this exponent and this exponent so that when I add the opposite of this it's going to give me positive one that happens all the time that's just created this whole nasty thing into a linear differential equation in terms of the notice we had to replace the dy/dx we had to replace all of the Y's we did that and by the magic and our formula here that I sort of explained to you we now have a linear now to just do the technique for linear so we're getting a lot of practice on this we know that for our linear differential equations we need an integrating factor oh we know that we know that I'm not going to repeat that I spent a long time talking about linear equations linear differential equations so we need something that we can multiply that repeats itself but also gives us this and as far as the derivative of an exponential so our integrating factor here is going to be e to the 2x let's multiply it everywhere if we multiply everything by e to the 2x I had the DV DX and the 2v and the negative two and if we multiply everything by e to the 2x well on the left hand side what we've done and you can check it right here the derivative of this piece is - that's exactly making four we've created the result of a product rule the product of which would be e to the 2x times B this goes back to our linear differential equation so if you don't remember that you need to watch that video or if you haven't watched a video that's really important because this can look really confusing if you don't understand what's going on I spend a lot of time explaining that so that's the result of a product rule here's the product this was not taking the derivative of and neither was that V so we're going to pull that together and now we just need to take an integral of both sides so the derivative of H with respect to X if we take an integral on both sides integrals undo derivatives on the right hand side we've got negative 2 e to the 2x so you could do a little sub sub or or think about in your head we get negative 2 e to the 2x but divided by the derivative of that because that's what you use those du plus C so our T's are going to go away so we get our e to the negative 2x times B equals negative e to the 2x plus C now there's two things we got to do that happen in all of our substitution techniques we're going to solve for V as much as possible but then we have to replace V whether we substituted for V so to be out so soon as V so let's go ahead and multiply both sides by e to the negative 2x if we multiply it by e to the negative 2x this will give us a B this would be negative 1 hey even then 2x times even the names she likes things you either zero is there's one negative one but then the constant gets multiplied times e to the negative 2x well we're pretty close oh maybe right like that but then we're gonna have to go and substitute back to get in terms of why'd we use our substitution to get away from lies we're gonna use to get back to wise my recommendation don't do it here don't do it here do here I'm gonna substitute back for this so y equals plus IV equals 1 over y squared it just allows me to think of this as a reciprocal suffocation idea so if we have 1 over Y squared equals an expression then Y squared equals 1 over that expression I want to talk through it one more time because it's so really important to us all right this is realize sound like little Bunney it's a really important idea for us to master to have down so the idea one more time we're still working on substitution techniques and a Bernoulli equation is one that has a very specific substitution to it it says I want something that looks pretty close to linear so we write something that's pretty close to linear we make sure our dy/dx plus or minus a function that has Y to the first power that's what I want equals something that has a wide to any other power that's not 0 over 1 then what these this renew the equation idea does is just fantastic it says let's let's call V not a direct substitution for one of these but something that when I take a derivative of it is going to let me replace V with something and the derivative of which would have an expert on the that's exactly one different so an of the number one different from whatever this is going to be well that's the substitution that's take take one minus this number and when we take a derivative of it it's going to subtract one from it obviously if I'm taking a derivative of this it's subtract one it's gonna be one different so this piece when I undo it when I multiplied by whatever the opposite of x1 is it can be positive it can be negative I can't say all the time multiplied by the positive because we'll have will have positive zip that we'll have to use negatives so whatever this power is it's one different from that based on the fact that we have to do a derivative obviously we're going to anyway we got to get you by DX so insulted for dy/dx we're getting a power that's exactly one different from this when we start putting this in our associating this into our differential equation this is going to be and because we basis on that power this is going to be exactly the same this is going to have a power that's one different in absolute value it's one less than whatever these powers over it's always different by one I can't say one less or more because a positive or negative but this is always different by one so that when we multiply to get rid of that exponent this number two but that exponent well here and here I'm always gonna get no V's if I get rid of this one I meant be get rid of this one that's pretty cool but this one when I multiplied by the appropriate term to get real pro growth factor to get rid of these two V's on the both sides it's going to leave me with a V to the first power so Bernoulli says linear pretty close make a substitution and make it exactly we want a linear in terms of V and then we use our linear techniques so then your differential equation techniques to solve the rest of the way down make sure we're substituting back in at the very end for V into terms of Y I hope that makes sense we have like five more examples that we're gonna do so we'll pick this up as we keep going I'm not going to spend a whole long time explaining after this the main idea is find one term that has Y to the first find one term that doesn't and then do what V equals y to the one minus that power it'll create something where you have a difference of 1 in your exponents when you undo this one you ought to make me undo this one and you automatically get a power one to force this thing to be linear awesome let's try someone let's try our next one so we have this differential equation we're looking at it we're going that that doesn't doesn't look like anything that's nice I certainly don't want to try a homogeneous on oh that's crazy it doesn't look separable I have lies in a couple places oh wait I've got lies in a couple places do I have a term something's that are subtracted with Y to the first power do have a term with Y to another power yes that right there when you get that when you see then you go I want your head to just flash for me I have a dy/dx great no lies that's important the cavity why is there I've got Y to the first I've got Y to the third so this is what we're looking for now it's not written perfect so when you get these these equations that have Y to the first on one term lined up to another power on a different term we do have to write them as much as it's possible in terms of linear remember what Bernoulli's doing Bernoulli has taken a substitution that's going to force this to be linear and how it does it it creates a difference