WEBVTT mathematics/differential-equations/murray
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Hi and welcome back to www.educator.com these are the lectures on differential differential equations, my name is will Murray and today we are going to talk about big topic, it is kind of a favorite 1 for students to hate.
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It is applications modeling and word problems, those are all kind of different words for the same thing applications means you are using differential equations to study real world scenarios.
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Modelling means the same thing, you have a real-world scenario and you try to set up the differential equation to describe it and of course everybody knows what we word problems mean.
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Let us go ahead and get started with the lesson overview, I'm going to give you 3 situations that random themselves very nicely to differential equations.
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Because of that they are very popular problems in all differential equations courses basically in every college in the world.
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You will probably see problems very like these in your own differential equations course, the first 1 here is the classic mixing problem and there are sort of 2 forms this there is either air or water.
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The air version is that you have a room with usually with pure air and then you have some smoke or pollution coming into the room, maybe somebody is smoking a cigarette, maybe there is some pollution coming in through a window.
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The idea here is that smoke is flowing into the room and that is introducing some kind of pollutant into the air and then you also have a vector window at the other side of the room and air is flowing out on the other side.
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The point is that the air that is flowing out is evenly mixed, as the pollution comes in, it becomes evenly mixed through the air and then mixed air flows out on the other side.
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There is another variant of this problem which involves water and it is basically the same idea except as water instead of air, but the idea is that you have either a water tank or a lake or some kind of enclosure with water.
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Again there some kind of contaminant coming in, so it might be saltwater is flowing in, it could be pollution is coming in, maybe there is a pipe dumping some kind of pollution into the water, the other variant is that you could have clean water come in.
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Somebody is trying to clean the lake by pouring in clean water and then on the other side of the lake again, just like with the air example.
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We had evenly mixed water flowing out, what happens is the pollution comes in on one side, it gets evenly dispersed through water in the lake and evenly mixed water flows out.
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We will study how that turns into a differential equation, second example that we are going to study is population growth.
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We will have a population of usually animals and they grow according to their natural reproductive rate and then there is also often some kind of control on their growth.
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For example, there might be some predation by a predator or the species might be hunted and so there is something that is restricting their natural growth, we will see an example of that and you will see how that turns into a differential equation.
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The last example is the financial example, there are 2 ways this can play out, 1 is that you have a savings account and your savings account is earning a certain amount of interest based on whatever amount of money you have in there at the current time.
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And then you are also making either deposits or withdrawals from your savings account, that is affecting the growth of your savings account as well and that sort of parallel counterpart to that is if you take out a loan.
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Instead of putting money into a bank you take out a loan and then you make regular payments back to the bank to pay it off, this is like the student loan that you might be paying off for college.
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It is basically the same idea because whatever you owe at any given time there is interest accruing on that, that increases the amount that you owe but you are also making payments so that decreases the amount that you owe.
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We will figure out how that turns into a differential equation, each one of these 3 situations turns into a differential equation and we are going to do in this lesson is study examples of each one and you will see how to convert the words into differential equations.
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And how to solve the differential equations , let us keep moving here, there sort of this same strategy works for all of these situations, it is a strategy that you can apply to any of these situations and also to other modelling problems as well.
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I'm going to illustrate it with these 3 different types of situations, the first thing you want to do is to set up your variables and I use T to be time, that is going be true in all of these problems and then y(T) will represent some quantity that is varying over time.
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In the mixing problems, the quantity will be the amount of the contaminant in either the air or the water, for example if you have somebody smoking a cigarette then the y(T) might represent the amount of carbon monoxide produced by the cigarette smoke.
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Or the other amounts of some other kind of contaminant in the air from cigarette smoke, if it is a water problem if you have saltwater pouring into a lake or a water tank, then it would be the amount of salt in the tank at any given time or the lake at any given time.
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In the biological example, y(T) would represent the population of the species that you are keeping track of at any given time and in the financial examples, y(T) would be the amount of savings you have in your savings account.
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For doing the other version where you are taking out a student loan, y(T) would be the amount that you owe to the bank at any given time.
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In all 3 of these, y(T) is the amount of stuff that you are keeping track of, it could be money, it can be the number of animals, it can be the amount of salt in a water tank.
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Once you set up your variables, you want to write the differential equation, you want to figure out an expression for Y′(T), remember Y′ is the derivative, that is the rate of change, that is what we learned calculus 1.
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The rate of change is determined by the amount of increase, let me highlight this, the amount of increase - the amount of decrease, in each of these situations there is something that is causing the quantity to increase.
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And there is something that is causing the quantity to decrease, for example in the mixing problems at somebody smoking a cigarette that is increasing the pollution in the air but then also you have air flowing out of the window.
