Oregon State University Mathematics Department

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1 Oregon State University Mathematics Department Study Guide for Mth Applied Differential Equations to accompany the text Elementary Differential Equations by William F. Trench 150 y t written by Dr. Filix Maisch and Dr. Stephen Scarborough Summer/Fall

2 Contents 0 Preliminaries Schedule Introduction and Notes to Students Basic Integral Table Table of Elementary Laplace Transforms Lesson 1: Introduction to Differential Equations 9 Lesson 2: Direction Fields 13 Lesson 3: Linear First Order Equations, Use of Integrating Factors 15 Lesson 4: Separable Equations 20 Lesson 5: Existence and Uniqueness Theorems 22 Lesson 6: Transformations 23 Lesson 7: Exact Equations 25 Lesson 8: Autonomous First Order Differential Equations 27 Lesson 9: Applications 35 Lesson 10: Homogeneous Linear Equations 43 Lesson 11: Constant Coefficient Homogeneous Equations 44 Lesson 12: Euler Equations 47 Lesson 13: Nonhomogeneous Linear Equations 50 Lesson 14: The Method of Undetermined Coefficients 52 Lesson 15: Reduction of Order 56 Lesson 16: Variation of Parameters 59 Lesson 17: Spring Problems I 62 Lesson 18: Spring Problems II and the RLC Circuit 64 Lesson 19: Introduction to the Laplace Transform 70 2

3 Lesson 20: The Inverse Laplace Transform 73 Lesson 21: Solution of Initial Value Problems 75 Lesson 22: The Unit Step Function 77 Lesson 23: Const. Coeff. Equations with Piecewise Cont. Forcing Functions 79 Lesson 24: Convolution 81 Lesson 25: Constant Coefficient Equations with Impulses 83 Lesson 26: Euler s Method and Runge-Kutta Method 86 3

4 0 Preliminaries 0.1 Schedule Study Guide Textbook Suggested Number Lesson Section(s) Topic of Lectures 1 1.1,1.2 Introduction to 1 Differential Equations Direction Fields Linear First Order 1 (use of integrating factors) Separable Equations Existence and Uniqueness 1 Theorems Transformations ,2.6 Exact Equations 1 8 Autonomous Equations 1 (material not in text) 9 4.1,4.2,4.3 Applications Homogeneous Linear Equations Constant Coefficient 1 Homogeneous Equations Euler Equations Nonhomogeneous Linear Equations ,5.5 Undetermined Coefficients Reduction of Order 1 (homogeneous case only) Variation of Parameters ,6.2 Spring Problems I Spring Problems II and 1 the RLC Circuit Introduction to 1 Laplace Transforms The Inverse Laplace Transform Solution of IVPs The Unit Step Function Constant Coefficient Equations with 1 Piecewise Continuous Forcing Functions Convolution Constant Coefficient Equations 1 with Impulses Euler s Method and the 1 Range-Kutta Method Catch-up and Review 1 Exams 2 Number of Lectures 29 4

5 0.2 Introduction and Notes to Students Introduction: This study guide is set up for a ten week term with 29 lectures and 9 to 10 recitations. Please check with your instructor for a detailed syllabus for your section of the course. Prerequisites: It is expected that you are comfortable with calculus. This includes familiarity with material from previous courses, such as basic algebraic manipulations, including those involving fractions and exponents, as well as trigonometry, differentiation, integration, partial fractions, multivariable chain rule. The most common difficulty for students in this course is with old concepts such as algebra and basic calculus; if your skills are rusty, practice now, and get help early. Comments and suggestions on study habits: There is much new material presented in this course. It is very important to develop good study habits and to keep up with the course material right from the beginning of the course. Listed below are a number of things to watch out for. Time: It is very important to devote enough time on a daily basis for the course. You should plan on spending two to three hours going over the material and assignments for each hour of lecture. If you have seen some of this material before, it is easy to fall into the pattern of not devoting enough time to the material at the beginning of the course. When more difficult material is presented in a few weeks you then will find yourself too far behind. Algebra skills: The most common difficulty for students in this course is remembering the correct algebraic manipulations from pre-calculus courses. If you find yourself having difficulty, get help early in the course. Don t wait until it is too late. The material in the course is based on the assumption that you remember the algebra and trigonometry that you learned in your previous courses. Attendance: Many students make the mistake of thinking that they can learn all the required material by just working the homework problems without attending the lectures. Often, your lecturer will present an additional viewpoint on the material that is not presented in the text. The exams for the course are based on the text material and on the presentations in lecture. Make sure that you attend all the lectures. If for some reason you miss a lecture, get notes for the missed class from one of the other students in the course. Homework: Often, students think that they have mastered the material if they can get the answers in the back of the book. You should look at the answers in the back of the book only after you have completed the assignment. Relying on the solutions can give you a false sense of security. Your homework assignments will include problems that are not available through the online homework. Do not make the mistake of not doing these problems. They often cover important concepts that are difficult to practice in an online environment. Other Resources: The Mathematics Learning Center (MLC) provides drop-in help for all lower division mathematics courses. The MLC is located on the ground floor of Kidder Hall in room 108, and is normally open Monday through Thursday from 9 a.m. to 6 p.m. and on Fridays from 9 a.m. to 5 p.m., from the second week of the term through the Dead Week. The MLC also provides evening tutoring in the Valley Library, in general Sunday through Thursday from 6 p.m. till 9 p.m. Current hours can be found at the MLC homepage, 5

6 0.3 Basic Integral Table k is any nonzero constant 1. kf(x)dx = k f(x)dx 2. (f(x) + g(x)) dx = f(x)dx + 3. (f(x) g(x)) dx = f(x)dx g(x)dx g(x)dx 4. x p dx = xp+1 p C, p dx = ln x + C x 6. e kx dx = 1 k ekx + C 7. ln(x)dx = x ln(x) x + C 8. sin(kx)dx = 1 k cos(kx) + C 9. cos(kx)dx = 1 k sin(kx) + C 10. tan(kx)dx = 1 ln sec(kx) + C k 11. cot(kx)dx = 1 ln csc(kx) + C k 12. sec(kx)dx = 1 ln sec(kx) + tan(kx) + C k 13. csc(kx)dx = 1 ln csc(kx) + cot(kx) + C k 14. sec 2 (kx) dx = 1 k tan(kx) + C 15. csc 2 (kx)dx = 1 k cot(kx) + C 16. sec(kx) tan(kx)dx = 1 k sec(kx) + C 17. csc(kx) cot(kx)dx = 1 k csc(kx) + C dx 18. x 2 + k 2 = 1 ( x ) k tan 1 + C, k > 0 k dx ( 19. k 2 x dx = x ) 2 sin 1 + C, k > 0, x < k k dx 20. x x 2 k dx = 1 x 2 k sec 1 + C, k > 0, x > k k 6