in your exponents by one so that when you undo one exponent it creates a power one to create a linear so when you do one exponent on both sides it's going to create one in the middle well let's get it as close as possible so as close as possible means you need a dy/dx by itself you need to have Y to the first power being added or subtracted to that as a term and you need Y to another power on the other side as close to possible as a linear so that when we do our substitution becomes a linear that means that this thing's got to go so if we divide everything by x squared we get this - XY over x squared equals five over x squared y cubed I'm also going to do one more thing I'm going to simplify this but I always leave my y's hanging out a little bit like this two over x times y equals 500 x squared Y cube for two reasons number one I want to really emphasize the fact that this is pretty close to linear Hey look at that again dy/dx I get plus something Y to the first power that right there is pretty close to what I'm gonna have as far as my my P of X for linear that's why it's even written as P of X Q of X it's really close to linear that guy screwed it up so this is what the padule-- equation does it says this looks really close to linear except for that guy sucks so let's take care of that let's let's use our let's use our substitution so that we create this we create this this thing that we're going to multiply when we take a derivative that's gonna get rid of this so when we when we take our 1 minus our power we're gonna get V - V equals Y to the negative 2 and much like the last example we're gonna have to solve this for y looks like every substitution technique you're gonna have to solve that for y because every substitution that we're doing is replacing the Y's with something and the dy/dx with the derivative of Y with respect to X in terms of V and X this is different because it's not a direct substitution like we've done it's a smarter substitution it's one of the same pretty close to linear let's multiply by something that makes it linear so we do have to solve for y we do have to take a derivative of Y with respect to X and we do have to make our substitution and now I'm going quickly through the substitution because we've done it so many times that we should know at this point I'm going to be replacing my y's I'm going to be replacing my dy/dx and it's all right here so every instance of Y becomes B to the negative 1/2 every instance of dy/dx which there's only one of them in a linear becomes this so let's replace that right now so dy/dx no dy/dx was negative 1/2 B 2 negative 3 halves DV DX well let's see what's the next part of it well we got a 2 over X okay so that that we don't replace with the why we replace Y we replace with let's see what what is y oh here it is B the negative 1/2 on the right hand side 5 over x squared we don't replace so we're only replacing our wise here but when we do it we get this be either negative 1/2 cubed replace D by DX left the x's replace the Y left X's replaced the why why we're doing that now fine as far as linear it was this but I looked pretty good too it was this piece that says this doesn't look linear or really only trying to replace that it's just so happens that when we place one why we're gonna replace all the Y's you can have 3 variables so this one just kind of plays along after we take a derivative it's creating for us because we subtract one from it it's created for us something that when we replace Y with this power and take a derivative it has something that's one less than this power there I can say less it's one off it's one off so then when I undo it Wow like that that looks crazy well how would I make it better I can get rid of this because my goal remember my goal it's trying to get linear we want to multiply so that we have the dy/dx by itself let's do that so when we multiply everything by the same exact thing from the last example this V 2 negative 2 V to the three-halves I'm not going to show it like I did last time because it's very very similar but negative 1/2 times negative 2 is positive 1 v e to the negative 3 halves plus 3 halves is 0 V 2 0 is 1 this whole thing is gone we just get DV DX this is what we're trying to do this is what Bernoulli equation says it's gonna happen we're gonna get a linear or just trying to make it look lennier so when we multiply you go yeah okay I'm gonna multiply by that but the way we structure our substitution is that because the derivative always subtract 1 from this when I multiply by the opposite term of the opposite sign and I distribute I'm going to be something that gives me positive 1 there so I have a I'm going to write my 2 over X and the red might be the negative 1/2 times this negative 2 V to the three-halves we distribute that on the right hand side we have a negative 10 Casey where the negative 10 comes from I have 5 times negative 2 over x squared but this is V to the negative 3 halves feed in May of three-halves times V to the positive three-halves is V to the 0 that that multiplication right there in getting rid of this factor it's always going to get rid of your V factor on the right hand side that's because we structured this substitution that lay on the in the middle we get this DV DX that looks pretty good but we're gonna now get minus 4 over X 2 times negative 2 is negative 4 over X V to the this is the magic this is what happens every time as we structured correctly negative one half plus three halves gives us positive one now that's a linear in terms of V and X so use your technique for linear we know that our row our integrating factor here would be e to the Ln of negative 4 over X done this many times that would be e to the negative 4 Ln X or if we want to think about this way e to the Ln of let's say X to the negative 4 or e to the Ln of 1 over X to the 4th that would work or just simply 1 over X to the 4th because if we get rid of our DN a laburnum and there's a composition of inverse functions so I know that our row of X is 1 over X to the 4 so you've taken this we've looked at and said that's pretty close to let's and 90 plus lunar but I see a y the 1st power Y to the 3rd let's make it as linear as possible let's make a substitution that when I find a derivative and I plug in this negative 1/2 it's going to give me the exact same power on both side so when I get rid of one side I get rid of both sides also it's giving me something that is 1 power off so when I multiply this times the thing that gets rid of that it's giving power 1 we do all that we get rid of the term that we don't want the fact that we don't want we get something that's linear and then we do use our integrating factor do these problems take a lot of work you better yeah they take a long time and now we're ready to finally start using our integrating factor so that we find the result of a product rule write it as a product rule and integrate the other side so we had DV DX we had minus 4x times V and then equals negative 10 over x squared but now we're going to multiply by our integrating factor multiplied by this 1 over X to the fourth it's gonna distribute and now we can simplify some stuff so 4 over x times 1 over X to the fourth is that 4 over X to the fifth and then the right hand side negative 10 over X to the sixth this is 4000 Linear's do they create the result of a product rule so when we when we get to this point we need to realize that what we multiply we multiplied was to create the result of some sort of product rule few videos ago this wouldn't have been a piece that was not that is not the derivative of anything this was like the leave the first