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Differential Equations Exam 1 Review Problems and Solutions

hey everybody in this video i’m going to gothrough a bunch of differential equations exam 1 review problems with solutions that will helpyou whether you are in calculus 2 differential equations or differential equations with linearalgebra these are problems that i’ve given in the past i might give them in the future i’msure they might be on your test too so we’re going to start with some separation of variablesproblems we’re going to move on to some slope field problems we’re also going to look at euler’smethod some common kinds of models like newton’s law of heating and cooling predator prey freefall problems and also a little bit of theory with existence and uniqueness starting with theseparation of variables problem you can see that we have this problem right here d y dt equals sineof t times y y of 0 equals negative 2 you can see the right hand side function can be writtenas a product of a function of t alone times a function of y alone that means this differentialequation is separable and we might have success in separating variables to solve it how do youseparate variables you try to get all the y’s on one side and all the t’s on the other y and t arethe variables t is the independent variable y is the dependent variable don’t worry about the d’sthose go with the dy and the dt on the left side what you do is you pretend this is a fractioneven if your teacher told you hey it’s not a fraction it’s the derivative go ahead and pretendit’s a fraction pretend that you’re multiplying both sides by dt and cancel the dt on the leftside we also want to get this y in the left side side and so we divide both sides by y tocancel it on the right side and when we do so we get d y over y equals sine of t dt okay and that’san equation in differential form that’s equivalent to the to the original differential equation wecan solve this by integrating integrate both sides when we integrate the left side we get the naturallog of the absolute value of y plus an arbitrary constant but i’m going to subtract that constantfrom both sides so i won’t bother writing it on the left side when i integrate sine of t i getnegative cosine of t plus an arbitrary constant i will include that arbitrary constant on theright side i’m going to call it c1 because we’re going to have like a c2 and a c3 in the end we’llhave an arbitrary constant that we’ll just call c but just leave it like this as is right now weneed to solve for y so we need to get rid of that natural logarithm we need to exponentiate bothsides and when we exponentiate the right side we can use a property of exponents to write that ase to the c1 times e to the negative cosine of t then we can call that c2 for example this thingbecomes the absolute value of y can we just get rid of this absolute value of y sign let me callthis c2 actually instead of c it turns out you can though really you should double checkthat by checking the final answer we can get rid of the absolute value signcall c2 just a c in a sense c is plus or minus c2 you might say so we have ctimes e to the negative cosine of t that would be what you might call yourgeneral solution we should probably take the time to check that check that it solves thedifferential equation find its derivative d y dt differentiate this where c isarbitrary we will need the chain rule we’ll get c e to the negative cos t then thederivative of negative cos t is going to be sine t and this is the same as y that’s our functionso this product on the right side is the same as y times sine t the same as the right handside of the differential equation that we had to begin with okay so that checks that itsolves the differential equation what about the initial condition y of zero equals negativetwo well just go ahead and replace t with zero and y with negative two and solve for c we getthis equation negative two equals c e to the negative cosine of zero cosine of zero is one thissimplifies to c times e to the negative one or if you prefer c over e solve for c by multiplyingboth sides by e we’ll get that c is negative two e that means the final answer the uniquesolution of the original initial value problem is y equals negative 2 e times e to the negativecos t which by properties of exponents we can also write negative 2 times e to the 1 minus cos t thatwould be another way that we can write the unique solution of the initial value problem okay we canalso just double check the initial condition maybe in the simplified form if i plug in t equals zerohere cosine of zero is one one minus one is zero e to the zero is one we also get negative twotimes one is negative two we do satisfy that initial condition as well all right let’s moveon to another separation of variables problem d y dt equals y squared times in parentheses tsquared minus one once again we have a product of a function of y alone and a function of t alonewe can separate variables divide both sides by y squared to cancel the y squared on the rightside and multiply quote unquote both sides by dt to get rid of the dt on the left sideto cancel on the left side when you do so the equation becomes d y over y squaredequals in parentheses t squared minus 1 dt this is the same as y to the negative 2 d y andnow we can integrate great both sides i’m going to go ahead and put the integral in all threespots here to help help us ultimately solve for y i should remark that sometimes you canintegrate but you might not be able to solve for y and so you might have to leave your answerin implicit form you can actually solve for y in this case and let’s go ahead and see what we getwhen we integrate y to the negative two we get y to the negative one over negative one when weintegrate t squared minus one we get one third t cubed minus t plus an arbitrary constant i’llgo ahead and call it c one i can multiply both sides of this by negative one i could alsothink of y to the negative one as one over y i get one over y equals negative one third t cubedplus t plus c two where really c two is negative c1 now take reciprocals of both sides carefulthis is a common place where people make mistakes the reciprocal of the right side is one over thisentire thing you don’t take the reciprocal of each piece individually so i need to write 1 overthe entire thing negative 1 3 t cubed plus t plus c 2 i could maybe just go ahead and callthat c that’s probably fine well maybe i should go ahead and multiply the top and thebottom by 3 for example to cancel the 3 there either way is ok i can multiply the top and thebottom by 3 to get a 3 up there to cancel the 1 3 there to get a negative t cubed plus 3t plus sayjust c where c would be really three times c two