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An introduction to differential equations and their applications. A differential equation, or DE, is an equation involving an unknown function and one or more of its derivatives. More simply, you can just think of it as an equation that involves at least one derivative. Mathematical models often take the form of a differential equation because many real-world problems involve rates of change, and rates of change are derivatives, and we use those rates to predict future behavior.
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So here are some examples of differential equations. So notice they all have equal signs. They're all equations, and they all have at least one derivative. So dy/dx is a derivative. Here, y double prime, y prime, so there's actually two different derivatives involved here. I have a second derivative of y with respect to t. Here, I have a first derivative of p with respect to t. And then here I have v prime.
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So the order of a differential equation is the highest order derivative that appears in the equation. So we had a second derivative, a second derivative. So those would be second-order differential equations. But the ones that involve just dy/dx or v prime, so that's just a first derivative. So that is going to be a first-order differential equation. Given a differential equation, we often try to solve it. We're looking for its solution.
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A function f is called a solution of a differential equation if the equation is satisfied when y = f of x and its derivatives are substituted or plugged into the equation. In this example, we want to verify that y = 2 e to the x is a solution to the differential equation y double prime minus 3 y prime minus 4 y = 0. So what we have to do is substitute y into the equation, but also y prime and y double prime into the
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equation to make sure that it's a solution, that the left side equals the right side. Well, we have y, but we need to find, through differentiation, y prime and y double prime. So I'm going to start the problem by finding y prime, the derivative of y, using the chain rule. So a negative is going to come out. We're going to have negative 2 e to the x as y prime, and y double prime is now positive 2 e to the x.
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Now we're going to go ahead and substitute those in. We're going to substitute in y double prime, y prime, and y into our equation. So y double prime was 2 e to the x minus 3 times y prime was negative 2 e to the x. So I'm substituting it after this minus 3. And then we're going to have minus 4 y, and y was given. That was 2 e to the x, and then we're checking, does that equal zero? So I'm going to put a question mark here. This is what we're trying to verify. So
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what we have is 2 e to the x plus 6 e to the x. So we're multiplying these coefficients. And then here, negative 4 times 2 is negative 8 e to the x. Again, does this equal zero? Well, now these are like terms. They're both e to the x. So we can add their coefficients. We're going to get 8 e to the x minus 8 e to the x. So does that equal zero? Yes, it does. So we have verified that this function, y = 2 e to the x, is a solution to this
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differential equation, to this equation that involves at least one derivative. In a class dedicated to differential equations, there's so many strategies and other things to go into, but we're going to focus on solving first-order differential equations, which only have first derivatives in them. And not even all kinds of first-order differential equations, just really specific kinds. And often, they can be written in the form dy/dx equals f of x,y.
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So basically, the derivative on one side of the equation, and on the other side, a function of x and/or y. A general solution to a differential equation is a family of solutions containing arbitrary constants. So basically, these have that plus c. Now our first example is asking us to find the general solution to the first-order differential equation dy/dx equals cosine 2x minus 4x. So recall from earlier that
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a solution was a function such that when you substitute it in, it makes the equation true. So here, the derivative is dy/dx. I'm looking for the function y, the original function. So remember, dy/dx is like y prime. So this is a first derivative, first-order differential equation, and we're solving for y, the original function. So an integral is involved. An integral is needed because how do you go from a derivative to the original function? Well, you integrate.
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So this is a really nice setup here because in general, I have dy/dx, the derivative, on one side, and then the other side is strictly a function of x. So cosine of 2x minus 4x. So I have dy/dx equals just f of x. So in order to solve a differential equation in this form, I can simply say that y, the original function, the one that I'm looking for, the solution, is the integral of that function of x. The integral of cosine 2x minus 4x dx.