7 0.4 Table of Elementary Laplace Transforms f(t) = L 1 {F (s)} F (s) = L{f(t)} 1 1 (n = 0, 1, 2,...) 1 s s > 0 2 t n (n = 0, 1, 2,...) n! s n+1 s > 0 3 t p (p > 1) Γ(p + 1) s p+1 s > 0 4 e at 1 s a s > a 5 t n e at (n = 1, 2, 3,...) n! (s a) n+1 s > a 6 cos (ωt) s s 2 + ω 2 s > 0 7 sin (ωt) ω s 2 + ω 2 s > 0 8 e λt cos (ωt) s λ (s λ) 2 + ω 2 s > λ 9 e λt sin (ωt) ω (s λ) 2 + ω 2 s > λ 10 cosh (bt) s s 2 b 2 s > b 11 sinh (bt) b s 2 b 2 s > b 12 t cos (ωt) s 2 ω 2 (s 2 + ω 2 ) 2 s > 0 7

8 13 t sin (ωt) 2ωs (s 2 + ω 2 ) 2 s > 0 14 sin (ωt) ωt cos (ωt) 2ω 3 (s 2 + ω 2 ) 2 s > 0 15 ωt sin (ωt) ω 3 s 2 (s 2 + ω 2 ) 2 s > sin (ωt) t arctan ( ω s ) s > 0 17 e at f(t) F (s a) 18 t k f(t) ( 1) k F (k) (s) 19 f(ωt) 1 ( s ) ω F ω ω > 0 20 u(t τ) e τs s s > 0 21 u(t τ)f(t τ) (τ > 0) e τs F (s) 22 t 0 f(τ)g(t τ) dτ F (s)g(s) 23 δ(t a) e as s > 0 24 f (t) sf (s) f(0) 25 f (t) s 2 F (s) sf(0) f (0) 26 f (n) (t) s n F (s) s n 1 f(0) s n 2 f (0) f (n 1) (0) 8

9 Lesson 1 Introduction to Differential Equations Sections 1.1 and 1.2 A common model assumes that the rate of growth of some quantity N is proportional to itself. dn dt = kn, where k is the proportional constant. To solve this we separate the variables: ( ) dn dt = k N ( ) dn dt N dt = k dt ln N = kt + C 1 N = e kt+c 1 N = e C 1 e kt N = C 2 e kt where C 2 = e C 1 Notice that with this model if N is ever 0 then it is always 0. So N is either positive or negative. In most models N will always be positive. C 2 if 0 N Let C = C 2 if N < 0 N = Ce kt If k > 0 this model is called exponential growth. If k < 0 this model is called exponential decay. 9

10 A differential equation is an equation that contains derivatives. Examples: dy dt = t2. In this example y is the dependent variable and t is the independent variable. dy dx = xy. 3. y + y = cos (t). 4. dv dt = g βv. Ordinary vs Partial A partial differential equation (p.d.e.) is a differential equation that contains partial derivatives. Examples: u 1. t = u a 2 is the heat equation. x2 2. u tt = au xx is the wave equation. Both of these equations have a wide range of applications. An ordinary differential equation (o.d.e.) is a differential equation that only contains ordinary derivatives. In this course we will restrict our studies to ode s. One of the major reasons to study ode s is to gain the background needed to understand and solve pde s. Systems Often differential equations make their appearance as systems. One famous example are the Lotka- Volterra equations that are used to model predator prey interactions in biology. x = number of hares (prey) y = number of lynx (predators) dx dt dy dt = ax αxy = cy + γxy We will not be studying systems of differential equations in this course. 10

11 Order The order of a differential equation is the order of the highest derivative that appears. Examples: dy dt = t2 dy dx = xy first order first order 3. y + y = cos (t) second order 4. dv dt = g βv first order Linear vs Nonlinear An n th order ordinary differential equation is linear if it can be written in the form: otherwise it is nonlinear. Examples: These are all linear dy dt = t2 dy dx = xy 3. y + y = cos (t) 4. x 2 y = 5xy 6y + sin (3x) a n (t)y (n) + a n 1 (t)y (n 1) + + a 1 (t)y + a 0 (t)y = g(t) 5. y + t 2 y cos (t) + y = sin (t) 6. y y = x These are all nonlinear 7. y y + ex = y 8. 5 dy dt + y = t + y2 A linear differential equation is homogeneous if g(t) = 0; otherwise it is nonhomogeneous. In the examples above 1., 3., 4., and 5. are nonhomogeneous, while 2. is homogeneous. 11

12 A solution of a differential equation is a function that solves the differential equation on some open interval. For example, the function y = e x2 2 is a solution to the differential equation dy = xy. To dx verify this just notice that dy dx = xe x 2 2 = xy. The function y = 5e x2 2 is also a solution. The function y = Ce x2 2 is a solution for any constant C. It can be shown that any solution to this differential equation is of the form y = Ce x2 2. We call this a general solution to the differential equation. To determine C we need a data point, called an initial condition. Example: Find a solution to dy = xy, y(2) = 7. dx y = Ce x2 2 y(2) = Ce 2 = 7 C = 7e 2 y = 7e 2 e x2 2 y = 7e x2 2 2 We call this type of problem an initial value problem, often referred to as an IVP. Suggested Homework: Note: The Assignment given by your instructor may be different. Read Sections 1.1 and 1.2, Pages 1-14 Section { 1a, 2a b c d e, 3a b c e, 4a } 12

13 Lesson 2 Direction Fields Section 1.3 Consider a differential equation of the form dy = f(x, y). At each point (x, y) we are given the dx value of the rate of change of y. We call f the rate function. A plot in which short line segments are drawn giving the slope at each point (x, y) is called a direction field or a slope field. As a practical matter, we draw these line segments at enough points to get a representation of the direction field. Here s a direction field plotted on R = {(x, y) 1 x 4, 1.5 y 4.5}: slope field for y x 2 y The graph of a solution to a differential equation is called an integral curve. The slope of an integral curve for dy dx = f(x, y) at (x 0, y 0 ) is f(x 0, y 0 ). 13

14 Here s a direction field for dy dx = x2 y with several integral curves plotted: y x In general plotting direction fields by hand is tedious at best. There are many applications and Java apps that can plot direction fields. Suggested Homework: Note: The Assignment given by your instructor may be different. Read Section 1.3, Pages Section { 1 - plot the integral curve where y(0) = 0, 2 - plot the integral curve where y(0) = 0.5, 3 - plot the integral curve where y(0) = 0.6, 4 - plot the integral curve where y(0) = 0, 7 - plot the integral curve where y(0) = 1, use an online tool ( such as ) to create direction fields for 12, 13, 13 } 14