alone derivative of the same so this would have been the result of something with the 1 over X to the fourth this would have been a derivative of that well you double check that if you really want you bring up the negative 4 that would be negative 4 next to negative 5 that's right there but that piece would have been the result of not taking a derivative that's what I get that would be every time on the right hand side we have something that should only be in terms of X if you have these over there you've done something wrong or this is but this technique will work so these powers should always cancel at the same time well if this is a derivative of and if this is the result of a derivative of this piece with respect to X and integral will undo that we get 1 over X to the fourth times V equals well on this side we can think about that as negative 10x and they you see to pull out the negative 10 so we have negative 10x to the remember adding 1 here negative divided by a new exponent making 5 plus C so this gives us this one or X to the fourth let's look at good times me all right negative 10 divided by negative 5 is positive 2 if you want to move that X to the negative 5 back back down to X to the fifth you can let's see I hope you're still with me on this one we're just doing some algebra now what's the next thing you would do what's next thing we have to do well we have to plug in whatever our substitution was for V but then we also have to get rid of our X to the fourth so I'm gonna solve for B as much as possible first I'm going to multiply everything by X to the fourth so X to the fourth gone thanks to the fourth gives me 2 over X X is or give me C X to the fourth and now we're ready to substitute back so we use our substitution to get into these we're going to use it to get back to wise I don't see what was it that's way over here so all the way down we said that V equals this yeah V equals V equals this or the V equals 1 over Y squared I like that substitution better the reason why I like it better is because it tells me what I need to do it tells me I'm gonna have to reciprocate this to get not 1 over y squared but Y squared that means in order to reciprocate this I need a common denominator I need one fraction so I'm going to take just a moment I'm going to multiply C X to the fourth by x over X s to give it a common denominator that's gonna give us 1 over Y squared equals 2 over X sure let's see X to the fifth over X which means that we can write that as one fraction if one over Y squared equals 2 plus C X to the fifth over X then Y squared itself well that's got to be x over 2 plus seen X to the fifth or C X to the fifth plus 2 that's the idea with these these Bernoulli equations is we're trying to make it linear we're going to be using linear techniques it's just we have to get rid of Y to some other power that's not 0 or 1 how it works is we make this substitution V equals y to the 1 minus that power what's going to create for us because we're typically because it's based on that power because we're basing on the power we want to get rid of what it's gonna do for us it's gonna create something that when I undo one side very read to one side B to the power on one side it's gonna get rid of both sides it's also creating something for us that when I make my substitution I have a power that's exactly 1 off from that guy when I multiply it it's going to give me power 1 it's gonna create a linear we're gonna try for more examples to really make this sink in I want to explain something just a little bit more I'm gonna say a lot of times that in getting rid of something you have over here this factor you're also getting rid of this one I want to show you that that that happens all the time so so real quick if you're gonna make this substitution V equals y to the 1 minus n well if you if you solve this for y this is gonna be a little bit of a theory behind this if you solve this for y you're always going to get y equals v to the 1 over 1 minus n so you can think about this as multiplying both sides are taking both sides with excellence so multiplying exponents by the reciprocal of this so 1 minus n okay how do I get rid of that one over one minus n all right cool if you take a derivative what would you do well you'd bring down the exponent we don't really care about that so much but you would take that exponent minus one this is what I want you to look at when we simplify that when we get one or when we subtract it actions we would end up getting 1-1 give you zero one minus negative n this would give you one minus one plus n over 1 minus n or n over one minus n which means this when you gain this is DVD n when you replace your dy/dx you will always be getting this 1 over 1 minus n who cares that's a coefficient no problem I'm V to the N over 1 minus n DP DN or DV DX you're always gonna want to get rid of this piece this is always gonna be the fraction of X once you want to get rid rid of keep that in mind but not watch when you substitute this in for this because this n is based on this watch what happens when you take this as y equals V to the 1 over 1 minus n you would get V to the 1 over 1 minus n to the N that's V to the N over 1 minus n here's what I was saying because of this Bernoulli substitution whenever you make this substitution and you take a derivative you're going to get this piece on the left right there you're going to get this piece on the right due to your substitution which is why I've been saying last two examples that when you can get rid of this piece you inherently get rid of this piece because that's what the substitution is based upon more than that when we take a derivative we're always going to get the derivative see this is always going to be 1 let one more than whatever your substitution is so when we do that when we multiply and get rid of these two pieces here and here so on the left side the right side it automatically creates V to the first power in the in the middle throat to make that linear so I hope that's a little bit more thurible explanation about why this works that it works but I never showed you that we work all the time so this is going to work all the time for us that's really really cool that when we substitute and we get rid of the left piece we get the right piece because it's based on that exponent and our derivative makes the same exact exponents what we're getting when we do our substitution so let's move on to the less videos with that in mind okay so next one we have differential equation but it doesn't look that great in fact it doesn't look quite like the badou the equation we think it might be it's definitely not linear right now doesn't even look close to linear doesn't look separable doesn't look like anything substitution so maybe what we can do is we can try to make it Bernoulli make it into what we want to be by dividing by one squared in fact what I see here is I see Y to the third and one in a second maybe if I'd 2 by I get a Y now this guy's gonna play along give us a different exponent but let's go ahead and multiply both sides by Y to the negative 2 power the idea is I'm trying to get rid of that Y squared and I see but that's just one power higher so when we multiply by Y to the negative 2 power now you might be asking why aren't you just divided by Y squared I really am divided by Y squared dy DX divided by Y squared dy divided by Y squared you get 6x over Y squared but now we look at it and we go oh oh man that's really close to linear this wasn't here that'd be close to linear do you have two terms one of which has Y the first power and one of which has lied to not the first power