Introduction to Differential Equations

Transcription

1 Math 56: Introduction to Differential Equations A differential equation is an equation that relates not onl a dependent variable and an independent variable, but also one or more derivatives of that dependent variable. For example, x + + = 0 is a differential equation. After completing this project, ou should be familiar with basic terminolog related to differential equations, and should also be able to solve some simple differential equations. Introduction: Wh are differential equations useful? Differential equations allow us to model situations in which the rate of change of some quantit depends directl on the quantit itself. Here are some examples that should illustrate the difference: When ou drive our car, ou have the abilit to independentl choose our speed. The speed of the car doesn t depend on how far the car has alrea traveled. Therefore, we would not want to use a differential equation to model this. When ou inflate a balloon, there is a probabilit that the balloon will get too full and burst. That probabilit changes as ou add air, and that rate of change depends on how much air ou have alrea added previousl. Therefore, we would want to use a differential equation to model this. You invest $0,000 in a long term, interest-bearing fund. Since our interest is compounded, the amount of interest ou make in future ears depends on how much interest ou have alrea made. Therefore, we could use a differential equation to model this. (See Example 4.) Next we re going to look at some examples of differential equations. Example : What is a differential equation, and what does it mean to solve one? onsider the equation =. This is a differential equation, because some of the variables are derivatives of other variables. In order for a function to solve a differential equation, the equation must be true when the function and its correct derivatives are substituted into the equation. The function = e x is a solution to the differential equation = : = e x = 9e x Substituting these into the original equation gives us: 9e x (e x ) = (e x ) 9e x 6e x = e x e x = e x Since the left-hand side equals the right-hand side, the function = e x is a solution. The function = x is not a solution to the differential equation = : = x =

2 Substituting these into the original equation gives us: (x) = (x ) 4x = x Technicall that last equation is true for some isolated values of x, (x 0.874, for example), but it s not generall a true statement like e x = e x obviousl is. Therefore, we determine that = x is not a solution to this differential equation. Example : What do we mean when we talk about the order of a differential equation? The order of a differential equation is the highest derivative that appears within the differential equation. For example, + 8x = x + is a third-order differential equation (because it contains a term.) For the remainder of this project, we focus exclusivel on first-order differential equations that can be solved using a technique called separation of variables. Example : How do we solve a first-order differential equation? In this example, we ll solve the differential equation like this one requires four steps: x =. Solving differential equations Step one: Rewrite the derivative as. This gives us x =. Step two: Rearrange the terms of the equation so that all the terms containing are on the lefthand side and all the terms containing x are on the right-hand side. x = = x (Note: It is not alwas possible to separate the -terms from the x-terms like this. If that isn t possible, the equation is called a non-separable differential equation. Solving non-separable equations is beond the scope of this course.) Step three: Find the antiderivative of both sides. Be sure to include on the right-hand side. Step four: Solve for. = x = x Technicall, we get a constant on both sides, and then combine them into the one constant ou see here.

3 = x = x = ± x So, an function in the form = ± x will be a solution to the differential equation. Q: When we multiplied b, wh didn t the change to a? A: There s nothing wrong with writing here, but times an unknown value is still an unknown value! Some people will use a new letter here, such as K, to show that the value of the constant has changed. Example 4: How about a word problem? ompound interest means that the rate of change in our balance is directl proportional to the da balance. That is, if ou invested $0,000 at 6% interest, then = 0.06A. (Literall, that equation reads The rate of change in our balance over time, is equal to 0.06 in other words, 6% times our current balance. ) Solve this differential equation. Note that Step one from the previous example has alrea been done for us. Step two: Rearrange the terms of the equation so that all the terms containing A are on the left and all the terms containing t are on the right. da = 0.06A da A = 0.06 Step three: Find the antiderivative of both sides. Be sure to include on the right-hand side. Step four: Solve for A. da = A 0.06 ln A = 0.06t The antiderivative of A is ln A. But since we know that balance must be positive (ou can t invest negative mone!), we can safel omit the absolute value signs here. ln A = 0.06t e ln A = e A = e 0.06t+ 0.06t+