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Maybe on the other side of the room so that is decreasing the amount of pollution, you have an increase factor and a decrease factor.
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The same with population, the animals are breeding so that is causing an increase in the population and sometimes you also have either predation or hunting and that is causing the population to decrease.
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You want to keep track of both of those and finally if for example you take out a student loan, then whatever amount you owe at any given time is accruing interest, that is causing the amount you owe to increase.
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But you are also making payments every month and so that is causing the amount you owe to decrease.
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Each one of these situations, there will be an increase and decrease, and the key thing to know about this is that you want to write down what the increases and what the decreases .
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Usually one of those will be a constant, I want to emphasize that one of them will be a constant, usually for example in the pollution, somebody will be smoking that is introducing a constant amount of pollution into the air.
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In the population example, the predation is usually constant, somebody is killing a certain number of animals per year, in the student loan example when you take out a student loan, you will make a certain payment each month.
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That will be $100 a month or $500 a month but that is a constant, that will be constant and then the other term will be dependent on y(T), that is what makes it a differential equation is Y′ (T) depends somehow on y(T).
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Usually one of the increase and decrease is constant, the other one depends on y(T), for example in the pollution, you have evenly mixed air flowing out of one window or evenly mixed water flowing out of the lake.
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The amount of pollution you are losing at any given time depends on the current amount of pollution that there is because for example if the air is completely clean and then you flush some air out the window then you are not really losing any pollution.
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But if the air is already heavily polluted and you flush some air out the window, then you lost more pollution, that depends on the current value of y(T).
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Same thing with the population growth, the amount of reproduction in a species depends on the number of animals you have right now, that depends on the current value of y(T).
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And finally when you make savings for or you are paying off a debt, the amount of interest that accrues is dependent on the current amount that you have in your savings account or that your paying off or you owe at your debt.
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Because the more you borrow, the more interest accrues each month, one of those terms will always be dependent on y(Y), you will see a y(T) on the right hand side in either the decrease or the increase.
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And then usually the other one will be constant but it is not always constant but usually in most of these examples the other one will be constant.
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We will practice actual examples of each one of these, you get the hang of it yourself, in the meantime once you have written the differential equation that is usually the hardest part, then you have to solve it.
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To solve it you are going use the techniques that we have been learning in the previous lectures here, they are usually linear, most of these examples are going to be linear.
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Sometimes they are even separable, those are not the hardest differential equations in the world to solve, once you get them written up they are not too hard to solve.
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You will also have some kind of initial condition, for example you initially take out loan of $20,000 or the initial population of a certain species is 1000 individuals, you have some kind of initial condition.
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If you do not have an initial condition, you can just establish one yourself, you can just say y(0) is Y(sub)0 and then just keep that as a variable in your equation.
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You will solve the differential equation and then often these word problems come with kind of capper question which you have to use your solution to answer this capper question.
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For example will say if you are putting money in a savings account when does your initial investment double? To answer that you would set y(T) = 2y₀, remember y₀ was your original investment.
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And so you would set y(T) = 2y₀ to get your investment to double and then solve for t, sometimes we will say, when is the salt diluted to 1% that would be an example of saltwater in a water tank and maybe we are pouring in clean water to try and clean the tank.
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We would set y(T) to be 1/100, that is 1% of the initial amount of salt which will be y₀, here is another financial example, what should your initial investment to be to get $1 million in 20 years.
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If you putting money into a savings account and you want to have $1 million in 20 years, what you would solve for is y(20) to be $1 million and then you try to figure out what y(0), what your initial investment would have to be in order to make that work.
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Each one of these will see comes often with some kind of capper questions, we use our solutions to the differential equation after we solve them to answer these capper questions.
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Let us go ahead and get started with some examples, first examples in mixing problem, we have a 100 gallon tank that is initially full of pure water and what we are going to do is pour saltwater into the tank.
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Water that is 3% salt enters at a rate of 10 gallon per minutes, every minute we are adding 10 gallons of water to the tank and then at the other end mixed water is leaving the tank.
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The saltwater gets poured at one side of the tank, it gets all mixed through very thoroughly mixed together and then the evenly next solution leaves out at the other end of the tank at the same rate.
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What we are asked to do here is to write an initial value problem to describe the situation and then to solve it, let us remember method here, we are going to start by setting up variables.
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We will start by saying T is = time, this notation that I'm using here with the colon equals, colon equals means defined to be, I'm saying T is defined to be time.
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That is what the colon equals means, that is a notation borrowed from computer scientists and then I'm going to define my y(T), y(T) will be defined to be the amount of salt in the tank at any given time, amount of salt at time (T).
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y(T) represents the amount of salt in the tank at time (T) and now following my strategy I want to figure out what Y′(T) is and that is the rate at which the amount of salt is changing.