go ahead and use your initialcondition now do a different color here there’s my initial condition when t is 0 yis 3 so 3 equals what i get when i plug in 0 here which is just going to be c over 3 and thatwill imply that c is 1 and that will imply if i have not made a mistake that the unique solutionof the initial value problem is one or is a three over negative t cubed plus three t uh plus c wasone one like that that’s probably worth checking just to make sure didn’t make a mistake maybewriting it like this would help with checking just to make sure i don’t make a mistake herego back to the differential equation find d y dt the left hand side i’ll label that lhs here isthe derivative of this thing which i could use the quotient rule for in this form or the chainrule in this form let’s go ahead and try that form bring down the negative 1 to get negative 3times in parentheses t cubed plus 3t plus 1 to the negative 2 power then because of the chain ruletake the derivative of the inside to get negative 3t squared plus 3.I could try simplifying itbut let’s just leave it like that for the moment the right hand side of the differential equationgoing back up here is y squared times t squared minus 1 you need to replace y with the functionthat’s really key replace y with the function that we think is the solution right here so i’mgoing to need to square that thing i’m going to get 3 squared which is 9 times in parenthesisnegative t cubed plus 3t plus 1 to the negative 1 squared which really gives me negative 2 powerit’s kind of looking pretty good here i’ve got the 3 there and the 3s there that could be factoredout to match the 9 times t squared minus 1. yeah i think it’s going to work here let’s goahead and distribute the negative 3 through here and we’ll also distribute the 9 through there soup here if we distribute the negative 3 through we still have the negative t cubedplus 3t plus 1 to the negative 2 power times distribute the negative 3 here throughthe parentheses parentheses here to get positive 9t squared minus 9 and if i distribute the 9 through here i do indeed get the same thing yayi’m very happy that i didn’t make a mistake there all right double check the initial condition ifyou plug in t equals zero here you are going to get three three over one which is three it matchesthe initial condition it looks like this is good okay so there’s a couple of separation ofvariables problems that’s symbolic let’s go out move on to qualitative problems withslope fields where we are thinking purely about graphs we’re not trying to get the slope fieldsexactly right we’re just trying to get the idea so in this problem it says suppose the graph ofa right hand side f of t notice the t here that’s important of this differential equation is shownbelow so this graph is a function of t it’s the right hand side of a differential equation it’sreally important to realize this is a function of t i call these pure antiderivative problemswe want to give a rough sketch of the slope field and some representative solutions okay allright so this is really a calculus problem it’s not really differential equations so much in theusual sense we’re really looking for a function whose derivative is little f of t we’re reallylooking for an antiderivative of little f of t commonly called big f of t this function isgoing to be the derivative of the function of the functions that we want to draw that aresolutions of the differential equation another way to think about thatif i try to sketch the slope field in a t y plane not a t d y dt plane but at y plane is that the slope marks on the slope field are going to be constant in slopealong vertical lines where t is constant the right hand side doesn’t depend on y so alongany vertical line it doesn’t matter what y is these slope marks only depend on t themost important things to mark first are where this function equals zero and thereforewhere the slopes in the slope field equal zero the slopes in the slope field will be zero whent is negative one 2 or positive 3.Let’s go ahead and mark those on here negative 1 2 or positive3. along vertical lines through those values of t the slope field is going to havemarks that have zero slope like this that’s where the derivative of solutions are zerobecause the derivative of the solutions has to be the f of t function which you see here and thatfunction is 0 again when t is negative 1 when t is 2 or when t is 3. then you need to think aboutthe sine positive or negative for the solutions in between these different spots in between thesevertical lines when t is less than negative one this function has negative values so the slopesin the slope field must have negative slopes and they get more and more extreme thefurther you get from t equals negative 1.When t is between negative 1 and 2 maybei shouldn’t use red for these numbers here when t is between negative 1 and 2 thisfunction the derivative of our solutions has positive values so these solutionshave positive slopes maximized in slope near t equals zero maybe slightly higher than tequals zero is where our slopes will be maximized but they are always positive so you’re goingto get something about like this when t is between 2 and 3 the value of the f little f oft function is negative but only slightly negative so our slopes in the slope field are goingto be negative slopes but only slightly negative maybe i’ll just go ahead and writedraw something about like this and finally when t is bigger than 3 the slopes in the slopefield are positive again because this function is positive and gets steeper and steeper higherand higher slopes the further you get from t equals 3. now draw in solution curves that followthe slope field for example something like this approximately about like thatmaybe something like this try to get the zero slopes rightand the positive slopes right the negative slopes for that matter hey andlook at this this is illustrating something important actually all these solutionsare vertical translations of each other that’s what happens for pure anti-derivativeproblems because two anti-derivatives of the same function over a given interval differ by aconstant you already knew that from calculus two all right this problem is similar except noticeit’s an f of y on the right side not an f of t that matters this is an autonomous differentialequation which basically means it’s independent of time it represents say a physical law thatis not changing over time which is very common so it’s a very look at the graphs are actually thesame they are the same cubic functions both here and here this cubic function generated solutionsthat are actually if you thought about it fourth degree polynomials in t this cubic functionfunction however because it’s a