4 At this point, we could stop and sa we were done, but our answer is not et in simplest form. There s a special trick we can appl when ends up in the exponent. Step five (onl when is in the exponent): Use the laws of exponents to put the constant out front. If ou re rust on the laws of 0.06t+ A = e exponents, the main point is 0.06t that a in an exponent can A = e e be moved to a c in front of 0.06t A = ce (define c = e.) the term containing the exponent. At this point, we ve solved the equation. But in fact, we have enough information in this problem to figure out the value of c as well. Since our initial balance (when t = 0) was $0,000: 0000 = ce 0000 = c() c = 0000 Therefore, A = 0000e 0.06t. This shouldn t be a surprising result, since we learned this formula back in 55 for solving compound interest problems. Now ou know where it comes from! A pair of values that allows us to solve for c for example, the pair t = 0, A = 0,000 in the previous example is called an initial value condition. Not all problems will have initial value conditions, however. Example 5: One more application: Newton s Law of ooling Three hundred ears ago, Isaac Newton discovered the Law of ooling, which gives a formula for the rate at which objects cool down to room temperature when removed from an oven. This law can also be used in marketing applications, and here is an example: A business normall enjos sales of $6000 per da. The owner decides to invest in a two-week series of television spots, and while those spots are running, her sales increase to $0000 per da. Three das after the advertising campaign ends, however, her sales decline to $8000 that da. What sales level should she expect seven das after the campaign ends? sales per da (in thousands of dollars) usual sales level Example das since end of ad campaign Here s one wa to visualize the problem: Her usual sales were at one level: $6000 per da (we ll call this usual level S). Advertising caused her sales to heat up to a higher level of $0000 per da. Once the advertising ends, those sales start to cool back down to their usual

5 level (see the graph). Newton s Law sas that the rate of cooling over time is directl proportional to the difference between the current sales level and S. In other words, = k( S) where k is the rate of cooling, is the current (temporar) sales level, S is the usual sales level, and t is the number of das since the end of the promotion. Note that k and S are constants in this problem; t and are our onl variables. Let s solve this differential equation. Once again, step one has alrea been done for us: Step two: Rearrange the terms of the equation so that all the terms containing are on the left and all the terms containing t are on the right: = k( S) = k S Step three: Find the antiderivative of both sides. Be sure to include on the right-hand side. ln( S) is an antiderivative of whenever S is a S constant. (You can check this b taking the derivative of the answer.) = S k ln( S) = kt Be careful! Don t write k on this side. k is a constant in this problem! Step four: Solve for. ln( S) = kt S = e kt+ = S + e kt+ Step five (onl when is in the exponent): Use the laws of exponents to put the constant out front. = S + e = S + ce kt+ Now we know what formula to use, = S + ce kt. Here are what the different letters in the formula stand for: kt

6 S represents her usual sales level, $6000 per da in this problem. Let s choose thousands of dollars as our unit to make the numbers smaller and set S = 6. c represents the amount of boost that the advertising created. Since the advertising temporaril raised her sales to $0,000 per da, that s a $4000 boost over her usual level, so c = 4. t and are the variables, representing the number of das since the end of the advertising campaign, and the amount of sales that da, respectivel. k represents the rate of cooling. We need to solve for k in order to finish this problem. To solve for k, we ll use the fact that on the third da (t = ), the sales were onl $8000 ( = 8). You will alwas need to use an intermediate value to solve for k the starting or ending sales levels won t work! ln = S + ce 8 = 6 + 4e = 4e = e k k = k kt k() ln k = 0. Now that we know what k is, we can answer the original question: What sales should she expect on the seventh da after the campaign ends? = S + ce = 6 + 4e kt (0.)(7) Therefore, her sales on the seventh da should be approximatel $6,794.

7 Differential Equations: Problems to turn in: Work through the example problems in the introduction before attempting these problems. Do not turn in the worked examples from the introduction. Also, do not hand in this document with our lab. Work all problems in order. Do not sa see attached or other notation referring the grader to another location in the lab for part of a problem. Label problems and parts of problems with appropriate numbers and/or letters. You ma write our answers neatl b hand for this project.. You are investing $50,000 in an interest-bearing account at 8% interest. Write a differential equation that models the balance in our account after t ears (Hint: See Example 4), and solve that equation completel.. Solve the following differential equations. You do not need to solve for in these problems, but ou do need to remove from the exponent when appropriate. In part c, ou have an initial value condition, which will allow ou to solve for the constant. a. x '= c. = 9x. For part c, the initial value condition b. x = is that when x =, = 0.. A locall owned grocer store has an average sales level of $8000 per da. One week the owner announced that he would split that week s profits with the local school sstem, and sales that week skrocketed to $0000 per da. However, four das after the promotion ended, sales were alrea back down to $000 per da. What level of sales should the owner expect ten das after the end of the promotion? (Use thousands of dollars for our units, as in the example).