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I can figure that out by finding the increase - the decrease.
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Let us figure out how the amount of salt is increasing and how it is decreasing, it is increasing because this water that is 3% salt is entering at a rate of 10 gallons per minute.
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What that means is we got 10 gallons per minute and in each of those gallons 3% of them are salt, that 3% is 3/100, that is 10×3 / hundred is 3/10, the increase right here is 3/10 and then the decrease is coming from the mixed water leaving the tank.
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Since 10 gallons of water is entering the tank at every minute and 10 gallons of water is leaving the tank that means that the amount of decrease is going to be 10/100 is the proportion of water that is leaving the tank .
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Since there is 100 gallons total and 10 gallons are leaving every minute, that is 10/100, we are losing 1/10 of the water in every minute and that means we are going to be losing 1/10 of the current amount of salt in any minute.
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The current amount of salt is y(T), the decrease here, this is probabaly the trickiest part is going to be 1/10 of the current amount of salt, what we have here now is the differential equation Y′(T) is = to 3/10 - 1/10 y(T).
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That is my differential equation that describes the situation, let me point out one thing here, remember how I said when we are looking at increase and decrease, usually one of them is constant and one of them is dependent on the current value of y(T).
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The increase is constant and the decrease is dependent on y(T) , that is one partial way to check your work is when you write increase and decrease, you should always see y(T) in at least one of those terms.
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That is what makes the differential equation because you have to have Y′ and y in the equation, here we see y(T) in the decrease and then the other term is constant.
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There is one more piece of information I need to make it an initial value problem, which is the initial condition, the initial condition here is that the tank is initially full of pure water, that means there is no salt.
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y(0) is 0, that is my initial condition that describes the fact that there is no salt in the tank at the very beginning, we have our differential equation, we have our initial condition, that means we have written our initial value problem.
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Now let us solve it, to solve it this is actually one of the easier differential equations to solve because it is a linear, I'm going to move the Y term over to the other side and I'm going to write Y′ + 1/10 y is = 3/10.
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That is a linear equation, it has the form Y′ + PY = Q and remember the way you solve that is you have an integrating factor, i(T) which is e⁺the integral of p, well here P is 1/10 d(T) and that is equal to e⁺the integral of 1/10 d(T) is just T/10.
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We are going to multiply e⁺T/10 by both sides here, we get Y′ x e⁺T/10 + 1/10 e⁺T/10, y is equal to 3/10, e⁺T/10, if this is not looking familiar to you, you might want to look back on our lecture on linear equations.
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I'm following the techniques that we developed during that lecture, you can check that out and get a little refresher on linear equations if you do not number how to do this.
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The point of this is that the left hand side is the derivative of (y)e⁺T/10.
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To solve this what we can do is integrate both sides and we get (y)e⁺T/10, now I have to integrate 3/10 e⁺T/10, that is just 3/10 e⁺T/10 ÷ 1/10 so that is the same as multiplying by 10 and that is 3 e⁺t/10 + a constant.
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If I want to solve for y now of the variable side by e⁺T/10, I will get Y = to 3 + c, now I'm dividing by e⁺T/10 is the same as multiplying by e⁻t/10, ce⁻t/10.
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I need to incorporate my initial condition at this point to figure out what c should be, I'm going to use this initial condition that means I plug in 0 for both t and y, I get 0 is = 3 + (ce)⁰.
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That tells me that c must be -3 because e⁰ is 1, so c is -3, I will plug that back in and I will get Y is = 3 - 3 e^- T/10 and that is my solution both to the differential equation and to the initial connection.
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I have solved the initial value problem, just a quick recap of what we did here, we set up our variables, we set T is = time, y(T) was equal to the salt at time T, by way you can use the same strategy for a lot of mixing problems.
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Instead of salts, it might be the amount of carbon monoxide at time T, it is usually the amount of whatever kind of pollution or extra element is in the air or the water that you are discussing.
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And then we set Y′ to be the increase - decrease, the increase was constant because we have salt water pouring in at a constant rate, the decrease depends on y(T) that reflects the fact that it is mixed water coming out.
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The amount of salt that we are losing at any given time is dependent on the current amount of salt in the water, once we get the differential equation we also throw in an initial condition that reflects the fact that it was initially full of pure water.
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That is where we get the y(0) = 0 from, then we solved the differential equation, it is a linear so we can use our linear techniques, we use integrating factor, multiply that by both sides, integrate both sides.
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Make sure when you integrate you have the constant right then and then after you do the integration, you keep track of that constant, in particular when we divide both sides by e^- T/10 or multiply both sides by e⁻T/ 10.