that's why I wrote that is y to the negative 2 instead of 6 x over y squared so that we can fit the Bernoulli equation model that's cool right now what I would like you to do if you can make your substitution so Bernoulli always has the same type of substitution it says I want you to replace Y - well I want you to make the substitution v equals y to the 1 minus whatever that exponent is don't get your signs wrong screw the whole thing up I just went through a proof of that if you don't mess this up you're going to get something on the derivative that goes here and something when you take the negative second power matches up exactly but if you mess up your signs it's not going to work so V equals y to the 1 plus 2 that's the third power so a writing that says linear as possible we're making our substitution Y v equals y to the 1 minus that exponent and then like every substitution ever that we've done we solve for y well if V equals y to the third power take both sides to the 1/3 power y equals B to the 1/3 like every substitution ever we solve for y and we know that we're replacing two things every instance of Y hey we've got that that's going to all the Y's you review to the 1/3 and dy/dx so we did it find that I went through that proof just a little while ago to show you that when this happens when you do your derivative we always get the same exact exponent as when we take this to that power notice 2/3 1/3 to the negative 2 is negative 2/3 I proved it just a while ago on why that happens so our substitution is going to be very nice for us we're gonna take and replace our wise with this our dy/dx with what we just solved for so I'm gonna do that right now so dy DX no this thing plus 2x yeah that'll still be there it's no close to it but the Y well we're gonna replace that as well that's what the substation does back so times of e to the 1/3 power notice right after that 1/3 and negative 2/3 are separated by 1 so when I get rid of this I'm going to get power 1 here on the right hand side get six things still but then we're gonna have why all right so we replace that why but we're gonna keep the negative 2 power so B to the 1/3 to the negative 2nd power these have to be the same exponent every time they have to be off by one so that when I get rid of these two exponents that's got to give me a power one so let's do that now what we're trying to do is we're trying to make this linear it's pretty close or come with a substitution that when I get rid of this thing it's going to force them to be a power run however one when I get rid of this thing it's going to force this V to disappear as well that's how this bonier the equation works so we need to get rid of this so that we have our DV DX then we're gonna power one linear we've no more B's think about what you would have to multiply by so we're going to multiply both sides by three for the 1/3 V to the positive 2/3 and the same thing there so 3 V to the positive 2/3 3 kills the 1/3 me to the 2/3 V the native 2/3 this is all gone that's going to be DV DX plus remember this is going to distribute so 3 times 2 gives a 6 sure X you still have it X but then we have V to the 2/3 times V to the 1/3 2/3 plus ones are up for you the first on the right hand side 6 times 3 gives us 18 we still have an X but since this is V to the negative 2/3 and this is V to the positive 2/3 negative 2/3 plus positive 2/3 is 0 it gets rid of our V I proved it that that's going to happen ever time so he has something that's linear nice let's go ahead and find an integrated factor because that's how we solve linear so we're going to find our role of X we know that role of X says one thing it's got to be e it's got to be an integral of this right here it has to be the result of a derivative of your exponent so undo that to find exponent itself when we integrate 6x see bad 1 is 2 divided by 2 so 3x squared let's multiply everything by that so we have a DV DX plus 6x times B equals 18x and we're gonna multiply it both sides so every single term by that e to the 3x squared check it out what's the derivative of 3x squared that's what you want on the left left hand side I don't even care if you rewrite that it doesn't matter what you've done inherently is you've created the result of some sort of product rule implicitly with respect to X where this would have been the first factor and that V would have been the second factor you can see right here leave the first alone here take the derivative of the second oh this is the derivative of the second the second is B on the right hand side we have an 18 X e to the 3x squared not too bad so we know how to undo derivatives we know that integrals with respect to whatever very we just took a derivative with can go on both sides on the left hand side we know that we have e to the 3x squared times the V on the right hand side well we're gonna need a substitution so if you equals 3x squared D u equals 6 X DX d you over six would equal X DX and so our integral here says well you got an 18 but our X DX becomes this D u over six and then we have e to the U so if 18 stays e to the 3x squared no we called three squared u so e to the U yep sure X DX you not only solve for that now we have D over six so when we simplify this just a bit 18 divided by 6 gives us three integral e to the UD u man we're almost done we know the integral of e to the U is going to be e to the U with respect to U we'll put a plus C up there and then we're gonna have to go ahead and replace that u we know we made a substitution we're gonna have to get back to that so U is 3x squared tell me have another substitution to undo we're gonna have to to replace a B but before we do that I want to solve for B as much as possible so I'm gonna multiply both sides by e to the negative 3 x squared this is gonna go away so is this one because 3 x squared + remember multiplying a dredge shows 3x squared plus negative 3x squared 0 3 squared plus negative 3 squared 0 but then this is going to get that e to the native 3x squared we have to distribute to every term on the left hand side just V on the right hand side we get 3 times what's going to be e to the 0 in 0 is 1 3 times 1 is 3 plus C e to the negative 3x I hope you're sticking with me here folks now it's just some algebra but it's important stuff now lastly we should know what to do we know that we've got to replace the V with terms of why some we're sort of looked way backward okay there's beef there's that's a better B to choose so I'm going to replace a B with Y to the third and you know what I'm gonna leave it just like that can you take a cube room yeah you probably could does it really super matter not really I would leave it just like that so after understanding the the proof that I gave you that this is gonna work every single time the idea is write a linear base your substitution on that number you're trying to get rid of this piece to make it linear and I showed you why that works just a minute ago what it's gonna have what's gonna happen is it's on your two riveted you're gonna get a piece that you want to get rid of but that exponent is always going to match up to this exponent when you get rid of one side you get rid of the other side that's fantastic also because you're taking a derivative of this piece right there you're you're creating this exponent upon your substitution that is one off one more or one less than whatever excellent you're getting on your derivative so when you get rid of this exponent you're multiplying by something that's different than one here that's gonna create that's gonna create something that's V to the first power it's creating your veneer after that find