function of y will not generate fourth degree solutions thisis an autonomous differential equation the slope field is going to be constant along horizontallines and that’s going to generate solutions that typically involve exponential functionsthough we’re not going to bother trying to solve for the solutions here so let’s go ahead and drawthe slope field again it is still in a t y plane just as this is still a well this is ano not a t d y dt plane it’s a y d y d t plane but here we make the y axis vertical forthe slope field this is tricky now these points where the right hand side function is zero arestill important but they no longer correspond to vertical lines where the slope is zero butinstead horizontal lines where the slope is zero the slope field will be constant alonghorizontal lines going back and forth because the value of t is irrelevant for the slopefield it only depends on y when y is negative 1 or 2 or 3 that’s where your slope field is goingto be horizontal and in fact because of that because the slope field is horizontalalong these horizontal lines that means those three horizontal linesare actually going to be solutions where the slope stays constant at zero no matterwhat t is those are called equilibrium solutions equilibrium solutions constant functionsthat solve the differential equation their derivative is always zero no matter whatwhat t is and when you plug those values of y into the right hand side function f of yyou get zeros you still look at the sign of this function but now for differentvalues of y instead of different values of t when y is less than negative one the values hereare negative so the slopes in the slope field are negative close to zero when you’re just barelyless than negative one but then getting steeper and steeper further away from zero as you getdown further instead of over further this time positive in between y equals negative one and yequals two maximize when y is slightly bigger than zero something like this is where the slopes aremaximized and closer to zero down here closer to zero up here somewhere in between and here um okaymy sketch is not perfect they’re steepest in here on negative slopes in here but when y isbetween 2 and 3 but just barely negative something about like this and very close to zerohere and then positive when y is bigger than three close to zero here and getting steeper and steeperas y gets further and further away from three and your solution functions representativesolutions you’re going to have when y starts out less than negative 1 you’regoing to have solutions that look like that when y is bigger than 3 you’re goingto have solutions that look like that in between when y is between negative 1and 2 solutions look about like this and in fact they are horizontal translationsof each other for autonomous equations you get solutions that lookabout like that if you know about phase lines you might want to tryto draw the phase line based on this i also will say even though it’s not asked forthat this solution is called this equilibrium solution is called unstable because solutionsmove away from it as t increases this one is called stable because nearby solutions go towardit as t increases and this one is unstable so we got unstable stable unstable for a phase line thatcorresponds to a source sink and a source okay for my students you aren’t quite to phase lines yetbut you will be soon all right so there’s a few qualitative problems let’s move on this oneis a mixture of qualitative and quantitative it’s got an ordinary differential equationwe are going to draw the slope field but it is a mixed problem the right hand sidefunction depends on both t and y and that is going to mean you’re going to need a little bit of helpif you’re going to draw the slope field nicely which i do based on something called the contourmap of the right hand side function here and we are also going to apply ultimately euler’smethod to approximate a solution in this case so it’s a mixed ordinary differential equationthe right hand side function depends on both t and y it’s not a pure anti-derivative problem it’snot an autonomous equation could we solve it by separation of variables well you could try but youwould not be successful this is not the product of two functions that where one function is afunction of t alone and one function’s a function of y alone it’s a difference of two such functionsyou can’t use separation of variables for it is there some other method you could use to solveit it turns out there is a couple other methods in fact one’s called the method of integratingfactors and one is called the method of undetermined coefficients and the key thing thatmakes those work is the fact that this is linear in y however my class my students we are not tothose methods yet and so you want to hang on that will be on exam two but we can do what’s requestedhere part a says show brief work involving the left hand side function and the right theleft hand side and right hand side of the ode to confirm that this particular function is asolution okay this is pretty easy to do just do the left hand side right hand side thing again theleft hand side of the differential equation is d y d t i need to simplify that for this particularfunction hey it simplifies very nicely to one half no matter what t is how about the right handside of the differential equation it’s one half t minus y replace the y with the function which iscareful with parentheses one half t minus one half and distribute the minus sign through theparentheses the one half t’s will cancel negative of negative one-half is positiveone-half we get one-half again no matter what t is these match you don’t always get a constantwhen you simplify sometimes you get something that depends on t but as long as they match forall allowed values of t that will confirm that it’s a solution of the differential equation andby the way you pick some other function here it probably won’t work okay you’d have to be verylucky to pick something else that would work so it’s not a equilibrium solution to thedifferential equation but it is a fairly simple solution it’s a linear function that solvesthe differential equation and that will be helpful all right on to part b in this picturebelow which is a contour map of the right hand side function showing what are calledlevel curves which in this case are lines where this function is constant that’s goingto help us draw the slope field this function one half t minus y when does it equal a constantcall the constant m well solve for y as a function of t you’ll get y equals one half t minusm for various values of m for all for any fixed value of m that’s a line with the slopeof one-half and a y-intercept of negative m and so for example