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Then that gets attached to the c, you cannot just write + c anymore, you have to keep track of the fact that C is multiplied by e^- T/10, then we introduce the initial condition, plug that into to figure out what the c should be.
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Figured out that c should be -3 and finally we plug that back into our general solution to get our specific solutions the initial value problem 3 - 3 e^- T/10.
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We are going to keep working with this in the next example, consider the water tank described above and then we have some follow-up questions, remember I said that you have some capper questions on this.
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How long until the tank is 1% salt until the water in the tank is 1% salt and then what is the long-term concentration of salt? In order to solve this, we are going to have to remember the equation that we had before.
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Let me write that down, y is 3 - 3 e^- T/10 that is what we worked out in the previous example and we want to figure out how long until the tank is 1% salt? Remember it is a 100 gallon tank.
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100 gallon tank for it to be 1% salt that means 1/100 of it is salt, that would mean we would have a total of 1 gallon of salt in the tank, we are trying to figure out when will y be equal to 1.
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We are going to set y(T) = 1 and then we will just solve that for t, 3 - 3 e^- T/10 is = 1, if I subtract 1 from both sides 3 - 1 is 2 and then I will add the 3⁻T/10 over to the other side, 2= 3, e⁻T/10.
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Now I divide both sides by 3, because I'm trying to solve for t, so 2/3 is equal to e⁻T/10 and to undo the e, I will take natural log of both sides, natural log of 2/3 is equal to -T/10.
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And I will take the negative of both sides, T/10 is = - natural log of 2/3, now - natural log of 2/3 is the same as natural log of 2/3 to the - 1, that is the same as natural log of 3/2 is equal to T/ 10.
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T is equal to 10 x the natural log of 3/2 and that is really as far as I can go without a calculator, I'm going to plug that into my calculator, I have got natural log of 3 ÷ 2 and it is says that natural log of 3/2 is .405.
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10 x that would be 4.05 and I believe our units here were minutes, that tells us that the tank will be 1% salt 4 minutes after we start this experiment or 4.05 minutes after we start this experiment.
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That is the answer to part A, part B what was the long-term concentration of salt, what that is really asking us for is the limit as T goes to infinity, we are going to take T going to infinity here and we get Y of infinity is 3 - 3 e⁻infinity/10.
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But that is 3 divided by e⁺infinity/10, since I put it down in the denominator because it was a negative exponent, 3/e⁺the infinity that just goes to 0 because it is 3 divided by a very large number.
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Y of infinity is equal to 3, what that tells us is that the long-term amount of salt in the tank is 3 gallons, the long term concentration, remember y(T) is the amount of salt at any given time, the long term concentration of salt is 3 gallons inside of 100 gallon tank.
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That is .03 3/100 is 3%, in the long term you are going to find you that your water has 3% salt and that is actually not surprising at all, in fact that is a good check on the work that we have been doing all the way through this problem.
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Because remember that the water that was entering the tank water is 3% salt, it really makes sense that even if we start with pure water initially, if we pour in water that is 3% salt and then wait a very very long time we would expect to come back and see that tank to be very close to 3% salt.
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That is in fact what the equations justify here, just to recap what we did there, we started with our solution to the differential equation, remember we got that from the previous example, that is where we resolve that.
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And then we had to answer these 2 questions, to figure out how long until the tank is 1% salt, 1% salt was 1 gallon, we solve y(T) = 1 and the we went through the algebra here to solve for t and we came up with T was equal to 4 minutes.
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That tells us that after 4 minutes the tank will be 1% salt and then to find the long term concentration, long term means what will the concentration be after you wait for a long time.
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We look at taking T going to infinity while we had e^- infinity so that becomes e⁺the infinity in the denominator which goes to 0, the long term amount of salt was 3 gallons and the long term concentration was 3 gallons divided by a 100 gallons.
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We got 3% salt and that checked with our intuition because we are pouring water into the tank that was 3% salt.
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Our next example is a population example, we got a deer population in a national park.
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And these deer are reproducing at a rate that is proportional to whatever the current population of deer, that makes perfect sense because however many deer you have this year, a certain percentage of them will have babies.
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And so the more deer you have, the more baby deer you will have next year and what we are given is that the population doubles every 69 years and there are 1,750 deer today.
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We asked to write an initial value problem to describe the situation and to solve it, let us work that out, the normal growth of the population, I'm going to write it up here is Y′ is equal to the increase - the decrease in the population.
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In this case there is no decrease in the population because we have not been given any information on any predation in the deer or any deaths in the deer, we are ignoring the fact that deer might die after a long time.
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We are just worrying about the increase and the increase is given by the fact that the deer are reproducing, it is some constant x the current number of deer, Y′ is = to KY.