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So the integral of cosine of 2x is one half sine of 2x. The integral of that minus 4x, that's going to be negative 2x squared. So if I add a power and divide by the new power, right, I'm going to get 2x squared. And then I'm going to put a plus c because I want the general solution. It's a family of solutions with the arbitrary constant. So it's off by a constant potentially. So the idea is that when you take the derivative of y and you substitute it back into dy/dx,
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does the left side equal the right side? Is the equation satisfied? Yes. Yes, it is. So this is my solution to this first-order differential equation. Now let's look at a particular solution. A particular solution is the single solution obtained by identifying the constants. So instead of leaving generic constants involved, we're going to find numerical values for them. Now an initial condition is a given value of the function or its derivative used
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to determine those constants. So in order to find a particular solution, I need some more information. I need an initial condition or conditions. So in this example, we're going to find the general solution to the first-order differential equation y prime equals 1 over x squared plus 9. But then we're going to go farther and find the particular solution that satisfies the initial condition y of 3 = 0.
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So this is a two-part problem. I have y prime is equal to 1 over x squared plus 9. So I have a differential equation in the form of that dy/dx equals f of x only, right? y prime and dy/dx are interchangeable. This right-hand side is a function of x only. So it's quite nice to find the general solution. I can simply integrate. So to find y, I'm going to find the integral of 1 over x squared plus 9 dx. And let's
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recall that that is arc tangent. So one third arc tangent of x over 3 plus c. So this is my general solution because it involves the arbitrary constant. In this example, we're going to go a step further and find the particular solution that satisfies this initial condition. So, in other words, a particular solution doesn't have a plus c. We're going to solve for c. And in order to solve for c, I need an extra piece of information called an initial condition.
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So I go ahead and substitute in 3 for x and 0 for y into my general solution in order to solve for c. So the input inputs are the independent variable x, and then the output is y. So this is an x. This is a y. I'm going to substitute it in here. So in other words, I'm going to bring this work up here. So I'm using the fact that y of 3 is equal to 0. So 0, I'm replacing the y right here with a zero, is equal to one third arc tangent of 3 over 3. So
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I'm replacing the x with a 3 and then plus c. So I replace both x and y so that I can solve for c. All right. So I have 0 is equal to one third arc tangent of 1 plus c. Now arc tangent of 1. What do I substitute into tangent to get out one within the restriction of arc tangent? That's going to be pi over 4. So we're going to have 0 is equal to one third times pi over 4 plus c. So 0 is equal to pi over 12 plus c. How do I solve for c? I subtract
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the pi over 12 to the other side. So negative pi over 12 is equal to c. Now unfortunately, a common mistake for students at this point is for them to box in c equals negative pi over 12 as the answer. But that is not the answer. The answer to a differential equation is a function. So not just this constant. What I have to do is substitute this constant into the general solution to get the particular
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solution. So we're going to have y = one third arc tangent of x over 3 and then minus pi over 12. So this is now my particular solution that corresponds to that specific given initial condition. Before moving on, when we're given a differential equation together with an initial condition, that's called an initial value problem.
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In this example, we want to determine if y = x ln x plus 2x is a solution to the initial value problem x dy/dx minus y = x with the initial condition y of 1 is equal to 2. So this is another type of verify problem, but there's going to be an extra piece to verify because it has an initial condition. It is an initial value problem.
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We're given y = x ln x plus 2x. So we're able to substitute that into the y of the differential equation. But I'm going to need to substitute in dy/dx. So I need to find dy/dx. I need to find the derivative of y. So let's do that first. dy/dx is going to be equal to, well, first we need a product rule. x times ln x is a product rule. So we'll leave one function alone, x. The derivative of ln x
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is 1 over x. Plus now we'll leave ln x alone, and then times the derivative of x, which is 1. So I've done the product rule in the first term, and then plus the derivative of 2x is just two. So plus two. So let's clean this up a little bit. dy/dx is equal to just 1 plus ln x plus 2. I can make this a little nicer. dy/dx is 3 plus natural log x. And now we're going to substitute dy/dx into my differential equation and y into
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my differential equation and verify that I get a true statement that this is a solution. So we're going to have, let's see, x times 3 plus natural log x minus y, which is two terms. So I need parentheses, x ln x plus 2x. Does that equal, question mark, x? Right? We're trying to verify this. So let's distribute in the x. And we're going to have 3x plus x ln x. We'll distribute the negative. So negative x ln x minus 2x. Does that equal x?