you can see in this picturewhen m is negative four this line has a slope of one-half and a y-intercept of positive four whichis negative of negative four when m is negative 2 it’s got a slope of 1 half and an interceptof positive 2 which is negative of negative 2. when m is positive 2 we got a slope of one-halfand an intercept of negative 2 etc i’ve made the one where m is 0.5 bold face because thatactually corresponds to that linear solution linear function which solves the differentialequation back from part a this is at an actual solution m equals 0 is important as wellwhen m equals 0 we get a function that has a slope of 1 half and a y intercept of 0 goesto the origin now the only the bold face line is a solution of the differential equation herethese other lines are not solutions they only help you draw the slope field that’s the purposeand that is the task first draw the slope field along these lines and sketch at least foursolution curves okay at least two should be above this particular solution and at leasttwo should be below that particular solution again i’m calling these things mbecause they are going to be slopes along the m equals 0 line i want to make a bunchof slope slope marks that have a slope of 0. when i’m along this line my i want my slopes tohave slope negative two so maybe about like this okay it’s hard to get this perfect but reasonableis okay along this line the slope marks have slope negative four which is going to lookpretty steep not vertical but pretty steep along this one where m is 2 that’s we’re going tohave slope of 2 positive 2 maybe about like this you don’t have to get it perfect justpretty good when m is four we want the slopes to be positive four sopretty steep but not vertical okay so that’s again re-emphasizing thesethings are to help you make the slope field they are not solutions only this dark one isa solution along there the slope field has slopes of one-half which exactly matches theslope of all these lines so now drawing some representative solutions of the differentialequation a couple above the one the linear function one may be looking about like this andmaybe one looking about like this the key thing is making sure you’ve got negative slopes over herethat are getting less steep and you’ve got zero slope when you cross this line at m equals 0 rightthere these other ones never have zero slopes but they do approach that linear solutionabout like this so the linear solution is not an equilibrium solution because it’snot constant but it is a linear function that solves the differential equation andit is sort of attracting you might say even though it’s not an equilibrium solution allright on an euler’s method with delta t equals 0.5 we are saw approximating the solution ofthe differential equation with this initial condition y of zero equals two so we’dbe trying to approximate the solution going through this point right there let me goahead and just draw it approximately about like that for values of t uh from zero up to t equalsone we’re trying to approximate the value of the true solution at t equals one all right so witheuler’s method it’s good to make a little table k is the step so to speak we start with thezeroth step which is the initial condition we have t sub k unless pen is running out giving us the different values of t startingwith t equals zero and going up by delta t each time so when k is zero t k is zero when kis one at the first step tk is 0.5 and when k is two after two steps we add another 0.5to that we get 1.06 we have y k as well y sub 0 is the first value of ythat is the initial value of y 2. and the key thing is we need to figureout the different values of y that we get by using euler’s method which is based on not onlyincrementing t up by delta t each time and using this recurrence relation or difference equationbut also this one here y k plus 1 the next value of y is the preceding value of y plus thechange in y which is given by the right hand side function evaluated at the previous point timesdelta t almost writing out a room on my paper here so this is the change in y youmight call this delta y sub k uh so this one we need for the first one weneed to evaluate the right hand side function at t zero comma y zero zero two at thispoint right here actually you can see its value is going to be negative 2.That’s theslope in the slope field which comes from plugging in the val the points on thisline into the right-hand side function for extra emphasis i’ll rewrite itdown here f of 0 2 times point five f the right hand side function is one half tminus y plug in t equals zero and y equals two you’re gonna get zero minus two is negativetwo negative two times point five is negative one hopefully i don’t make a mistake here doublechecking things that is the change in y so y is going to go from two down to one after onestep of euler’s method we go down by one unit the slope times the change in t gives theapproximate change in y next i need to evaluate the right hand side function at thispoint t is 0.5 y is 1 multiply times delta t so i need to plug in t equals 0.5 and y equals1 into here 0.5 times 0.5 is 0.25 minus 1 is negative 0.75 don’t forget to multiply by 0.5and again hopefully i’m not making mistakes if i am i’ll maybe try to make a note on the videohere this is going to be negative 0.375 that’s going to be the new change in y so y is goingto go from 1 down by this much to positive 0.625 yeah that would be the right answer andthen that would be your approximation to y2 here would be your approximate approximationto the solution function when t equals 1.Okay t equals 1 because t keeps going up by 0.5 at eachstep and that’s so it’s the second step that we approximate the solution at t equals 1. 