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We do not know what k is because we have not figured out the exact rate of reproduction, we are going to use the 69 at some point to determine what K would be but that is going to come later on.
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We also have been given that there are 1,750 deer today, we will say that, that is times 0, y (0) is = to 1750, there we have a differential equation and an initial condition, we got an initial value problem to describe our situation.
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Now the harder part is to solve that, let us go ahead and solve that, that is actually a separable differential equation, it is one of the easier ones, I'm going to write DY by DT is equal to KY and then I will pull.
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Remember the idea of separable differential equations is you try to get all the on one side, all the t or x whatever the other variable is get all those on the other side, I will pull that y over under the DY.
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DY/y is equal to, I will pull the t over on the other side, KDT and now I can integrate both sides, the integral of DY/y is natural log of y is equal to the integral of K is just KT + a constant.
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I got at my constant right when I do the integration and now I'm going to solve for y, y is equal to e⁺KT + c, remember that + c gets pulled up into the exponent of the E, you can not leave it as a + c down below anymore.
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That is equal to e⁺KT x e⁺c and I'm just going to call e⁺c that is a constant raised to another constant, I'm going to call that D if representing another constants, DE⁺KT, we got y = DE⁺KT.
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And now I'm going to use my initial condition which is that y(0), that means I plug in 0 for t, that is DE⁺K x 0 , e⁰ is 1, y(0) is just D and we are given that y(0) is 1,750 that means D is = 1750.
00:36:43.000 --> 00:37:10.000
My differential equation is now y = 1750 E⁺KT, that is not my differential equation, that is my solution to the differential equation including the initial value condition, what I do not know yet is what K is.
00:37:10.000 --> 00:37:28.000
I need to solve for K, I need to figure out what other information I have in the problem I can use to solve for K and of course the thing I have not used yet is the fact that the population doubles every 69 years.
00:37:28.000 --> 00:37:56.000
The way I'm going to use that is that I know that 69 years from now we should have twice the population we have now, y of 69 is twice the population we have now, 72 × 1750 which is 2 × 1750 is 3500.
00:37:56.000 --> 00:38:29.000
I'm going to use that, I'm going to plug that back in and solve for K, plug that back in I will get 3500 up here, 3500 = 1750 x e⁺K and now remember this was at time 69 so K x 69.
00:38:29.000 --> 00:38:55.000
Now I'm going to try to solve that and figure out what K should be, if I divide 17500 away from both sides, on the left hand side I will just get 2 is equal to e⁺69K and from there I can take the natural log of both sides.
00:38:55.000 --> 00:39:16.000
Undo the E, natural log of 2 is equal to 69K and natural log of 2, you can check this on a calculator is approximately 0.69, that is actually the problem was ready to make the numbers work out nicely at this step.
00:39:16.000 --> 00:39:35.000
And 0.69 is 69K, K must be 1/100, of course if the numbers did not work out so nicely you can still solve for K, you just pull out a calculator at this step but I wrote this problem to try and avoid the need for a calculator.
00:39:35.000 --> 00:40:10.000
And so I'm going to plug this K back in to our solution which is right there, y = 1750 x e⁺K is 1 / 100, 1/100 T and that is my final answer to the initial value problem, it satisfies both the differential equation and the initial condition.
00:40:10.000 --> 00:40:31.000
Let us recap what we did here, we started out with Y′ = KY that indicates the fact that the population is increasing at a rate that is proportional to its current population, remember the increase is Y′, the current population is y there.
00:40:31.000 --> 00:40:53.000
Proportion of the current population is K x Y and at that point we did not know what K was and then we use the fact that there are 1750 deer today to give us an initial condition y(0) = 1750, that was our initial value problem.
00:40:53.000 --> 00:41:10.000
Then to solve it that was a separable differential equation, we separate the variables, get the y on one side, I get the t on the other, integrate both sides and then we solve for y in terms of t and you have to be very careful to add the constant of integration .
00:41:10.000 --> 00:41:29.000
Right when you do the integration, that is we did here and then you keep track of the constant, the constant goes up into the exponent of e after that as we solve for y that turns into e⁺KT x e⁺c here.
00:41:29.000 --> 00:41:37.000
e⁺c I just called that D, since that is another constant, we ended up with y = DE⁺KT and then we use the initial condition y(0) = 1750 we plug that in and we figured out the D was 1750.
00:41:37.000 --> 00:42:10.000
We got our equation y = 1750 E⁺KT and then finally in order to figure out what K was, we had to use this condition that it doubles every 69 years, we plug in T = 69 and we plug in the fact that the population had double.
00:42:10.000 --> 00:42:27.000
It had grown to 2 x the initial population 2× 1750 was 3500, we plug that in for y and we plugged 69 in for T and then we work through and we solved for K and we worked out that K is 1/100.