15 Lesson 3 Linear First Order Equations, Use of Integrating Factors Section 1.3 Trench uses the method of variation of parameters to solve first-order linear differential equations. While this non-traditional method works, the traditional method of integrating factors is, in our opinion, simpler to learn and use. The following can be used to replace pages 31 through 35 in the textbook. Consider a first-order linear differential equation a 1 (x) dy dx + a 0(x)y = g(x). By dividing through by a 1 (x) we get the standard form: dy + p(x)y = f(x). dx ( ) The left hand side almost looks like the product rule applied to y(x) and some function µ(x): d dx [µy] = µy + µ y. Multiplying both sides of ( ) by µ yields µy + µpy = µf. Matching the left hand side to the form of the product rule for µy, we want a function µ such that µ = µp. From here we get µ µ = p, and by integrating both sides ln µ = p dx. So the possibilities for µ are These are called integrating factors. µ = ±e p dx. 15

16 Since we only need one we will always choose µ = e p dx and set the constant of integration to 0. With this choice µy + µpy = µf becomes [ye p dx ] = fe p dx. From here we can integrate both sides and solve for y. Examples 1. y + 3 x y = 4 x 5. We can use an integrating factor of µ(x) = e 3 x dx = e 3 ln x = ±x 3. By multiplying both sides of y + 3 x y = 4 x 5 by µ = x3 we get x 3 y + 3x 2 y = 4 x 2. This equation can be rewritten as (x 3 y) = 4 x 2. By integrating both sides, x 3 y = 4 x + C. So y = C x 3 4 is the general solution. x4 2. xy 3y = x 4 e x. First we rewrite this in the standard form to get y 3 x y = x3 e x. So an integrating factor is µ(x) = e 3 x dx = e 3 ln x = ±x 3. Notice the integrand s sign here. A common mistake is to leave it behind. By multiplying both sides of y 3 x y = x3 e x by µ = x 3 we get x 3 y 3x 4 y = e x. This equation can be rewritten as (x 3 y) = e x. By integrating both sides, x 3 y = e x + C. So y = x 3 e x + Cx 3 is the general solution. 16

17 3. y + 2xy = x 3. An integrating factor is µ = e 2x dx = e x2. By multiplying both sides of y + 2xy = x 3 by µ we get e x2 y + 2xe x2 y = x 3 e x2. This equation can be rewritten as (e x2 y) = x 3 e x2. By integration by parts, Hence y = 1 2 (x2 1) + Ce x2 e x2 y = 1 2 ex2 (x 2 1) + C. is the general solution. 4. y cos (t) + y sin (t) = 1, y(0) = 2. First we rewrite this in the standard form to get y + y tan (t) = sec (t). So an integrating factor is e tan (t) dt = e ln sec (t) = ± sec (t). By multiplying both sides of y + y tan (t) = sec (t) by µ = sec (t) we get sec (t)y + sec (t) tan (t)y = sec 2 (t). This equation can be rewritten as (sec (t)y) = sec 2 (t). By integrating both sides, sec (t)y = tan (t) + C. Then y = sin (t) + C cos (t) is the general solution. Imposing the initial condition of y(0) = 2 yields C = 2 and a solution of y = sin (t) + 2 cos (t). 17

18 Sometimes you have to leave your answer in an integral form and use numerical integration. 5. y 2xy = 1, y(1) = 2. An integrating factor is µ = e 2x dx = e x2. By multiplying both sides by µ we get e x2 y 2xe x2 y = e x2 This equation can be rewritten as (e x2 y) = e x2. But the integral of e x2 does not have an expression in terms of elementary functions. Looking at the initial condition of y(1) = 2 we will use the integral function F (x) = antiderivative of f(x) = e x2. So we may write that e x2 y = x 1 e t2 dt + C. Then the general solution is x y = e x2 e t2 dt + Ce x2. 1 x 1 e t2 dt as an Imposing the initial condition of y(1) = 2 yields C = 2/e. So the solution is x y = e x2 e t2 dt + 2e x Try doing the homework for this section using the method of integrating factor instead of the method of variation of parameters shown in the text. 18

19 Existence and Uniqueness Theorem for Linear First Order Differential Equations: Suppose that p and f are continuous on an open interval (a, b) that contains the point x 0, then the initial value problem: y + p (x) y = f (x), y(x 0 ) = y 0, has a unique solution that is valid on the interval (a, b). The proof for this result can be found in the textbook at the end of section 2.1. Example: What is the largest interval on which the initial value problem (x + 1) dy + xy = tan x, y (0) = 7, dx is guaranteed a unique solution? The first step is to put the differential equation into standard form so that p and f can be identified. dy dx + x x + 1 y = tan x x + 1 Then p (x) = x x + 1 and f (x) = tan x x + 1. We see that p has continuity problems at x = 1 and f is not continuous at x = 1 and x = π 2 + nπ, n = 0, ±1, ±2, ±3,. The largest interval on which both p and f are continuous that contains 0 is ( 1, π 2 ). This is the largest interval on which we are guaranteed a unique solution. Suggested Homework: Note: The Assignment given by your instructor may be different. Section 2.1 { 1, 2, 5, 6, 10, 16, 18, 19, 20, 21, 31, 33, 38 } 19

20 Lesson 4 Separable Equations Section 2.2 A first order differential equation is separable if the rate function can be factored into pure singlevariable functions of the independent variable and the dependent variable as shown below. dy = g (x) h (y) dx To solve a separable differentiable equation, separate and integrate. 1 dy h (y) dx dx = g (x) dx 1 h (y) dy = g (x) dx Examples of separable differential equations include: y = y dy dx = x y dy = t 1 + ty y = (t 1) + y (t 1) = (t 1) (1 + y) dt Example: Consider the non-linear first order homogeneous (since every term involves y or a derivative of y) separable o.d.e. (x 2 + 1)y y = 0. Find the general solution, and then solve the IVP with the initial condition y(π/4) = 0. (x 2 + 1) dy dx = y 1 dy y dx = x 2 1 dy y dx dx = x 2 dx dy y = tan 1 (x) + C 2 y = tan 1 (x) + C 20

21 y = ( 0.5 tan 1 (x) + c ) 2 Let s solve the IVP with y(π/4) = 0. 0 = (0.5 + c) 2 implies c = 0.5 so y = ( 0.5 tan 1 (x) 0.5 ) 2. When you read the textbook, please play special attention to the section on page 51, Differences Between Linear and Nonlinear Equations. Suggested Homework: Note: The Assignment given by your instructor may be different. Read Section 2.2, Pages Section 2.2 { 1, 2, 3, 11, 17, 18, 19, 21, 26 } 21