0.625 sowhen t is 1 on the solution we’re approximating the solution to be 0.625 which is going to bedown further i didn’t draw this super well i guess it’s going to go down more like this well itis a concave up graph so euler’s method is going to be under the true graph so the true value ofthe solution is going to be higher than 0.625 but um don’t worry about it okay that’s that’s beyondthe scope of what we’re doing here in this problem all right so that was an euler’s method problemnow we’re going to go on to focusing on some applications before we get to some theoreticalstuff with um existence and uniqueness all right hot cup of coffee with a temperatureof 55 degrees celsius at t equals zero cools to 53 degrees celsius at one minute assumeroom temperature is 22 degrees celsius let y be the temperature in degreescelsius of the coffee at time t in minutes the temperature coffee at timet in minutes we’re using newton’s law of cooling says the rate of change of y thederivative with respect to t is proportional to the difference between the coffee temperatureand the room temperature that’s key to getting the differential equation right part a says writedown a differential equation initial value problem to model the temperature of the coffee i shouldremark that we are assuming the coffee is well stirred here so we don’t have to considerspatial effects you’ll need a proportionality constant identify its sine positive or negativeproportionality constant identify its sign and the sign actually depends on which differenceyou take but you just need to be consistent is it the temperature of the coffeeminus the room temperature or vice versa using any method you want find out how long tothe nearest minute it will take for the coffee to be a drinkable 40 degrees celsius sureyou work all right that’s a lot to take in so you need to practice thiskind of thing ahead of time the rate of change of the temperaturewith respect to time is proportional to the difference between the coffee temperature andthe room temperature if i write that difference as a y minus 22 y is the coffee temperature 22 isthe room temperature that is a positive difference when i have coffee that ishigher than room temperature and it’s cooling therefore i need a negativeconstant of proportionality call it negative k where k itself would be positive so thatnegative k is negative and it’s really negative k that is the constant of proportionality if i write it this way constant of proportionality iswhat i’m trying to write here okay so that when when i have cooling coffee thatdifference positive times negative k is a negative quantity making dy dt negative indicating thecoffee is cooling okay so that’s important to realize if you had written the differences22 minus y then 22 minus y would be negative and your constant of proportionality would bepositive you actually can write it either way but it sort of makes more intuitivesense to think about it this way all right um so we need tosolve this differential equation we’ve done part a here and we’re going to need to use thisinformation and this information it’s the second bit of information that’s goingto tell us k and it’s this bit of information that’s going to give us the integrationconstant you might say this is separable let me go ahead and separate variables k is aconstant so it can be just left on the right side let me divide both sides by y minus 22 andpretend that i’m multiplying both sides by dt and write the differential equation that wayin differential form now integrate both sides get a natural log of the absolute value ofy minus 22 on the left side plus c1 but i’ll subtract that from both sides and get a negativekt plus a constant here i guess i’ll go ahead and call that constant c1 we want to solve for ywe get rid of the logarithm by exponentiation and then use a property ofexponents to write this one this way here the e and the ln of this quantity cancel so we’re left with justthat quantity the absolute value of y minus 22 i can get rid of the absolute value signs it turnsout i’m not really fully justifying that here i could call this c3 or maybe it’sfine just to call it c now i think and ultimately y is going to be add 22 toboth sides 22 plus c e to the negative kt and that is a general solution its derivativewould be well the derivative of that is zero it would be negative k c e to the negativekt that would help be how the left hand side simplifies the right hand side simplifies tothe same thing the y minus 22 the 22s cancel and then you multiply this part by negative kyou get the same thing as the derivative of it negative k c e to the negative kt it’s goingto work double check that the initial condition y of 0 equals 55 will help us solve for c e to the zero is one solve for c c is going tobe 55 minus 22 is going to be 33.And then the fact that the temperature is 53when t is 1 helps us solve for k 53 is going to be 22 plus c is 33. e to thenegative k times 1 is just e to the negative k we’re going to be able to solve this for k by first subtracting 22 from both sidesso we get 31 equals 33 e to the negative k e to the negative k is going to be 31 over 33 take the natural log of both sides andthen multiply both sides by negative 1. k is going to be negative natural log of 31 over33 which is actually a positive number because 31 over 33 is less than one ln of that is negativebut the negative sign out here makes it positive i could also write this as ln of 33over 31 by a property of logarithms and that is about about point zero six two five two okay carrya fair number of decimal places right away and and carry those along and just round at the veryend so that the errors don’t build up so to speak there’s k so now we can solve the problem uhhow long to the nearest minute will it take for the coffee to cool to 40 degreesso now we want to set y equal to 40.Set c equal to 33 k equal to 0.06252 don’t forget the negative sign therebecause it’s a negative k up there and solve that for t we’ll have to uselogarithms again subtract 22 from both sides and get an 18 on the left side equalsthat divide both sides by 33 we’re going to get 18 thirty thirds equals e to the negativepoint zero six two five two t take the log of both sides and then divide both sides by negative0.06252 you’re going to get t is 1 over negative oops i can put the negative sign ineither spot 1 over negative 0.06252 natural log of 18 over 33 that’s going to be a negative number times thisnegative number is going to be positive in the end ln of 18 over 33 divided by the previousanswer but then i’m going to have to put a negative sign in front ofthat the answer is about 9.695 to the nearest minute that’sgoing to be about 10 minutes okay so hopefully i didn’t make a mistake i don’tknow for sure i think i did it right uh to get when the coffee is going to be 10minutes double check these things before i post the video and i’ll makelittle annotations if i do make mistakes plus newton’s law of cooling newton’s law ofheating is actually the same differential equation you could write it this way and if if there’ssomething heating up y would be less than negative less than 22 that difference would be negativetimes another negative would make the derivative positive and the thing would be heating up inthat case another model is a predator prey model um the goal is not to solve the predator preymodel in fact it’s impossible to find the general solution of this it’s a non-linear system ofdifferential equations in general it’s impossible to solve such systems we want to think about itqualitatively dx dt is the negative 2x plus 4xy which does factor like that you can checkand dydt is 2y times 1 minus y over 3 minus xy which can be factored like that you cancheck this is predator prey