00:42:27.000 --> 00:42:48.000
And we plug that back into the K here and we finally got our solution 1750 e¹ / 100 T, keep that in mind we are going to be using this solution to answer some follow up questions in the next example.
00:42:48.000 --> 00:43:04.000
Here is the next example, we are going to control the deer population and the way we are going to do that is we are going to allow some hunters in to hunt a certain number of deer per year, the hunters are going to take 20 deer per year.
00:43:04.000 --> 00:43:15.000
And the question is will the population increase or decrease from then on and then there is some follow up questions if it increases when will it get to 3000?
00:43:15.000 --> 00:43:26.000
If it decreases, when will the population completely die out? This is actually going to change our differential equation, let me remind you the differential equation we started with before.
00:43:26.000 --> 00:43:41.000
We had Y′ was KY and we figured out what the value of K was, that was our initial differential equation reflecting the increase in the deer population just due to natural reproduction.
00:43:41.000 --> 00:44:06.000
What we are doing now is we are introducing a decrease in the deer population due to hunting, we are taking 20 deer per year and I'm going to subtract off a 20 from that, let me remind you that this reflects the whole increase - decrease idea that we mentioned before.
00:44:06.000 --> 00:44:24.000
That is the increase in the deer population and it depends on y because the amount of baby deer that will be born each year depends on the number of adult deer that you have each year.
00:44:24.000 --> 00:44:43.000
The more adult deer you have the more babies will have next spring, the decrease was constant and that is because that is not dependent on how many deer there are right now.
00:44:43.000 --> 00:44:53.000
The park authorities just simply said the hunters are allowed to take 20 deer per year and that is a fixed number it does not depend on how many deer there are.
00:44:53.000 --> 00:45:02.000
Now we have to solve this new differential equation, we can use the solution from before because we got a different differential equation, even though it started out the same.
00:45:02.000 --> 00:45:17.000
We are going to solve this differential equation, I'm going to move the KY over the other side and I will get Y′ - KY is =-20
00:45:17.000 --> 00:45:30.000
Now, I want to remind you we did figure out what the value of K was in the last example, it was 1/100 but it would get a little cumbersome if we included that right now, I'm just going to leave it as K.
00:45:30.000 --> 00:45:48.000
We will remember later on that K actually was 1/100, we will use that when we need to, but it is probably easier now just to leave that out, this is a linear differential equations, we are going to use our technique of linear equations.
00:45:48.000 --> 00:46:08.000
i(T) is e⁺the integral of the P function, now P is whatever in front of the Y and the key step here is you got to be careful to include that negative sign, let me highlight that negative sign.
00:46:08.000 --> 00:46:28.000
That is part of the P, we have to include that when we are finding the integrating factor, e⁻KDT or e⁺the integral - KDT, is e⁻KT, we are going to multiply that by both sides.
00:46:28.000 --> 00:46:51.000
Y′ e^- KT - Ke⁻KT x Y= -20, e⁻KT, remember here we are using our strategy for linear equation, if what I'm doing right now seems a little foreign or a little rusty to you.
00:46:51.000 --> 00:47:03.000
What you might want to do is go back and look at the lectures here on www.educator.com for linear differential equations and you will see the strategy worked out carefully.
00:47:03.000 --> 00:47:24.000
The point of multiplying by this integrating factor is what we have here on the left is y⁻KT, the derivative of that using the product rule is equal to -20 e⁻KT and we are going to integrate both sides.
00:47:24.000 --> 00:47:50.000
The integral of the derivative is Ye⁻KT integral of the right hand side is -20 e⁻KT divided by -K, that just turns into + 20/K e⁻KT and we just did the integration, I got at the constant right now.
00:47:50.000 --> 00:48:09.000
Tuck that constant on the right now and then I got to keep track of that constant as I solve things from here on, in order to solve for y I need to divide by e⁻KT, which means multiplying by e⁺KT.
00:48:09.000 --> 00:48:39.000
I will get y = to just 20/K + CE⁺KT and the I think might be easy now to incorporate the fact that K is 1/100 and we figure that out the previous examples, you can go back and check that out if you do not remember where that comes from.
00:48:39.000 --> 00:49:09.000
We get y = 20 divided by K, 2 0÷ 1/100 is the same as 20× 100, y is 2000 + Ce⁺KT, I'm just going to leave this other K now because I do not think we are going to need that immediately.
00:49:09.000 --> 00:49:24.000
We want to use our initial condition, our initial condition which we got from the previous problem, I will wirte it up here was y(0) = 1750, we got that from the previous problem that was given in this problem.