22 Lesson 5 Existence and Uniqueness Theorems Section 2.3 Consider the following initial value problem: y = y, y (0) = 0. It is easy to see that y = 0 is a solution. However, if we did not notice this and proceeded to solve this separable differential equation we see that: y 1 2 y = 1 2 y = t + C Apply the initial condition to see that C = 0. So 2 y = t or y = t2 4 is also a solution. We have found two different solutions to this initial value problem. The first thing to notice is that this differential equation is not linear. The existence and uniqueness theorem for nonlinear equations is not as strong or useful as the one for linear equations. The proof for this theorem is beyond the scope of this course. Please see the textbook (section 2.3) for a statement of this result and some examples on applying this theorem. Suggested Homework: Note: The Assignment given by your instructor may be different. Read Section 2.3, Pages Section 2.3 { 1, 2, 3, 17, 19 } 22

23 Lesson 6 Transformations Section 2.4 A first order differential equation of the form y + p(x)y = f(x)y r is called a Bernoulli equation. For r = 0 or r = 1 the differential equation is linear, so we will restrict our attention to the case where r 0 or 1. These can be reduced to a linear equation by the substitution u = y 1 r. Example: xy 5y = x 3 y 3. We can reduce this Bernoulli equation to a linear equation by the substitution u = y 1 3 = y 2 : Then u = 2y 3 y which gives y = 1 2 y3 u. Substituting into the differential equation gives 1 2 xy3 u 5y = x 3 y 3. This simplifies to xu + 10y 2 = 2x 3. Then recalling that u = y 2 we see that we have a standard first order linear differential equation: xu + 10u = 2x 3. This can be solved using integrating factors. Please verify that the solution is Then u = 2 13 x3 + C 1 x 10. y 2 = C 1 x x3. Which simplifies to y 2 = 13x10 C 2x 13 where C = 13C 1. 23

24 Another type of first order differential equation that can be solved by use of a transformation is a so-called nonlinear homogeneous equation. A nonlinear first order differential equation dy = f (x, y) is homogeneous if f(x, y) only depends dx on the ratio y dy x, which means dx = f ( y x). The substitution u = y x will turn such homogeneous differential equations into separable differential equations. We can see this since y = ux and by the product rule dy dx = u + xdu dx. The differential equation them becomes which is separable. u + x du dx = f (u). Suggested Homework: Note: The Assignment given by your instructor may be different. Read Section 2.3, Pages and Section 2.4, Pages Section 2.4 { 1, 2, 7, 9, 15, 16, 22, 23 } 24

25 Lesson 7 Exact Equations Section 2.5 A convenient way to write a first order o.d.e. is with differentials: M(x, y) dx + N(x, y) dy = 0. This is equivalent to M(x, y) + N(x, y)y = 0 for y is a function of x. We will say that F (x, y) = c is an implicit solution of M(x, y) dx + N(x, y) dy = 0 if every differentiable function y(x) such that F (x, y(x)) = c is a solution of M(x, y) + N(x, y)y = 0. By the Chain Rule F (x, y) = c is an implicit solution of F x dx + F y dy = 0 where F x, F y denote the x, y partials respectively. This motivates the following important definition: M(x, y) dx + N(x, y) dy = 0 is exact on an open rectangle R if there is a function F (x, y) such that F x, F y are continuous on R, and F x (x, y) = M(x, y) and F y (x, y) = N(x, y) for all (x, y) in R. It turns out that M(x, y) dx+n(x, y) dy = 0 is exact on an open rectangle R if and only if M y = N x (assuming M y, N x are continuous on R). Finding an implicit solution for an exact equation M(x, y) dx + N(x, y) dy = 0 is straightforward: Integrate M with respect to x, producing a constant of integration that can be a function of y. Then take the y-partial of the result to determine the constant of integration. Note: One can do the above with the roles of M and N switched and the roles of x and y switched. This gives F (x, y) and an implicit solution of F (x, y) = c. If possible, find a solution y(x) from F (x, y) = c. 25

26 Examples: 1. (y 3 1)e x dx + 3y 2 (e x + 1) dy = 0. ( (y 3 1)e x) = 3y 2 e x and ( 3y 2 (e x + 1) ) = 3y 2 e x. y x So this equation is exact. Integrating (y 3 1)e x with respect to x yields F (x, y) = (y 3 1)e x + g(y). Then F y = 3y 2 e x + g (y). We want g (y) = 3y 2. So set g(y) = y 3. Thus, (y 3 1)e x + y 3 = c is an implicit solution. We can take this one further! e This implicit solution is equivalent to y 3 (e x + 1) = e x + c and therefore, y = x + c 3 e x + 1 explicit general solution. is the 2. 3x 2 y 2 dx + 4x 3 y dy = 0. The y-partial of 3x 2 y 2 is 6x 2 y and the x-partial of 4x 3 y is 12x 2 y. Hence there is no open rectangle on which this equation is exact. Now suppose M(x, y) dx + N(x, y) dy = 0 is not exact, but can be made exact by multiplying both sides by some function µ(x, y). Such a function is called an integrating factor. Example: Consider the o.d.e. (3xy + 2y + 2) dx + (x 2 + x) dy = 0. The reader is left to check that this equation isn t exact. Multiplying both sides by µ(x, y) = x we get (3x 2 y + 2xy + 2x) dx + (x 3 + x 2 ) dy = 0. The reader is left to check that this equation is exact. Now we can solve as before: Taking the integral of x 3 + x 2 with respect to y yields x 3 y + x 2 y + g(x). Taking the x-partial and equating with 3x 2 y + 2xy + 2x implies that we may set g(x) = x 2. So an implicit solution is x 3 y + x 2 y + x 2 = c which we can solve for y to get the explicit solution: y = c x2 x 3 + x 2. Suggested Homework: Note: The Assignment given by your instructor may be different. Read Section 2.5, Pages Section 2.5 { 1 9, 18, 19 } 26

27 Lesson 8 Autonomous First Order Differential Equations Not in textbook A first order differential equation of the form dy dt = f(y) is said to be autonomous. The rate function f does not depend on the independent variable t. Here are a few useful properties of first order autonomous differential equations 1. They are separable. 2. The slopes in the direction field only depend on y. 3. The general solution is invariant under horizontal translations. Example: Exponential growth or decay. Let q(t) be the quantity of something. This common model assumes that the rate of growth in q is proportional to itself: dq dt = rq. Examples include population growth (r > 0) and radioactive decay (r < 0). As we have already seen, with an initial condition q(t 0 ) = q 0, we have q(t) = q 0 e r(t t 0). 27