one of these variables represents the population of a predatorand one represents a prey predator prey predators eat prey predators are like foxes or wolves andprey or like rabbits or chipmunks or something the key to deciding which one is which is lookingat these two interaction terms and their sign when x and y are both positive whichthey would be for populations this 4xy positive 4xy is contributing positively to thegrowth rate of the x species so the interaction between predator and prey benefits x x has gotto be the predator it benefits from the actions the predator is eating the prey and thisone has a negative coefficient negative xy this has got to be the prey the interactionaffects the growth rate of the prey negatively it could make it negative for exampleinteractions between predator and prey are harmful to the praise populations sothat’s got to be the prey so x is the predator and the reason you could give if you’re asked for a reason is what i just saidverbally and why is the prey okay so that’s part a part b the points 0 0 and 0 3 aredefinitely equilibrium points where the vector field is it’s got what’scalled an equilibrium point just a dot for my students actually haven’t talked aboutvector fields in depth yet at all a little bit in some of the lectures show work show algebra workto verify this point is also an equilibrium point now actually the sort of quickest way todo that you might say is just plugging the points in the right hand side but i want youto be able to do this algebraically as well setting these right hand side functionsequal to zero and solving for x and y now it’s pretty clear if x and y are both zero becauseyou have the factor of two x there and a factor of y there that you’re gonna get zero you’re gonnasolve these equations it’s also pretty clear if x is 0 and y is 3 it’s going to work if x is 0that’s going to make the right hand side here 0 and dx dt will be 0 the x populationwill not change and if x is 0 here and y is 3 here well did i make a mistake oh no it doeswork when x is 0 here i get 2 minus 0 which is 2 minus 2 3 times 3 2 3 times 3 is 22 minus 2 is 0.It does make this 0. where else can this the right both right handsides be 0. if y is 0 actually x is forced to be 0 so there’s no other equilibrium pointswhere y is zero but how about if both this and this are zero and that’s a system oflinear equations that’s pretty easy to solve because the first equation only involves ay and that’s going to be 0 when a 2y is 1 so y is one half then you can substitute that intothis equation and solve for x you’re going to get 2 minus x minus 2 3 times 1 half is going to be1 3 equals 0 and so x is going to be 2 minus 1 3 5 3. and that is matching those twothings are matching the coordinates of that point right there x is 5 3 is one halfthat’s an equilibrium point and in fact that’s the equilibrium point that these solutions thatyou see here or the solution is spiraling toward there’s a dot right there that represents anequilibrium where the populations don’t change now of course populations in real lifedo change but effectively we imagine that as a being a place where the births anddeaths all cancel each other out so to speak the last part of the question saysthe graph of the solution is shown on the set of axes further below draw x of t andy of t take this parametric curve in the x y phase plane and draw how x and y depend on time focuson getting the general behavior and the limiting values right and also the max and min values rightin the first oscillation also make sure to label which graph is which okay so the key thing hereis we’re looking at this graph and we’re imagining that the point is moving along this curvein this direction indicated by the arrows as time goes by you need to imaginethat you need to use your imagination different colors here x is starting out at a value that is the first coordinate of thatpoint and is a minimum close to 0.25 maybe that’s going to be the lowest value of x so this green graph i’m about to make is going tobe the x graph as t increases x increases because this graph’s moving to the right and reachesa maximum value a little higher than 4 at some value t how fast does it increase that’s unclearthen it’s going to go back down again to about 0.6 a little higher here something about like this before going back up again to a value closeto 2.85 or something something about like this and ultimately i’m also trying toget the limiting values of x correct the first coordinate at this point that’s thefive thirds right there 1.67 approximately it’s going to have a horizontal asymptote thatit does touch it’s going to oscillate up and down about that horizontal asymptoteas t gets bigger and bigger so this is an approximate graph a rough graph of xas a function of time how about y as a function of time look at the second coordinate of the pointat the starting value of time it’s close to 1. it is going to increase a little bitto a maximum close to 1.6 when x is around 1.So y is going to have a maximumclose to 1.6 when x is about close to 1. so this is the x graph it’s going to goup just a little bit to right about there well maybe a little higher right aboutthere before coming back down again y then reaches a minimum say at thispoint close to y equals 0.3 or something when x has gone past its maximum and isdecreasing and x is approximately 1.6 there 1.7 um it’s going to be the minimum this is tricky here right aroundthere when x is back down to about 1.7 close to 1.67 that’s when y is going to be at aminimum but force going back up again and only reaching less than 1 here it’s not going togo back up very fast it reaches back up here when x has gone down to its minimum and then backup again to around 1.5 i’m probably not expecting this level of perfectionyou might say in your graph get the maximum values right get theget the long-term behavior right the long-term behavior for for y is thatit approaches uh well one-half as t goes to infinity one-half right around hereit’s got a horizontal asymptote so something about like this and i’m definitely not trying to beperfect after this part here and i’m not perfect even to begin with but this is approximatelyright okay that’s a pretty tough skill that’s definitely something that would take practiceto do that within a reasonable amount of time a couple minutes so once again it’s something thatmy students you should work on okay translating between the phase plane and the solution curvesthemselves as functions of t i got existence and uniqueness questions here i think i’m going tosave that for the end actually let me jump ahead to some somewhat theoretical questions but notquite as theoretical as existence and uniqueness um here’s a true false question any