your integrating factor row multiply to all three terms go ahead and do your integration and hopefully you're getting the picture here man I hope it's making sense I wish I could ask you right now but I can't so we're gonna do three more examples to make it sink in and then we'll be done okay here we go so something a little funky because we have a 4/3 power but I want you thinking through this in the following terms number one is it easy is it a basic integration no remember that even though we're on a video that says Bernoulli equations I don't want you to forget a second learned so if there's another way to do it we might explore that because these frankly take a lot of work I'm going to show you that on the last example that I'm alone say this can be done a different way I'm going to show to you in another video so we think about it a while is it something that's easy that it's as separable is it a direct linear is it a substitution that might be easy it doesn't look like it but what I do see I see a function term with Y the first enter without Y to the first and no otherwise that's great let's try to write this in the form of linear as much as we can so linear says this is that's a dy/dx by itself it has something to the term to the line of the first power right next to it so add or subtracted that X has to go so linear says I need this piece by itself I want a term that has can't that exit sure but it's got to have Y to the first and then over here I have all my other junk so X's and Y's great because we're gonna do a substitution if this fits the bernoulli model so if we divide by X so gone 6 over X and just 3 so we divide everything by X that's exactly what we want this is really close to linear it just has an Y to the 4/3 and we're going to want to make disappear for new the equations do that for us that substitution the bernoulli substitution does that so let's make V equal to Y to the 1 minus whatever the power is of the thing that you want to get rid of that would be let's see one that's three thirds minus 4/3 so negative 1/3 negative one negative 1/3 now we've got a saw just like every other substitution ever we've gotta solve for y because not only do you want to replace the Y which you can't do right now you need to be able to find dy/dx replace that - so maybe multiply both sides by negative three that would raise both sides to the exponent negative 3 that would allow you multiply negative 1/3 times negative 3 multiply exponents gives you 1 and then V to the negative third power alright can you take a derivative sure so dy/dx would equal negative 3b to the negative form DV DX we've created something that when I plug in this to this and raise it to the power we'll have exactly the same exponent that's awesome that's what we want to have happen so now we're ready to do our substitution we looked at and said linear almost let's divide by X linear that's got to go let's do a substitution let's solve for y so their substitution works let's find dy/dx so that our substitution works and now we're ready to rewrite this so dy/dx no I'm gonna replace it with this piece plus 6 over X yeah I like that y no I don't want Y I want the replacing Y with the B to some power that's one different from when I took a derivative so I'm gonna replace this with feed in negative 3 notice always happens because of the case of the derivative we talked about how many times when you take a derivative this exponent changes by 1 in undoing that you're going to create this idea that when you're adding exponents if you're off by 1 you're gonna get B to the first power on the right hand side we keep our 3 but I don't want the Y in fact that's the whole thing I'm trying to get rid of that's why I'm doing the substitution in the first place that's got to be V to the negative three days on the 4/3 and when you simplify this what I approved this power is going to be same as that power we still have our three negative three times four-thirds this negative for catholics good looks really good so what now well it doesn't look linear - you might become that doesn't look good your your writing but when we get rid of this it's going to force it to become good so when we get rid of this in other words when we multiply by negative one third the each of the positive fourth power the positive 4 and the negative 3 since these are off by one that's going to give us power 1 when I multiply this it's going to give us negative 4 and when I multiply by 4 it's going to when I multiply by negative 1/3 and be to the positive boards it's going to give us 0 it's going to give us V to the 0 that's 1 so let's do that so on both sides we're going to multiply by negative 1/3 reading out of the fourth to try to get rid of this side so negative 1/3 times negative 3 that's positive 1 V to the fourth times V to the negative fourth that's B to the 0 this is Lemos of DV DX when we distributed right there we'd have 1/6 sorry negative 1/3 times 6 over X that's negative 2 uh sure that you on the right hand side we should see what's happening three times negative one third is negative one and then feed a negative fourth times B to the fourth that's B to the 0 that's one so this whole thing is you can negative one here is where the awesome magic happens negative 1/3 times 6 is negative 2 so we're gonna get minus 2 over x times V ah V to the fourth times V to the negative 30 adding exponents we can view the first now that's something that's linear we like linear linear works really nice for us so when we get this we go well since we have something later we know our P of X we know our Q of X eated well function of X and now we're gonna find our integrating factor so rho e to the integral of whatever this function of x is which means for us we get negative 2 Ln absolute value of x we're going to give me to that and a lot of times what we'll do is we'll say keep X positive that way we get rid of our absolute value happens quite a bit they don't even show that in some books to go out as just as so much positive and then we get Ln X to the negative 2 that would be Rho of X equals e to the ln 1 over x squared composition of interest functions say they are multiplied our integrating factors just 1 over x squared that's what we're going to multiply everything by so let's do that if we take that and we multiply it by 1 or x squared we're going to create the result of some product rule on the left hand side you do t have to read you can rewrite it really want to if you want to write this as negative 2 over X to the third that's fine you'll just see this that would be the derivative of this piece this piece has to be the result of some sort of a product rule when we had the product of 1 over x squared times B now that we have that we're going to integrate both sides I hope you don't mind but I'm going a little quicker on these ones because we've covered the derivatives of linear differential equations sorry the solving of linear differential equation so much that we're going to kind of rip through these if we can this would be the integral of X to the negative 2 DX don't do it LS now Ellen we're gonna get let's see add add 1 is negative 1 divided by the new two so negative x to the negative 1 over negative 1 that's 1 over X plus C negative over negative is positive take that negative exponent back to the denominator if we multiply both sides by x squared we get V we also get X plus CX squared not too bad if we distribute that we're almost done we know with substitutions we replace wise with the beasts now we replace these back with flies so that's a back here V was you know what I don't like the negatives so what if we did