00:49:24.000 --> 00:49:52.000
We are assuming that we are talking about the same park as before that has 1750 deer in it, I plug in 1750 for y and 0 for t, 2000 + Ce⁺K x 0, e⁰ is just 1, I get 2000 + C is 1750.
00:49:52.000 --> 00:50:16.000
I get C = 1750 - 2000 is -250 and I will plug that back in to my solution and I will get y = 2000 - 250 I'm plugging in the value of C now.
00:50:16.000 --> 00:50:48.000
I'm plugging it back into here, e⁺KT is e¹/100T, what I got here is a solution to the differential equation, actually to the initial value problem which tells me how the population is going to behave at any given time.
00:50:48.000 --> 00:50:59.000
We are going to use this solution to try and answer these follow-up questions about whether the population is going to increase or decrease from here on.
00:50:59.000 --> 00:51:29.000
Let us keep moving with this, the solution that we have here is y is 2000 - 250 e⁺KT and remember the K was 1/100 and if we look at this what we can tell is that the population is decreasing because of that negative sign there.
00:51:29.000 --> 00:51:56.000
The population is decreasing because of that negative sign and we are being asked when the population will die out if it is decreasing.
00:51:56.000 --> 00:52:25.000
To figure out when the population will completely die out, we are going to set y(T)=0 and solve for T because we want to know when the population will be 0, we will say 2000 - 250e⁺KT = 0.
00:52:25.000 --> 00:53:02.000
2000 = 250 e⁺KT and if I divide both sides by 250,2000 ÷ 250 is 8, I will get 8 = e⁺KT and to undo that e, I will take the natural log of both sides, I will get KT = the natural log of 8.
00:53:02.000 --> 00:53:38.000
8 by the way is 2³, that is 3 x the natural log of 2 and if you work out 3 x the natural log up 2, that is 3 x about 0.69 which is 2.07 and T = 2.07 divided by K which is 2.07.
00:53:38.000 --> 00:54:10.000
Let us remember the K was 1/100 divided by 1/100 which is the same as multiplying by 100, that is 207 and our unit of time here was years.
00:54:10.000 --> 00:54:27.000
What we did here just to recap, is we took our solution to the differential equation, remember we found this in the previous slide and we notice that because that negative sign, the population would be decreasing.
00:54:27.000 --> 00:54:41.000
Since it is decreasing we are asked when the population will die out, let us set the population = 0 and then we just go through the algebra to solve for T, we are using the value of K that we learned last time.
00:54:41.000 --> 00:55:24.000
The K was 1/100 we solve for T, we get to 2.07 × 100 so 207, what this tells us is that if the deer are really breeding at the rate that we predicted and we are only losing 20 deer a year then the deer population will last another 207 years before they die out completely. 5506 Our last example is a financial example, we have a student borrowing $20,000 for college, she initially collects $20,000 and uses that to pay her tuition and then she agrees to pay off an annual interest rate of 5%.
00:55:24.000 --> 00:55:39.000
She is going to make her payments continuously at a rate of $1000 per month, the idea here is that every month she is going to pay $1000 that is a constant but at the same time she got this large balance which is building up interest.
00:55:39.000 --> 00:55:55.000
We have to write an initial value problem to describe this situation and we do not actually have to solve it in this example but we should determine how we could figure out when the loan will be paid off.
00:55:55.000 --> 00:56:20.000
Let us go ahead and set this up, I'm going to set up a couple variables, I will go ahead and work up here, my T is going to be time as usual and my y(T) is going to be the amount of money she owes at a given time.
00:56:20.000 --> 00:56:41.000
Amount owed at time T and I should put some units on here, there are 2 ways you can go with this because there is an annual interest rate but then she is making payments every month.
00:56:41.000 --> 00:57:01.000
We have to decide which one we want to use here, it will be a little easier if we work in terms of month, we will measure time in months and then the amount that she owes, we will measure that in dollars.
00:57:01.000 --> 00:57:16.000
We are going to write an initial value problem that means we need to write an expression for Y′(T) and remember our strategy for that Y′ (T) is the rate at which y is changing.
00:57:16.000 --> 00:57:39.000
That is always an increase and then there will be a decrease, let us think about what each one of those would represent, the increase what is making the amount that she owes increase, that is the interest.
00:57:39.000 --> 00:57:56.000
As she keeps the money interest builds up and she has to pay back more later, the increase is determined by the interest and the decrease is determined by the way in which she pays it back.
00:57:56.000 --> 00:58:29.000
The decrease is given by the payments that she makes, the interest that she pays is dependent on the amount that she owes at any given time, the amount that she owes at any given time is y(T).
00:58:29.000 --> 00:58:37.000
And then it says that she is borrowing at an interest rate of 5% but that is the annual interest rate, what that means is it every year she would have to pay back an extra 5% of what she owes.