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Well, in the middle here, plus x ln x minus x ln x is gone. 3x minus 2x is x. So does x equal x? Yes, it does. So the differential equation is verified. But is the initial value problem verified? Not until we check the initial value y of 1 = 2 in the original equation given or the solution that was given. So when we substitute 1 into x, do we get out two? So let's go ahead and check. So what is y of 1? Well, 1 times natural log 1 plus 2 times 1. Well,
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natural log of one is zero. So all I have left is two. So is y of 1 equal to two? Also, yes. So we have determined that this is a solution to this initial value problem. So that answers the question. It wasn't actually a verify problem. It was determine if it is. And so we have to check both. So yes, it is a solution to the initial value problem, which is often abbreviated IVP. In our next example, we're told to solve dy/dx = 2x
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over 3x squared plus 1 with the initial condition y of 0 = 1. So this is an initial value problem. We're solving for a particular solution. It's nice because I have my derivative is equal to just a function of x. So let's go ahead and start by setting up our integral. y is going to be the integral of 2x over 3x squared plus 1 dx. And we can solve this using a u-substitution. So I can go ahead and let u be 3x squared plus 1. du is 6x dx. I'm looking to replace
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a 2x dx. So I can just go ahead and divide by 3 or write this as one third du is 2x dx. So I can do my change of variables now. So I have my 2x dx is going to become a one third du. I'll put the one third in front, the du on top. And then 3x squared plus 1 gets replaced with u. So I have one third natural log of the absolute value of u plus c. So this is going to start as a general solution. Let's plug our u back in. So we get y = one third natural log. I don't need absolute value
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bars anymore because my u is positive, 3x squared plus 1. So I'll just put 3x squared plus 1 plus c. And now we're going to use the initial condition to solve for c. So what I like to do here is kind of draw an arrow because this is work off to the side to solve for c. And I'm using that y of 0 is equal to 1. So I'm going to let x, the input, be zero, and then the output be 1. So I end up with
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1 = one third natural log of 1 plus c because if I let x be zero here, then the inside is just one. Natural log of one is zero. So I end up with just 1 is equal to c. And so now I go ahead and go back and plug that c in. So my solution is one third natural log of 3x squared plus 1 and then plus 1. That is the solution to this initial value problem or a differential equation with an initial condition. In this example, we're told the function y of t satisfies the
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differential equation dy/dt = y to the fourth minus 6 y cubed plus 5 y squared. When we see y of t satisfies the differential equation, it really means that y of t is a solution to the differential equation. But a differential equation can have many solutions. In part a, we're asked what are the constant solutions of this differential equation? Let's take a moment to do some work off to the side and think about this. So that word constant. So the solution y of t being a constant,
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like a number a, would have to mean that the derivative of that constant is zero. So our constant solutions have to have this form. y of t is equal to a or just y = a. So then that must mean that dy/dt has to equal zero. And that's ultimately what we're going to solve to find the constant solutions. We have dy/dt up here. So we're going to go ahead and set dy/dt equal to 0. 0 = y to the fourth minus 6 y cubed plus 5 y squared.
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And we're going to factor out a y squared. And we have y squared times the quantity y squared minus 6y plus 5. And then we'll factor the quadratic there. We'll have the factor y minus 5 and then the factor y minus 1. By the zero product property, we'll set each factor equal to 0. So y = 0 makes y squared = 0. y = 5 makes y minus 5 = 0, and y = 1 makes y minus 1 = 0. So we have solutions y = 0, y = 1, and y = 5. So these are the constant solutions to my differential equation.
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If I take the derivative of each of these, that's 0. And then if I substitute 0 into each y, we can see that would equal zero. If I substitute in one into each y, that would make this equal zero. And then if we did the arithmetic of plugging five into each y, that would also make this equal zero. For part b, we're still working with the same differential equation, but we're being asked for what values of y is y increasing and then for what values of y is y decreasing.