I got rid of money over there the equals y to the negative 1/3 so we could write this as V equals 1 over Y to the 1/3 that might be a little bit nicer V over V equals 1 over Y the positive 1/3 and now this gives us the idea so I don't like this negative exponent because it doesn't give me the idea that reciprocal now I want to do that so 1 over Y to 1/3 equals this and why do the 1 clear itself equals x plus CX squared what else could wow that's not true I said reciprocal about reciprocal what else could we do now that we have reciprocal reciprocal that way we have 1/3 power maybe we raise both sides to the 3rd power just to make it look nicer so raising to the third power this would still give us 1 but 1/3 times 3 gives us Y to the first on the right hand side b1 put on the denominator we get X plus DX squared quantity that looks a little bit nicer so we don't have to deal with that 1/3 power is it making sense to you are you seeing the ideas kind of work behind or not do the equations at this point we're trying to get linear it's close we just need to get rid of one term that has wide to a power that's not one if that's the case we've determined that making V equal to Y to the 1 minus that power does some pretty cool things that proved to you throughout the course of these examples and with an actual approval or a while ago that this will always work it'll always create something that's linear for us then we may find an integrating factor and the rest is just history so let's do two more examples to put this thing to bed okay here we go so our next instance of Bernoulli equations here and what you're looking for again is you're looking for two terms one with a Y to the first won't they why not to the first but you're also trying to write them as linear and as possible to fit our model so that when we make our substitution it becomes linear we talked about why that is it forces it so when you deal with something like this the form is really important which means that that term and that term are on the wrong side with lynnie every one our dy/dx all by itself that's gonna have to go if we want our plus or minus a term with Y to the first that's got to move and then on the other side the term with y to another power that's got to move so we're gonna do two things we're gonna switch these terms subtract subtract and we're divided by 2x so we've subtracted those two terms on both sides that looks fine with that 2x has to go so we'll divide everything by 2 X and we'll get dy/dx no problem that's gone we'll also get minus y to the first that's exactly what we want look at that we want dy/dx you want - why - the term with y to the first power on the right hand side we'll do a couple things we want to show this Y last remember we're trying to make substitutions on Y so the function of X is okay I'm gonna write negative e to the negative 2x over 2x I'm going to show this Y to the third at the back end that is what's driving this substitution that says hey if you make this substitution of V equals y to the 1 minus 3 V equals Y to the negative 2nd power solve that for y we know the substitution we're always solving for y because we need to get rid of dy/dx after we take a derivative back so let's solve that for y we can use to make a substitution and get rid of that piece we'll take both sides to the negative 1/2 power that right there is going to let us substitute for Y here and here when I take a derivative of it it's going to subtract 1 creating this piece that when I get rid of it well so get rid of this piece with this razor third power and it's going to create an exponent upon a derivative that is one off so that when I get rid of it creates a B to the first power so let's find that derivative I think we've done this one like three times now so that derivatives not much different from anything you've done but this substitution is important be careful in this we're going to substitute dy/dx y and y we got to get rid of every instance of Y dy DX is now that derivative let's replace it - Sonic okay we'll get - but why we need to replace a B to the negative whenever equals this whole nasty junk stays but then Y to the third power we need to get rid of every instance of life that's what drove the problem to make us choose a substitutions that we create the appropriate fractions upon derivatives that allow us to get rid of both factors on the sides and one from the creative to the V of the first power in the middle so why now B didn't make it one half but we still have that third power let's simply just a bit here we're going to leave everything but that we're going to make that negative three halves so we can do the gates these two have to be the same if they're not you've done something wrong this one's gotta be off by one from that exponent if not you've done something wrong and now we're gonna try to make this linear that's a whole ideas make it linear so when we multiply to get DV DX by itself which is what we're trying to do here to make it linear we're gonna have to multiply both sides by negative two V to the three halves negative 2 V to the 3 halves well when we start distributing a negative two times negative one half is positive one lead in three paths times of even negative three notes gives us V to 0 that's just DV DX I'm gonna rewrite I'm gonna do this and write it from city citizen for you so that you see what's going on we have a minus negative two V to the three halves times being in the negative one half so I kept my - here's my negative 2 V to the 3 halves here's my V to the negative 1/2 let's let that hang on for second on the right hand side really see this negative times a negative or minus a negative gives you plus B to the if these are always different by one when I when I add the exponents together you're going to get positive one every time is just V on the right hand side well let's see negative times a negative is a positive I'm not doing anything get rid of that E but two times one half gives us positive one and then over X no V V to the negative three halves times B to the positive 3 house B to the zero so this is just e to the negative 2x over X man now what hopefully you know now what now that we have something that's linear and it is linear get DV DX by itself plus V the first power people's function of X we just need a gradient vector kind of cool that an integrating factor is this a really easy one well if I were to forget the two one hopefully you didn't do that so minus 1/8 I said positive I just forgot to do so 2 is our function of X when we find an integrating factor e to the integral of 2 DX well that gives us e to the 2x which is really convenient because when we take and multiply everything by e to the two x so x e to the positive 2x e to the positive 2x e to the positive 2x even the negative 2x times e to the positive x and direction on sets e to the 0 that's 1 these are gone on the left hand side we have the result of some sort of a product rule with that products already given to us e to the 2x and B so that when we take an integral on both sides on the left hand side fundamental theorem of calculus says that's that's going we just get e to the X e to the 2x times v on the right hand side the integral of 1 over X DX is just Ln X a lot of times you're gonna see no massive value then we'll say a little 7 I keep absolutely they keep X positive and then this will work with no absolute value makes it a little bit nicer for us well now what could we solve for B well yes we could all we have to do really is multiplying by e to the negative 2x on both sides so just multiply an e to the negative 2x on both sides and now we're