00:58:37.000 --> 00:58:57.000
5% is .05 but we are measuring things in terms of months, every month her interest rate would be 1/12 of that, I'm going to put .05 ÷ 12 there that reflects the monthly interest rate would be .05 ÷ 12.
00:58:57.000 --> 00:59:28.000
And then the decrease is coming from her payments, now the amount of payments she is making is $1000 per month, I will -1000 here and we have is y′(T) = .05/12 y(T)-1000.
00:59:28.000 --> 00:59:44.000
What will we do on their is rewritten the differential equation that describes the amount she owes at any given time and how it is changing, we also need an initial condition, that is the amount she owes initially which is $20,000.
00:59:44.000 --> 01:00:16.000
y(0) is $20,000, that together gives us an initial value problem because we have the differential equation describing the increase and the decrease and an initial condition describing the and initial amount that she owes on the loan.
01:00:16.000 --> 01:00:32.000
Let me point out a couple of things here, remember I said the very beginning of this lecture that you always have an increase and a decrease, generally one of them is constant and one of them is dependent on the current value.
01:00:32.000 --> 01:00:47.000
Or the current quantity that your keeping track of and this is a really good illustration of that because the payments she is making are constant, that means she is paying $1000 a month no matter what.
01:00:47.000 --> 01:00:50.000
Every month she pays $1000 irrespective of how much she actually owes but the interest that is building up on the loan is dependent on the amount that she currently owes.
01:00:50.000 --> 01:01:23.000
This interest depends on the current value of y, the more she owes the more interest builds up and at the very beginning of the loan she is building up a lot of interest then later on as she get some of it paid off she would not be building up interest so fast.
01:01:23.000 --> 01:01:35.000
The second part of the problem here says assuming you could solve the equation described how you would determine when the loan will be fully paid off, that was our answer to part A right there.
01:01:35.000 --> 01:01:58.000
For part B, what you do is you would solve, we want to figure out when the loan will be fully paid off, we want to figure out when she owes 0 dollars, y(T) is the amount she owes, you would solve y(T)=0.
01:01:58.000 --> 01:02:40.000
That will tell you what the time will be when she owes 0 dollars and for a little more explanation here, the answer for t is the number of, remember T was measured in months to completely pay off the loan.
01:02:40.000 --> 01:02:51.000
If you could solve this differential equation and actually you could, it is a linear differential equations, we could solve it using the techniques that we learned in the previous lecture on linear equations.
01:02:51.000 --> 01:03:05.000
But if you want to head and solve this differential equation then in order to figure out how long her loan is going to last, you would solve y(T) = 0 because you are trying to figure out what time it would be when she owes 0 dollars.
01:03:05.000 --> 01:03:15.000
And then whatever you get for T, you would say it is going to take that many months to pay off the loan, just to recap here, we wrote our initial value problem.
01:03:15.000 --> 01:03:25.000
We set up our variables T is always time, y(T) is the quantity we are keeping track of which is the amount she owes at time T.
01:03:25.000 --> 01:03:36.000
You got to be careful about your scale here because part of the problem was given in months and part of the problem was given in years, we said okay we are going to make our time scale to be months.
01:03:36.000 --> 01:03:49.000
And then Y′(T) is the increase - the decrease and we have to think what is making our quantity increase and what is making our quantity decrease, our quantity is the amount that she owes.
01:03:49.000 --> 01:04:08.000
That increases as interest builds up on it and that decreases as she makes payments, the interest depends on how much she owes at any given time and we said its 5% interest, that is .05 but that was an annual interest rate.
01:04:08.000 --> 01:04:21.000
Our monthly interest rate you just divide that by 12 ,that is where that 12 came from, that was the conversion from years to months and then her decrease is just the fact that she is paying off $1000 a month.
01:04:21.000 --> 01:04:38.000
We just subtract $1000 from Y′ there, that gives us the differential equation, the initial condition reflects how much she owes initially which was $20,000, we got our initial value problem that would be a linear differential equations.
01:04:38.000 --> 01:04:50.000
We could solve it using our techniques but the question just asked how we would figure out when the loan will be paid off? In order to do that, we set y(T) = 0.
01:04:50.000 --> 01:05:00.000
And then the answer wherever we get it when you solve that for t will be the number of months to pay off the loan because T was measured in months.
01:05:00.000 --> 01:05:14.000
That is the end of our lecture on applications which is also known as modelling and is also known as world problems, any of those 3 different phrases they all mean the same thing in the constant context of differential equations.
01:05:14.000 --> 01:05:19.000
These are the differential equations lectures here on www.educator.com, I'm Will Murray, thanks for watching.