28 Example: Population dynamics. Let p(t) represent the size of a population at time t. When resources are restricted we can employ a competition factor, bp 2, so that where K = r b. dp ( dt = rp bp2 = rp 1 p ), K This equation is called the logistic equation. This differential equation has equilibrium solutions at y = 0 and y = K. We call K the carrying capacity and r the intrinsic growth rate. Let s determine the stability of these equilibrium values. Assume r > 0 and K > 0 (negative values would make no sense for this model). To do this we plot dp dt dp versus p for (1 dt = rp p ) : K p t K p t The graph is a parabola with x-intercepts at x = 0 and x = K. 28

29 Note: If 0 < p < K then dp dp > 0 and p is increasing. If p > K then < 0 and p is decreasing. So if p is dt dt close to K then p is pushed towards K. A similar analysis shows that if p is close to 0 it will be pushed away from 0. (While we get dp dt < 0 when p < 0, it s not relevant to the model as p cannot be negative.) We can use arrows to indicate how p is pushed. Here it is with the logistic equation: p t 0 K p t The solution p = K is an asymptotically stable equilibrium solution, since small perturbations (changes) are pushed back towards K. The solution p = 0 is an unstable solution, since any perturbations will push p away from 0. With the arrows, the p-axis is called the phase line. 29

30 More Examples: Find and classify the equilibrium solutions of dy dt = (1 y)(2 y). y t 1 2 y t The equilibrium solutions are y = 1 and y = 2. Since y is positive outside of the interval [1, 2] and negative inside (1, 2), y = 1 is an asymptotically stable solution and y = 2 is an unstable solution Find and classify the equilibrium solutions of dy dt = y(2 y)2. y t 0 2 y t The equilibrium solutions are y = 0 and y = 2. y = 0 is an unstable solution and y = 2 is a semistable equilibrium solution since it is stable from one side (in this case from below) while unstable from the other side (in this case from above). 30

31 Find and classify the equilibrium solutions of dy dt = sin (y). y t -Π 0 Π 2Π 3Π y t Unstable equilibria exist at t = 0, ±2π, ±4π,... Stable equilibria exist at t = ±π, ±3π,... Let s return to and solve the first order logistic equation dy (1 dt = ry y ) for y, t > 0 where K, r K are positive constants. ( y = ry 1 y ) K y ( y 1 y ) = r K 1 ( y 1 y ) y dt = K r dt ( 1 K 1 y K + 1 y ) dy = rt + C 1 ln 1 y + ln y = rt + C 1 K ln y 1 y = rt + C 1 K y 1 y K = C 2 e rt (where C 2 = ±e C 1 ) 31

32 1 y K y = C 3 e rt (where C 3 = 1/C 2 ) 1 y 1 K = C 3e rt 1 y = Ce rt + 1 K (where C = KC 3 ) y = K 1 + Ce rt. Next we use an initial condition y(0) = y 0 0 to find C. If y(0) = 0 then y = 0. By setting y 0 = K 1 + C we have C = K y 0 y 0. So the solution to the IVP dy (1 dt = ry y ), y(0) = y 0 is K y = y 0 K y 0 + (K y 0 )e rt. Let s compare solutions when K = 100 and r = for initial values y(0) = 25 and y(0) = 125. Here are the graphs (shown within the slope field): 150 y t

33 References: Elementary Differential Equations, 9th ed., by Boyce, DiPrima (Pages ) Suggested Homework: Note: The Assignment given by your instructor may be different. Find and classify the equilibrium solutions for the differential equations (1) through (6): 1. y = 3 y 2. y = 5 + y 3. y = y 2 (3 y) 4. y = y 2 5y y = (1 y)(y 2)(y 3) 6. y = y y 3 7. Show that for the logistic equation, dy (1 dt = ry y ), that the steepest ascent occurs at K y = K Let r, K be positive constants. Consider the IVP dy dt = ry ln ( ) K, y(0) = y 0. y (a) Find and classify the equilibrium solutions. (b) Solve for y. 33

34 Answers to selected exercises: 1. y = 3; stable equilibrium. 3. y = 0; semistable equilibrium. y = 3; stable equilibrium. 5. y = 1; stable equilibrium. y = 2; unstable equilibrium. y = 3 stable equilibrium. 34

35 Lesson 9 Applications Section 4.1, 4.2, 4.3 We start with the classic exponential model: y (t) = ky, y(t 0 ) = y 0. The general solution (without an initial value) is y(t) = ce kt. If k > 0 the model is called exponential growth and if k < 0 the model is called exponential decay. The constant k is called a growth constant and decay constant respectively. Applying the initial value, y 0 = ce kt 0, so c = y 0 e kt 0. The initial value problem has the solution y = y 0 e k(t t 0). We assume that you have seen examples (in algebra and calculus) of exponential growth and decay, such as least population growth, compound interest, and radioactive decay. Let s consider a variation on the exponential model: Suppose that a radioactive substance has decay constant k > 0 (reported as a positive number, so use y = ky). At the same time the substance is being produced at a constant rate of a units of mass per unit time. Let y(t) be the mass at time t and y(0) = y 0 be the initial mass. Let s derive an IVP and solve it. Then we can take the limit of y(t) as t to find the equilibrium solution. First the IVP: y (t) = rate of increase in y(t) rate of decrease in y(t) = a ky(t), y(0) = y 0. Since y (t) = a ky is separable let s solve it by separating and integrating: 1 a ky y = 1 1 k ln a ky = t + c 1 a ky = ce kt y = a k + ce kt. Imposing the initial condition we get: y 0 = a k + c c = y 0 a k. So the solution to the IVP is y = a ( k + y 0 a ) e kt. k The equilibrium solution is y = a k. In particular, if y 0 > a k and if y 0 < a then y increases towards this solution. k then y decreases toward this solution, 35

36 Example: An immortal alchemist creates bitcoin at the rate of 1 bitcoin per day. Going unnoticed, a cyber thief has setup a script that is continuously stealing the aclchemist s bitcoin at a rate of 2.5 percent of whatever is there. What is the number of bitcoins b(t) the alchemist has at time t? Assume b(0) = 1. How many bitcoins will the alchemist have in the long run (as t )? b (t) = rate of increase in bitcoin rate of decrease in bitcoin. So b = b. Using a = 1, k = 0.025, and b 0 = 1 we get b(t) = 40 39e 0.03t. In the long run the alchemist will have 40 bitcoins. Newton s Law of Cooling: If an object of temperature T (t) at time t is in a medium of temperature T m (t) at time t then the rate of change in T (t) is proportional to the T = T (t) T m (t). When T (t) > T m (t) we have T (t) < 0 and when T (t) < T m (t) we have T (t) > 0. So T (t) satisfies T = k(t T m ) where k > 0. Note: You don t need to convert temperatures to Kelvin units as the model is about relative temperatures. Example: Suppose a cup of coffee brewed at 150 degrees Fahrenheit is cooling in a room of constant temperature 70 degrees Fahrenheit. After 10 minutes the temperature of the coffee is 100 degrees Fahrenheit. How hot is the temperature after 20 minutes? First we setup the IVP: T = k(t 70), T (0) = 150. Then ln T 70 = kt + c 1 implies T 70 = ce kt and hence T = 70 + ce kt. By imposing the initial condition, T (t) = e kt. Using T (10) = 100 we get that 100 = e 10k e k = 0.1 ln (3/8) = ln (8/3)/10. Then after 20 minutes the cup of coffee is at T (20) = e 2 ln (8/3) = (9/64) = degrees Fahrenheit. 36