ready to get back into our watch so let's look at it V equals y to the negative 2 or 1 over y squared I like that better if it's in my mind that I'm going to be reciprocating something so V is now 1 over Y squared okay we can deal with that ln x plus c even a 2x and now all we have to do is think that well if you have 1 over Y squared equals this expression then Y squared itself equals 1 over this expression one more thing you might want to do because you have a negative exponent on this factor in the denominator you can actually pull that to the numerator that's about as good as we can get we don't want to be taking square roots so Y squared equals e to the two x over Ln X plus C this is about as bad as it gets I'm going to show you one more example because I want to illustrate that the next example we talked about it can be done two different ways so are you getting it are you getting that we want to make linear out of this and Bernoulli forces it to happen by getting rid of something that we don't want a power that's not one for that y factor great substitution works every single time if we can write in this form then we solve for y I like always with every substitution solved for y use that to replace it later but that lets you take a derivative of Y with respect X and get something to replace dy/dx so let's move on to one more example okay so this looks crazy and you look like anything ever the only thing that looks pretty close is that that looks okay so we looking at this we're going that no please no why why would you because I want to but secondly because you're gonna give stuff like this it doesn't fit anything that we have so far as far as being easy and so one thing we know is that when we have a term of Y with a power one and something with y that's not a power one maybe we can fit in linear as much as possible and use it Bernoulli equation making it fit a Bernoulli equation so let's try how to do this what not to do do not distribute all this stuff you get Y to the third that ruins the Y that you have the idea would be well get all your X's on one side multiply by Y to the negative tooth to get the power of Y on the other side so let's do that let's leave us where it is let's leave this X let's divide by this and let's multiply both sides by Y to the negative 2 the idea is we're trying to make this into linear as much as we can there's one more thing that we got to do do you see if we got to get rid of that bag so we looked here we said I want I like the light of the first power but all this other garbage has to go let's divide we did let's get Y to the negative 2t rid of this we did and now let's divide everything by X including this so we'll have a dy/dx plus 1 over x times y remember I'm always going to write my function of X next to my Y just in front of it equals these X's are actually gone so we get 1 over the square root of 1 plus X to the fourth times y is a negative 2 my encouragement to you right now because the rest of it is just some stuff we've done already I would really want you to try this on your own try to go ahead and find the appropriate for new the equation substitution see if you can find it solve for y see if you can find the derivative see if you can substitute see if you can get rid of the piece that you're gonna have over here see if you can find you integrating factor see if you can find the linear differential equation in terms of B solve it and replace Y see if you can do that so let's see what we do we know that we can make a substitution where we have V equals y to the 1 minus whatever exponent this is on the right hand side that you don't really want that's making this not linear this will undo it when you take a derivative and you get rid of the piece that you get upon derivation of this after solving for y so we get y equals RV equals y to the third power if we solve for y that we have two on every substitution multiply both sides or take both sides to the 1/3 power we would get 1/3 let's see it's one 1/3 B to the 1/3 now this lets us take a derivative and we begin 1/3 V to the negative 2/3 notice how this exponent when I substitute it here and this exponent when I substituted here are off by one when I get rid of this exponent it will cause this one to become positive one creating for us this linear thing linear differential equation terms B that we want this to be also notice that when I take this power to that power I get exactly that power that means when I get rid of this thing with two great this linear in terms of e it's going to get rid of this whole entire factor as well so let's substitute hopefully you've done this and we can work this through as an integral in just a bit so let's see dy/dx know this junk plus 1 over X yes why no no we want to be to the one third equals all this garbage absolutely why to the negative second no B to the 1/3 that's our wife to the negative second let's simplify just a bit let's make sure that we can see that this exponent will be exactly same as that and then we're going to try to write this as linear which means that we're going to multiply everything by the reciprocal of 3 and the opposite reciprocal of that sorry the opposite of that fraction so both sides get multiplied by something that removes this factor because linear has just dy/dx in it and just some function of X V to the first power and just some function of X so when we multiply both sides hopefully you see it we're gonna have to multiply by 3 v to the positive 2/3 if we let that distribute 3 times 1/3 that's 1 V to the 2/3 times B to make 2/3 they cancel we just get DV DX Plus let's show this we'll have 1 over X V to the 1/3 times 3 V to the 2/3 off by 1 look how when I multiply by the opposite of that it adds up to 1 M times V to the negative two-thirds times 3 V to the positive 2/3 2/3 and 1/3 0 V to the 0 is 1 1 times 3 is 3 over square root of 1 plus X to the 4th here we get this DV DX plus we have a 3 in that and 3 into the Sun 3 over X V to the 1/3 2/3 that's one love it now you have something that's linear' you have DV DX you have function of x times whatever this variable is right there and you have a function of X find your integrating factor composition of inverse functions we have row of x equals x cubed just X cubed let's multiply that on all three spots so we're going to go back up here we're going to say let's take these things DB DX plus 3 over X B equals 3 over square root of 1 plus X to the fourth and let's multiply both sides so every single term by X cubed so we've got X cubed and we've got X cubed and we've got X cubed on the left hand side we have now created the result of some sort of product rule with respect to X on implicit differentiation so this whole piece and yeah you can simplify this this would be 3x squared we can actually get 3x squared it's the derivative of x cubed that's exactly right because if this is the result of a product rule then we leave the first alone we would have got X cubed from X cubed times the derivative of the second then if that's the derivative of the second factor V would have been in because the derivative of V with respect to X is DV DX and now we're down to something that we can just do with it nice integral with respect to X both sides X cubed times B looks fine on the right hand side well you're gonna want a substitution and thank goodness this came up with a substitution so if u equals 1 plus x to the 4 D u equals 4x cubed DX let's divide both sides by 4 and it looks like we're gonna get this um let's see we like three one three okay so we got a three let's keep the three let's notice that X cubed DX is do report over the square root of U let's