horizontaltranslation of a solution of a pure antiderivative problem d y dt equals f of t is anothersolution of the pure anti-derivative problem we talked about that at the beginning of the videothat would be false right because this is a pure anti-derivative problem vertical translationsof solutions are going to be other solutions it would be true if i changethe horizontal to vertical or it would be true if i change the horizontal toif i kept the horizontal as horizontal and made this autonomous with an f of y at the right sideif i made either of those changes it would be true if i made both changes made this vertical madethis y there then it would bite me back to false okay number 11 is a modeling problem freefall got an object of mass m in kilograms falling under the influence of gravity anddealing with air resistance g is the downward acceleration due to gravity and because it’sdownward i’m thinking of it as being positive because i’ve got the word downward in therethe positive direction is going to be upward meters per second squared ormeters per second per second negative k where k is positive so negativek is going to be a negative quantity is a constant of proportionality in here are theunits but you don’t need to worry about that newtons per cent meter per second for the airresistance when it is assumed to be proportional to velocity and it’s going to be because ofthe negative sign in the opposite direction the goal is to write down a first orderdifferential equation for this model when y is the height in meters and v the derivativeof y is the upward velocity in meters per second so y is a height so it’s positive going upwardand its derivative would be an upward velocity it’s going to be positive when the object’smoving upward and negative when the object’s moving downward it’s a first order differentialequation so it only involves the first derivative you need to use newton’s secondlaw here f equals m a which because a is a second derivative seems to be asecond order differential equation however we can write this as a first order differentialequation if we replace the second derivative of y with respect to t with just thefirst derivative of v with respect to t okay so that’s part of the differential equationthe other part comes from the force itself and it comes from two sources to the force there’s theinfluence of gravity and there’s air resistance the influence of gravity let me writethis initially as f grav plus f air is you should know negative mg and g remember isa positive quantity i’m taking it to be positive here so negative mg is a negative quantity thisis going to point downward no matter what v is no matter what y is it’s a it’s constant with respectto v and y the air resistance is proportional to the velocity it’s a constant multiple of thepropor the velocity with negative k in fact being the constant of proportionality minusk v careful here in thinking about this when v is positive that means the object is goingupward negative k is negative and so this force will be in the negative direction which makessense air resistance is opposing the direction of motion if v is positive the air resistance isgoing to be negative on the other hand if v is negative if the object’s falling downward thennegative k times v is going to be positive the air resistance is going to be in the positivedirection so this does make sense anyway this equaling this is your first order differentialequation it is common to divide both sides by m and you could write your answer d as dv dt equalsthe m would cancel here and get negative g minus k over m times v that would be your first orderdifferential equation modeling this situation i didn’t ask for initial an initial conditionbut if you if i asked for one it would be v of zero equals something you could call thatspecific something v sub 0 in general or maybe i give you a particular value of the initialvelocity which could be positive or negative positive if you throw the object up negativeif you throw it downward if you drop it from rest then v0 would be zero number 12 says findan abstract formula abstract formula for the unique solution of the pure antiderivative initialvalue problem d y dt equals cosine of t squared and y of 2 equals 5.Now why do i say findan abstract formula i mean can’t we just just integrate this to find the answer and theanswer is yes and no yes you can integrate it in theory but no you can’t really integrate itin practice because integrating cos t squared is just too hard in the usual senseit’s a non-elementary anti-derivative so i mean i could write yequals the integral of cos t squared but that reallyrepresents a general solution what i really want is a definite integral i wanty my solution function which i commonly with my students call phi of t here to be the integralof cos t squared except write it as a definite integral with the variable t in the upper limitof the integration to also go through the 0.25 i want to add 5 out of here and i want2 to be the lower limit of the integral it’s also good to use a dummy variable for insteadof t squared do like u squared or something this is the answer does it really work well certainly thefunction value in t equals 2 is indeed 5 because we’re integrating from 2 to 2 which of course would have to be 0no matter what function we have there and so the answer is five it satisfies theinitial condition what about the derivative of this function well thederivative of a constant is zero and the derivative of this i help you rememberfrom the fundamental theorem of calculus let me write it as d dt first d dt of thisintegral the d dt and the integral sign cancel leaving you with the integrand though you shouldreplace u with t okay some textbooks call that the first fundamental theorem of calculus sometext books call it the second fundamental theorem of calculus so which is it it’s eitherjust depending on the author’s preference um it is like the derivative and theintegral sign are canceling so to speak some books call it the antiderivative constructiontheorem that might be the best name for it but it does verify that the derivative is theoriginal right hand side function this is an abstract formula for the unique solution ofthe pure anti-derivative problem so that’s kind of a theoretical question and we’re endingwith a couple theoretical questions as well a bit more theoretical you might say based onexistence and uniqueness theorems that are more general than the fundamental theorem of calculusi’m not going to show you the statements of those theorems but in solving these problems i’lltry to remind you of the idea of these theorems these theorems are based on right hand sidewell based on first order differential equations where the right hand side is f of t