37 The following type of problem is usually referred to as a mixing problem (technically we have already seen an example) Example: A tank initially contains 10 kg of salt dissolved in 200 L of water. Brine that contains 0.5 kg of salt per liter is pumped into the tank at 5 liters per minute and at the same time water is drained from the tank at 5 liters per minute. Assume the tank is always kept uniformly mixed. Calculate the amount of salt in the tank in the long run (the equilibrium solution as t ). Let s pose this as an IVP: Let y(t) be the amount of salt in kg in the tank (which always contains 200 L of water). Then y = rate of salt in rate of salt out. So ( 0.5 kg y = L ) ( ) 5 L min ( y(t) kg 200 L ) ( ) 5 L, y(0) = 10. min Again, we could choose to apply the formula derived in general on the first page of this lesson, however it s good practice to solve y = 2.5 y/40 as separable equation: y/40 y = 1 40 ln 2.5 y/40 = t + c y/40 = ce t/40 y = ce t/40. Imposing the initial condition yields c = 90, so y = e t/40. Thus as t, y 100. So in the long run the amount of salt in the tank is 100 kg. 37

38 Here another mixing problem example with a wrinkle that was not present in the last one: A tank of maximum capacity 1, 000 gallons initially contains 10 lb of salt dissolved in 500 gallons of water. Brine that contains 0.5 lb of salt per gallon is pumped into the tank at 10 gallons per minute and at the same time water is drained from the tank at 5 gallons per minute. Assume the tank is always kept uniformly mixed. Calculate the amount of salt in the tank the moment it begins to overflow. The change here is the amount of water in the tank at time t isn t constant: V (t) = t is the volume of the tank in gallons after t minutes where V (0) = 500. We still set it up like before: Let y(t) be the amount of salt in the tank in lb at time t, where y(0) = 10. Then y = rate of salt in rate of salt out. So ( 0.5 lb y = gal ) ( ) ( 10 gal y(t) lb min t gal ) ( ) 5 gal, y(0) = 10. min So we have to solve the following IVP: y = 5 1 y, y(0) = t Let s rewrite the o.d.e. as y 1 + y = 5. An integrating factor is found by computing t 1 µ = e 100+t dt = e ln 100+t = ±(100 + t). Let s multiply both sides of the o.d.e. by t: (100 + t)y + y = 5(100 + t). This becomes ((100 + t)y) = 5(100 + t). Integrating both sides yields (100 + t)y = 5(100t + 0.5t 2 ) + c y = 500t + 2.5t2 + c t By imposing the initial condition y(0) = 10 we get 10 = c, so c = 1, 000. So 100 y = 1, t + 2.5t t Now to find the amount of salt at the moment the tank overflows: Set V (t) = t = 1, 000 to get t = 100 minutes. Then y(100) = 380 lbs of salt when the tank overflows. 38

39 Let s look at motion with resistance. We assume an object is traveling vertically through a medium (air or water for example) and that the only forces acting on the object are a constant gravitational force pulling the object down and the resistance of the medium (assumed to be directly proportional to the speed of the object). Suppose an object of mass m > 0 moves through vertically a medium under a constant downward gravitational force F g = mg (where g is the constant acceleration due to gravity) and the medium exerts a force of F med = k v, k > 0, in the opposite direction of the motion, where v is the velocity of the object. Let s derive a model for this scenario: Let v be the vertical velocity of an object (where moving up corresponds to a positive velocity). Newton s second law of motion asserts that F net = ma where a = v. Suppose the object is moving up. Then the resistance acts downward: F med = k v = kv. Now suppose the object is moving down. Then the resistance acts upward: F med = k v = k( v) = kv. Either way, the net force on the object is and F net = mg kv. So we get the following first order linear o.d.e. mv = mg kv. This can be expressed as v + k m v = g. While this is separable, we can solve it quickly by finding an integrating factor: µ = e (k/m)t. 39

40 e (k/m)t v + k m = ge(k/m)t (e (k/m)t v) = ge (k/m)t e (k/m)t v = mg k e(k/m)t + c. Hence v(t) = mg k + ce (k/m)t. (Exercise left for the reader: Solve it as a separable differential equation.) Finally, we see that lim v(t) = mg t k velocity of the object in the medium. is the equilibrium solution, which is called the terminal Given an initial value v 0 for the velocity, we get c = v 0 + mg k. So the solution to the IVP v + k m v = g, v(0) = v 0 is v(t) = mg k + ( v 0 + mg k ) e (k/m)t. Example: A 4 kg object is launched vertically into the air at 40 meters per second near the surface of the Earth (though we assume it s far enough up that the object will get close to its terminal velocity during the fall). Suppose the air resists motion with a force of 0.49 Newtons for each m/s of speed. Using g = 9.8 meters per square second, find a model for the objects vertical velocity and find the terminal velocity. We can just plug-in to the IVP solution we just derived: v(t) = 4(9.8) ( (9.8) 0.49 ) e (0.49/4)t. So v(t) = e 49t/400. The terminal velocity is 80. Meaning that in the end the object falls at a constant rate of 80 meters per second. 40

41 (Note to the instructor: Coverage of motion with resistance (previous pages) is typically included in the course, but coverage of escape velocity (pages that follow) is often skipped, and it is suggested that you do so if pressed for time.) More realistic models don t assume that the force due to gravity is a constant. Rather they assume that the force due to gravity that a massive object exerts on another object is inversely proportional to the square of the distance from the center of the massive object (assumed to be a sphere) to the other object. This is an example of an inverse square law. Assume a spacecraft launches from the surface of the Earth and exhausts its fuel as it reaches a height h that is sufficiently large that atmospheric resistance is negligible and can be assumed to be effectively 0. Let t = 0 be this time (known as burnout ) and consider that the force of gravity on the spacecraft at altitude y h is c F g = where c is a constant R is the radius of the Earth. (R + y) 2 Using F g = mg when y = 0 where g is the acceleration due to gravity on the surface of the Earth, we get mg = c R 2, so c = mgr 2. Thus F g = mgr2 (R + y) 2. Since F g is assumed to be the only force acting on the spacecraft for t 0, F g = ma = my = mgr2 (R + y) 2 y = gr2 (R + y) 2. This equation is second order, but we can transform it into a first order o.d.e. involving velocities: Let v(y) be the vertical velocity at altitude y. By the chain rule, d2 y dt 2 = dv dt = dv dy dy dt = v dv dy. Now we solve the o.d.e. v dv dy = gr2 (R + y) 2. This equation is separable, and integrating both sides yields 0.5v 2 = gr2 R + y + c v = 41 2gR 2 R + y + c.