clean that up just a just a tad here so let's pull out the three fourths that's right EU to the negative U to the negative one-half power did it we add one divide by the new exponent so you to the positive one half over one half this is going to give us and see four times what happens to this is going to give us three half three paths U to the one power let's see let's put it back so we know we have to be in terms of X let's go ahead and make our you back into 1 plus X to the fourth so I'm doing that I'm gonna make a square root one half as a square root but inside should be u so inside should be one plus X to the fourth plus C oh man we're so close you know we could do we just divide everything by X cubed let's try that let's divide everything by X cubed to really make this into one fraction so what I've done I've made a 3 square root of 1 plus X to the 4th over 2 and then C and if I divide that X cube have it divide this by X cubed and the C but X Q so to be to be mathematically almost like grammarly correct here even though that arbitrary constant could easily be two seats you want to do this you'd want to show that you're getting a common denominator you're gonna want to show that and you don't want to show that then you're gonna change to C sub 1 into just C so I'm have to move over here now we can say you know what that that's still a constant let's just call it C now we're gonna do two things right now so we're gonna call that C but also we know that V is equal to Y cube let's replace them so we're gonna have a 3 of course we have a square root of 1 plus X to the fourth no problem we're gonna have a plus C because that's the constant we found a common number so 2 C sub 1 made C and then we're gonna have to execute just leave it we can take everything to the 1/3 power if you really really want to I wouldn't leave just like that that's pretty good and that's the idea for Bernoulli equations is you're trying to make something that's linear it should be pretty close or you should be able to manipulate it to be pretty close that was a nasty problem because it doesn't look close at all in fact I want to say a little blurb about that we're gonna have some embedded derivative problems next and I'm gonna show you how this is the result of one of those embedded derivatives so I'm going to show you that we're gonna do this a different way coming up later in a few few videos so sometimes there are differ techniques that we can use we've seen that a couple times already this is one that maybe it's easier a different way so anyhow with Bernoulli the idea is you're going to try to find linear you're gonna want dy/dx something with a y so plus or minus something with y equals something with a different why your substitution always should be Y to the 1 minus that exponent of a thing that you don't even want what's going to happen upon your derivative you're gonna create some sort of a factor that when you get rid of it it gets rid of both of these terms and creates a B to the first power then we have something linear and we know in a deal with that I hope that it made this make sense to you I hope you see how it works I hope you're not just running into this like well I just do this in this and it works somehow hope you saw and the proof that the reason why is because we manipulate that that exponent in such a way that your derivative helps you and that's pretty that's and that's nice it's really nice and then sometimes our integrals can just be nasty but stick with it you know how to do it you guys are going to be defined on this stuff so I'll see you for the next video we're gonna talk about a slightly different substitution technique I call embedded derivatives time to look at some domain again so very similar them to keep this very short and sweet when we're doing these these Bernoulli equations or doing these differential equations by that method we're having to rewrite stuff we're gonna run into domain issues and we always know that we're getting linear so we're gonna get a lot of the same exact domain restrictions that we did in those linear first-order difference equations that we're gonna get right now so when we're rewriting this obviously we're gonna try to make this fit Bernoulli so we're dividing by x squared great we simplify awesome but you divided by X Y divided by x squared X can't be 0 but because we're going to get down to this integrating factor Rho x equals e to the negative for absolute value sorry integral of 1 over X to avoid that absolute value inside that Ln did and some domain restraints of domain issues that we might get with that any negatives inside Ln that would be a problem we avoid that right now so it's not going to simplify later very well if we don't do this so we say yep you know what X can't be 0 for sure because we're dividing by X but more than that to deal with this right here we go well let's just fix it all with this right there and do X is greater than zero that takes care of the dividing by zero part that also takes care of the no more need into the absolute value same thing happens here we're dividing by X so X can't be 0 to avoid the absolute value X has to be greater than zero this one is interesting this one it's not necessarily the integrating factor that causes an issue we're dividing by X sure we divide by 2x so I know X can't be 0 if it wasn't for this piece that's where we leave it but we end up running into this hey when you wrote your product rule and we take this integral on both sides well in order to avoid that absolute value any domain problems of having negatives inside of Ln no matter what we pick for our X well we're going to say that X has to be strictly greater than 0 so if you're understanding the idea to to divide by a variable you need to restrict the domains that you can't have that equals 0 to avoid absolute value we just say let's make it greater than 0 and that's what we're doing here so on the let's see second third fourth and the fifth problem here the last one I'm a telescope like it was fourth and did ya sorry second third six problems second third fifth and six problems last one was crazy but we run into the same exact idea when we start taking this and saying well you know what I'm gonna have to divide what am i dividing by well let's see it B divided by Y I do why can't be 0 I know that that can't be 0 so Y can't be 0 oh I have something over X can't be 0 but now I'm thinking I'm gonna have this integrating factor that has an integral of 1 over X I'm gonna try to avoid that absolute value that helps me simplify that avoids two main issues later on and so where'd you let all that up that's a greater than 0 so I hope that's making sense to you again when we're trying to modify our differential equations to fit a technique we are gonna run into some demand issues we're dividing by these variables so to sit to save yourself a whole bunch of work later on we're gonna restrict our domain right when we come up to it we have hopefully mastered these Bernoulli equation understand all about them and now we're going back and going yes but but we have some issues we're dividing by variables those variables can't equal zero to avoid absolute value to help us simplification and not having negative two-sided l ends in certain cases we're going to restrict it even more than that and say let's assume that variable is strictly greater than zero okay that is the idea I hope that makes sense you

Channel: Professor Leonard

Views: 108,159

Keywords: Math, Professor Leonard, Differential Equations, Solve, Bernoulli Equations

Id: NjIMGAIPbzg

Length: 103min 35sec (6215 seconds)

Published: Mon Dec 03 2018