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Let's remember from calculus one that in order for a function to be increasing, its derivative has to be positive. So for what values is y increasing? We ultimately want to know where is dy/dt greater than zero, and then where is y decreasing? That's asking when is dy/dt less than zero. Well, in part a, we figured out the values that make dy/dt equal zero. So what we're going to do is we're going to go ahead and create a number line to do a sign analysis.
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So in part a, we figured out that at y = 0, at 1, and at 5, that's when dy/dt was equal to zero. In other words, we found the critical values. That's when the derivative is equal to zero. And we want to know what's happening on the intervals in between. So we want to know what's going on between negative infinity and zero, between zero and one, between one and five, and between five and infinity. Is the derivative positive or is the derivative negative? So again,
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we're going to do a sign analysis for dy/dt. And if we remember from calculus one, we want to pick a test value in each interval. So between negative infinity and zero, let's go ahead and pick a nice number like negative 1. Between zero and one, I'm going to go ahead and pick a half. Between one and five, I'm going to pick two. And between five and infinity, I'm going to pick six. And we're substituting that into dy/dt to think about the sign,
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whether that sign is positive or negative. Now from part a, we know that dy/dt as a factored equation was y squared and then times y minus 5 times y minus 1. So when we substitute negative 1 into each factor, well, we can technically ignore y squared because that's always positive. So we just think about substituting negative 1 into the factor y minus 5 and the factor y minus 1. So negative 1 minus 5, that's going to be negative 6. So a negative value. And
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then negative 1 minus 1 is negative 2. So another negative value. But what's a negative times a negative? That is a positive. So dy/dt, the derivative at negative 1, will end up being overall positive. Now let's go ahead and think about one half. So again, one half squared, anything squared is positive. So we'll just focus on y minus 5, y minus 1. One half minus 5 is negative, and one half minus 1 is negative. And again, a negative times a negative is a positive.
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The next interval, we chose 2 as our test value. So 2 minus 5 is negative and 2 minus 1 is positive. So a negative times a positive overall is a negative. So we have our first interval where the derivative is negative. And then if I substitute in 6, again just focusing on y minus 5 and y minus 1, that is going to give me overall positive. 6 minus 5 is positive. 6 minus 1 is positive. So that last interval is also positive.
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Based on our sign analysis, based on the sign chart we have created, we can now answer this question. So y is increasing whenever dy/dt is positive. So the intervals where dy/dt is positive are from negative infinity to 0. So we'll summarize our results by saying y is increasing on the y interval from negative infinity to 0. At 0, we know that dy/dt was equal to zero. So dy/dt was not positive at zero.
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dy/dt was zero at zero. So this should not be included. So we put an open parenthesis. We're skipping over that value. And then we're saying all the values between 0 and 1. So all the values between 0 and 1 and all the values from 5 to infinity, not including five. So the last interval here where y is increasing is from 5 to infinity. So y values in each of these intervals is where y is increasing. And then y is decreasing only between 1 and 5. So
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we're going to say y is decreasing on the y interval from 1 to 5. So really, we are going back to our calculus one roots and remembering what it means for a function to be increasing is that its derivative is positive, and a function is decreasing when its derivative is negative. And we did the legwork in part a. We found the zeros, the critical values. Where is the derivative equal to zero? When we found those constant solutions. So let's go ahead and box in our answer here.
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The sign chart was really just scratch work. Our answer is the intervals. Before moving on, I want to talk about why we're interested in this type of problem. So we did not actually fully solve the differential equation. We found some solutions, the constant solutions in part a, and in part b, we just found where the solutions, the other solutions, are increasing and where the other solutions are decreasing.
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So why not just find those solutions? Well, sometimes differential equations are really hard to solve or not even possible to solve with kind of nice integration technique methods. So we would still be interested in finding out as much as we can about the solution set. So finding out properties about those solutions, like where are those solutions constant, where are those solutions increasing or decreasing, without actually fully solving the differential equation.