42 Now suppose we have an initial velocity of v(h) = v 0. Then v 2 0 = 2gR2 R + h + c. So c = v 2 0 2gR2 R + h and hence 2gR v = 2 R + y + v2 0 2gR2 R + h. If v 0 2gR 2 R + h then v 0 for all y h and the spacecraft continues on up into space! The expression v e = 2gR 2 R + h is called escape velocity. We will show that if v 0 < v e the spacecraft falls back to Earth: 2gR 2 If v 0 < v e then v = R + y + v2 0 2gR2 = 0 for y such that R + h 2gR 2 R + y + v2 0 2gR2 R + h = 0, which occurs at y = 2gR2 h + R(R + h)v 2 0 2gR 2 (R + h)v 2 0 h since 2gR 2 R + y = 2gR2 R + h v2 0 2gR2 R + h. Suggested Homework: Note: The Assignment given by your instructor may be different. Read Sections 4.1, Pages , 4.2, Pages , 4.3, Pages Section 4.1 { 1, 2, 6, 7, 11, 14, 16, 17, 18 } Section 4.2 { 1, 2, 4, 6, 7, 8, 9, 11, 14 } Section { 1, 2, 10 } 42

43 Lesson 10 Homogeneous Linear Equations Section 5.1 In this lesson be begin the study of linear second order differential equations. differential equation y = F ( y, y, t ) A second order is linear if it can be written in the form: a 2 (t) y + a 1 (t) y + a 0 (t) y = g (t). We can divide through by a 2 (t) to put the differential equation into standard form y + p (t) y + q (t) y = f (t). If g (t) = 0 the differential equation is homogeneous, otherwise it is nonhomogeneous. We call the function f a forcing function. In this lesson we will cover the existence and uniqueness theorem, the concepts of linear combinations, linear independence, linear dependence, trivial solution, fundamental set of solutions, and the general solution. Make sure you thoroughly understand these concepts from the reading. We will also learn about the Wronskian, a tool that is useful for testing for linear independence and has many other applications. A useful theorem related to the Wronskian is Abel s Theorem (Theorem in the textbook.) Suggested Homework: Read Section 5.1, Pages Section 5.1 { 1, 2, 3, 5, 7, 24, 37, 38, 39 } 43

44 Lesson 11 Constant Coefficient Homogeneous Equations and Euler Differential Equations Section 5.2 Let s consider some very simple, but important linear second order differential equations. If a, b, c are real numbers and a 0 then the linear second order o.d.e. ay + by + cy = f(x) is said to be a constant coefficient equation. If f(x) = 0 it is a homogeneous constant coefficient equation. Otherwise it is non-homogeneous. Suppose we are trying to find the general solution to the linear homogeneous constant coefficient second order o.d.e. ay + by + cy = 0. The quadratic p(r) = ar 2 + br + c is called the characteristic polynomial of ay + by + c = 0. The equation p(r) = 0 is the characteristic equation. By the quadratic formula we have the the roots of the characteristic polynomial are r = b ± b 2 4ac. 2a So we consider three cases: Case 1: b 2 4ac > 0. In this case the characteristic polynomial has two distinct real roots. Let r 1 r 2 be the real roots. Then y 1 = e r 1x and y 2 = e r 2x are both solutions of ay + by + cy = 0. The reader is left to verify this. Since y 2 /y 1 = e (r 2 r 1 )x is non-constant as r 2 r 1 0 we have that y 1, y 2 are linearly independent. Thus the general solution to ay +by +c = 0 is y = c 1 e r 1x + c 2 e r 2x. Case 2: b 2 4ac = 0. In this case the characteristic polynomial has one repeated real root. Let r = b/(2a) be the repeated root. Then y 1 = e rx and y 2 = xe rx are solutions of ay +by +cy = 0. The reader is left to verify this. Moreover, y 2 /y 1 = x is non-constant, so y 1, y 2 are linearly independent (on any interval). Hence the general solution of ay + by + cy = 0 is y = c 1 e r 1x + c 2 xe r 1x = (c 1 + c 2 x)e r 1x. 44

45 Case 3: b 2 4ac < 0. Then the roots of the characteristic polynomial are complex conjugates λ ± ωi, where λ, ω are real numbers and ω 0. Then y 1 = e λx cos (ωx) and y 2 = e λx sin (ωx) are solutions of ay + by + cy = 0. The reader is left to verify this. Since y 2 /y 1 = tan (ωx) is non-constant y 1, y 2 are linearly independent, and hence the general solution to ay + by + cy = 0 is y(t) = c 1 e λx cos (ωx) + c 2 e λx sin (ωx) = (c 1 cos (ωx) + c 2 sin (ωx))e λx. Examples: 1. Find the general solution of y + 5y 14y = 0. The characteristic polynomial is r 2 + 5r 14 = (r + 7)(r 2). The roots are r = 7, 2. So the general solution is y = c 1 e 7x + c 2 e 2x. 2. Solve the IVP y 6y + 9y = 0, y(0) =1, y (0) = 5. The characteristic polynomial is r 2 6r + 9 = (r 3) 2. The repeated root is r = 3. So the general solution is y = c 1 e 3x + c 2 xe 3x. Then y = 3c 1 e 3x + c 2 e 3x + 3c 2 xe 3x. So imposing the initial conditions yields c 1 = 1 and 3c 1 + c 2 = 5 c 1 = 1 and c 2 = 2. So the solution is y = e 3x + 2xe 3x. 3. Solve the IVP y + 4y = 0, y(0) = 1, y (0) = 2. The characteristic polynomial is r 2 +4, which has complex roots r = ±2i. Since ±2i = 0±2i, the general solution is y = c 1 cos (2x) + c 2 sin (2x). Then y = 2c 1 sin (2x) + 2c 2 cos (2x). By imposing the initial conditions we get c 1 = 1 and 2c 2 = 2, so c 1 = 1 and c 2 = 1. So the solution is y = cos (2x) + sin (2x). 45

46 Suggested Homework: Note: The Assignment given by your instructor may be different. Read Section 5.2, Pages Section { 1-12, 13, 14, 17, 22, 23, 24, 25 } 46