y so theyare mixed in general and based on supposing that right hand side function andits partial derivative with respect to y are continuous for all t and all y aremaybe on some rectangle they are continuous we’re going to in this problem suppose we alsoknow that this function and this function are solutions of this differential equation okayi’m not saying what f of t y is this is kind of abstract but i’m pretending i know these aresolutions the question is kind of general excuse me what can we say if anything about the solutionof this initial value problem where y of 0 is 1 and how about the limit of the solution ast goes to plus infinity explain your answers evidently we must be able to say somethingso how do you approach a problem like this it’s good to draw uh phase plane sorry about theextra loudness here people laughing out there a t-y plane can i draw the slope field well in alimited sense you can i can draw these solutions and they have to follow the slope field soin a sense by drawing these the graphs of these functions i’m drawing the slope fieldto some degree the graph of t over 1 plus t squared you can check with yourcalculator looks about like this and its value at t equals 0 is 2.Andthe graph of negative e to the negative t looks about like this and itsvalue t equals 0 is negative 1. these are two solutions of this differentialequation that’s what we are supposing in this abstract problem i’m also looking at now thedifferential equation with this initial condition y of zero equals one hey that means the solutionthat i’m after has got to go through that point i have no idea what the solution lookslike other than it goes through this point and because of well it’s going to exist becausethis is these are continuous but it’s also going to be unique because these are continuous inparticular this derivative is also continuous so what that means is it cannot cross these othertwo graphs there are solutions excuse me distinct solution curves cannot touch each other when youhave uniqueness so i mean the solution could be something crazy it’s got to pass the verticalline test so i shouldn’t get too crazy here but no matter what it is no matter how crazyit looks it’s got to stay between these other two functions that are solutions so as far aswhat you can say you can say that the solution um y of t has got to be between what did idraw something wrong i drew something wrong negative e to the negative t sorry it’sgoing to look like that cross that out make an annotation on the video i’msure i you saw the annotation already gonna look like this so thisis 2 this is negative 1.Here’s your initial condition yeahit could be crazy over here but it’s got to stay betweenthese two functions for all t negative e to the negative tis going to be less than y of t is less than 2 over 1 plus t squared fort that’s what i can say one thing i could say and i can also say what the limit hasgot to be because of the squeeze theorem these two other solutions approachzero as t goes to infinity so i could say i you don’t haveto quote the squeeze theorem but the red graph is getting squeezed between theother two graphs means the limit as t goes to plus infinity of the solution has got to be the same asthe limits of those two green graphs which is zero okay the limit as t goes to minusinfinity may or may not exist last question theoretical y one of t equals t tothe sixth power y two t is the constant function zero and equilibrium solution verify that theseare both solutions of this initial value problem the same initial value problem so do theleft hand side right hand side thing d y d y one d t not t one is six t to thefifth and right hand side 6 y 1 to the 5 6 is going to be 6 times t to the sixth tothe 5 6 and the 6’s will cancel when you multiply those exponents you’re going to get 6t to the fifth and these match match for all t therefore this is a solution ofthe initial value problem for for y 2 the constant function 0 both theleft hand side and the right hand side simplify to zero for all t these match for all t as well and therefore both of these functionssolve the differential equation how about the initial condition we should checkthat as well yes to get full credit you need to check the initial condition y 1 of 0is 0 to the 6th which is 0 and y 2 of 0 is 0.Okay so both these functions whichare different functions solve the same initial value problem why does thisnot contradict the uniqueness theorem i said distinct solution curvescan’t cross back in this problem what’s going on here the graph of t to the sixthand zero cross each other the zero function is just the horizontal axis and t to the sixth wellit’s very flat near zero it looks about like this it doesn’t equal zero except when t iszero but it’s very close to zero you know 0.1.1 to the sixth would be 10 to thenegative six which would be one millionth so it’s very close to zero when t isnot zero but it doesn’t equal zero why does this not contradict uniqueness well thehypotheses of the uniqueness theorem must not be satisfied this is your your right handside function which is an f of y function no t that’s autonomous but its derivativewith respect to y must not be continuous the derivative of f of y with respect to ybring down the 5 6 and get 5 y to what power 5 6 minus 1 is negative 1 6 power 5 over y tothe positive 1 6 power which is undefined sorry undefined at y equals zero that actually doesn’treally prove it’s not differentiable at zero because you really should do a limit calculationthis is really assuming y is nonzero to begin with but uh i’m not going to bother doing ityou’d have to do well or maybe i should this would be the limit calculation to confirmthat this function is not differentiable at zero it’s the limit definition of the derivative this is just an indication that it’sprobably not differentiable at zero and i guess i’m going to saygood enough for my students should be h okay let’s change this to a y usually use the letter h there but function right hand side functionagain is 6 y to the 5 6.And this limit does not exist does not exist as y goes to 0 okay and that’senough to say f prime of zero does not exist again usually you’d write an h there insteadof a y but it’s it’s not a real big deal that limiting nodding exists not existing means f isnot differentiable at zero this just is a hint that it’s probably not though i’ll take it to begood enough for my students and that means um it doesn’t contradict the uniqueness theorem becausethe hypotheses of the uniqueness theorem are not satisfied this derivative is not continuousand that’s what you would want to explain okay so that’s it i hope that will helpyou for your test thanks for watching

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