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So we're interested in finding out as much as we can about the solutions because maybe we can't actually explicitly find them. Differential equations have so many applications. And just to give us a glimpse, we're going to take a look at a simpler one here. We're going to look at a population that's modeled by a differential equation. So we have dp/dt is equal to 1.5p times the quantity 1 minus p over 2000. So a population
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is growing, and the rate of change of the population, dp/dt, is a derivative. That's a rate. So the rate of change of the population is given by this differential equation. And in part a, we're asked what are the constant solutions? Now we were asked the same question in the last example. What are the constant solutions? But now this represents a population model. And in a population model, we call these equilibrium solutions.
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So to find constant solutions, what we had to do was set the derivative equal to zero. So to find these equilibrium solutions, these constant solutions, we're going to set dp/dt equal to zero and solve the equation 0 = 1.5p times the quantity 1 minus p over 2000. And this is already factored for us. So we're just setting 1.5p equal to zero. And then we're going to go ahead and set 1 minus p over 2000 equal to zero. So 1.5p = 0 is solved by just p is equal to 0,
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right? I can divide by 1.5. I can just look at this and, you know, 1.5 times 0 equals 0. Over here, what I can do is add p over 2000 to the other side. And so I get 1 = p over 2000. And then to solve for p, multiply by 2000. So p is equal to 2000. So I have two equilibrium solutions or constant solutions, p = 0 and p = 2000. Now let's talk about what these mean in the context of the problem. If p equals 0,
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the population is zero. So that basically refers to extinction in this population. If p is equal to 2000, then the population is 2000, and that's a constant solution. So what that means is it's basically the limit of this population model. So in any environment, we can only have, you know, so many, let's say people, be sustained by the food and the space that's there. So every environment has a limit. And for
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this particular population, the limit is 2000. And it's actually called the carrying capacity. And there's some other pieces that we can go into with this, but this is just to give you an idea of why we're interested in these constant solutions, depending on the model that's given to us. In part b, we want to know for what values of p is the population increasing and what values of p is the population decreasing.
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So for the population to be increasing, the derivative dp/dt would need to be positive. And for the population to be decreasing, dp/dt would need to be negative. From part a, we already know for what values of p the population was equal to zero. So we're going to go ahead and put those values of p on a number line. So those values were 0 and 2000. Those are our critical values. And then we're going to pick test points in each interval. But for this example, since there's some
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context here, the population cannot possibly be negative. So we're not going to have a number line from negative infinity to infinity because, again, we can't have a negative population. So we're really just going to start this at zero. And so we're going to pick a test point between 0 and 2000, like one. And then bigger than 2000 and less than infinity, we'll just pick 2001. And we're interested in the sign of dp/dt. And dp/dt was that 1.5p and then times 1 minus p over 2000.
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If we go ahead and substitute in 1, 1.5 times p, that's going to be positive. And then 1 minus 1 over 2000, that's also going to be positive. So a positive times a positive is a positive. This first interval from 0 to 2000, dp/dt will be positive. So the population will be increasing. And then if I substitute in 2001, 1.5 times 2001, that's positive. But now 2001 over 2000, that's slightly bigger than 1. So 1 minus
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something slightly bigger than 1 is negative. And a positive times a negative is a negative. So the population is increasing. So p is increasing for p values between 0 and 2000. And then the population is decreasing for values above 2000. Oh, whoops. So 2000 to infinity. So what does that mean? Let's think about this for a second. So the population is going to be growing, growing, growing from zero all the way to 2000,
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let's say people. And we said that 2000 was like the limit, right? There's a fancy word called the carrying capacity that we talked about in part a because that's all the environment can sustain. So between 0 and 2000, the population is increasing, increasing, increasing. If we end up going above 2000, so let's say 2100, then the population starts decreasing. So let's think about that. As sad as it is, if this environment can only sustain 2000 people,
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then it makes sense that the population, once it goes above 2000, starts decreasing. So people start dying because there's not enough environment or food to sustain that population. So I hope this was kind of an interesting example and it gave you a glimpse into an application of differential equations. To wrap up, I want you to just note that this last example is a logistic differential equation, and it pretty accurately models population growth, and it has a general form.
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dp/dt equals kp times the quantity 1 minus p over capital k. And the lowercase k and the capital k represent growth rate and the carrying capacity of the population. And again, there's many additional applications of differential equations that are studied in a full differential equations course.
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