47 Lesson 12 Euler Differential Equations Section 7.4 A second order differential equation that can be written in the form ax 2 d2 y dy + bx + cy = g (x) dx2 dx where a, b, and c are real constants with a 0 is called an Euler equation. By the existence and uniqueness theorem we see that if g is continuous everywhere, except possibly at 0 that an Euler equation has solutions defined on (, 0) and (0, ). To see this put the differential equation into standard form and observe that we have a discontinuity at 0 for both p (x) and q (x). We will restrict our attention to the interval (0, ). To find the general solution of the homogeneous Euler equation ax 2 d2 y dy + bx dx2 dx + cy = 0 we will assume that there is a solution of the form y = x m. Then y = mx m 1 and y = m (m 1) x m 2. Substituting into the differential equation: ax 2 m (m 1) x m 2 + bxmx m 1 + cx m = 0 am (m 1) x m + bmx m + cx m = 0 Since we have x > 0 we can divide by x m to get the characteristic equation am (m 1) + bm + c = 0 am 2 + (b a) m + c = 0 As with constant coefficient equations we have three cases. Case 1: Two real solutions, m 1 and m 2. The general solution is y = c 1 x m 1 + c 2 x m 2. 47

48 Example: Find the general solution of 6x 2 y + 17xy + 3y = 0 on (0, ). The characteristic equation is 6m 2 + (17 6) m + 3m = 0 6m m + 3m = 0 (3m + 1) (2m + 3) = 0 Then y 1 = x 1 3 and y 2 = x 3 2 and the general solution is y = c 1 x c 2 x 3 2 Case 2: One real solution, m. One solution is y 1 = x m. Using the technique reduction of order which is covered in lesson 15 you can show that a second linearly independent solution is given by y 2 = x m ln x. The general solution is then y = c 1 x m + c 2 x m ln x. Example: Find the general solution of 4x 2 y + 8xy + y = 0 on (0, ). The characteristic equation is 4m (m 1) + 8m + 1 = 0 4m 2 + 4m + 1 = 0 (2m + 1) 2 = 0 Then y 1 = x 1 2 and y 2 = x 1 2 ln x. The general solution is y = c 1 + c 2 ln x x. 48

49 Case 3: Two complex solutions m 1 = α + iβ and m 2 = α iβ. y = x α+iβ y = x α x iβ To deal with x iβ we use Euler s identity: x iβ = e iβ ln x = cos (β ln x) + i sin (β ln x) From this we can see that the general solution is given by. y = x α (c 1 cos (β ln x) + c 2 sin (β ln x)) Example: Find the general solution for x 2 y + 3xy + 10y = 0 on (0, ). The characteristic equation is m 2 + 2m + 10 = 0. From the quadratic formula m = 2 ± (10) 2 = 1 ± 3i So the general solution is then given by y = c 1 cos (3 ln x) + c 2 sin (3 ln x) x Suggested Homework: Note: The Assignment given by your instructor may be different. Read Section 7.4, Pages 324 (Start with EULER EQUATIONS) Section { 1-10 } 49

50 Lesson 13 Nonhomogeneous Linear Equations Section 5.3 Let s consider the non-homogenous case where y + p(x)y + q(x)y = f(x). A single solution to this equation is called a particular solution. The equation with f(x) = 0 is called the complementary equation. If p, q, f are continuous on (a, b) and x 0 is a point in (a, b) then for any constants k 0 and k 1 the IVP y + p(x)y + q(x) = f(x), y(x 0 ) = k 0, y (x 1 ) = k 1 has a unique solution on (a, b). Furthermore, if y p is a particular solution to y + p(x)y + q(x)y = f(x) on (a, b) and c 1 y 1 + c 2 y 2 is the general solution to the complementary equation y + p(x)y + q(x)y = 0 on (a, b) then y is a solution to y + p(x)y + q(x)y = f(x) on (a, b) if and only if y = y p + c 1 y 1 + c 2 y 2 for some constants c 1, c 2. Example: Let s solve the following IVP: y + 9y = 9, y(0) = 0, y (0) = 4. The characteristic polynomial of the complementary equation y + 9y = 0 is r which has roots r = ±3i. So c 1 cos (3x) + c 2 sin (3x) is the general solution of y + 9y = 0. Obviously y 1 = 1 is a particular solution of y + 9y = 9. Hence y = 1 + c 1 cos (3x) + c 2 sin (3x) is the general solution of y + 9y = 9. Imposing the initial condition y(0) = 0 implies 0 = 1+c 1, so c 1 = 1. Imposing the initial condition y (0) = 4 implies 4 = 3c 2, so c 2 = 4 3. So the solution to the IVP is y = 1 cos (3x) + 4 sin (3x). 3 WARNING: Do not use the initial conditions to find the constants in the general solution of the complementary equation and then add a particular solution. Doing so yields an incorrect result. 50

51 Superposition Principle: Suppose p, q, f 1, f 2 are continuous on (a, b), y p1 is a particular solution to y +p(x)y +q(x)y = f 1 (x) on (a, b), and y p2 is a particular solution to y +p(x)y +q(x)y = f 2 (x) on (a, b). Then y = y p1 + y p2 is a solution to y + p(x)y + q(x)y = f 1 (x) + f 2 (x) on (a, b). Example: Let s find the general solution to y 7y + 12y = 5e 2x + 12x 2 2x + 5. Clearly the general solution to the complementary homogeneous equation is y = c 1 e 3x + c 2 e 4x. We can try to find a pair of particular solutions to y 7y + 12y = 5e 2x and y 7y + 12y = 12x 2 2x + 5 respectively. Let s begin with the former. Let s judiciously guess that there is a solution of the form y = Ae 2x and substitute it into y 7y + 12y = 5e 2x : 4Ae 2x 7( 2Ae 2x ) + 12Ae 2x = 5e 2x 4A + 14A + 12A = 5 A = 1 6 So y p1 = 1 6 e 2x is a particular solution. Now we proceed with the latter. Let s judiciously guess that there is a solution of the form y = ax 2 + bx + c and substitute it into y 7y + 12y = 12x 2 2x + 5: 2a 7(2ax + b) + 12(ax 2 + bx + c) = 12x 2 2x + 5 2a 7b + 12c = 5, 12b 14a = 2, and 12a = 12 So y p2 = x 2 + x is a particular solution. a = 1, b = 1, c = 0. By superposition, y p = 1 6 e 2x + x 2 + x, is a particular solution of y 7y + 12y = 5e 2x + 12x 2 2x + 5. So the general solution is y = 1 6 e 2x + x 2 + x + c 1 e 3x + c 2 e 4x. Suggested Homework: Note: The Assignment given by your instructor may be different. Read Section 5.3, Pages Section 5.3 { 1, 4, 16, 17 } 51