Chapter 9: Some Differential Equations

Chapter 9: Some Differential Equations
Introduction Order of Differential Equations Section 9.1 First-Order Linear Equations First-order Linear Differential Equation Solving Equations y´ + p(x)y = q(x) Example Initial Value Problem Newton’s Law of Cooling Properties of Newton’s Law A Mixing Problem Mixing Problem (continued) Section 9.2 Integral Curves; Separable Equations Separable Equations Functions as Solutions Example Example (continued) Logistic Equation Population Growth Differential Notation Section 9.3 The Equation y´´ + ay´ + by = 0 Homogeneous Second-Order Linear Differential Equations The Characteristic Equation Solutions of Differential Equations Case 1 Case 2 Case 3 Linear Combinations of Solutions Existence and Uniqueness Theorem Definition: Wronskian Theorems: Wronskian The General Solution Example Salas, Hille, Etgen Calculus: One and Several Variables Copyright 2007 © John Wiley & Sons, Inc. All rights reserved.

Introduction An equation that relates an unknown function to one or more of its derivatives is called a differential equation. (1) [there written f (t) = k f (t)] to model exponential growth and decay. We previously used the differential equation (2) the equation of simple harmonic motion, to model the motion of a simple pendulum and the oscillation of a weight suspended at the end of a spring. The order of a differential equation is the order of the highest derivative that appears in the equation. Thus (1) is a first-order equation and (2) is a second-order equation. A function that satisfies a differential equation is called a solution of the equation. Finding the solutions of a differential equation is called solving the equation. Salas, Hille, Etgen Calculus: One and Several Variables Copyright 2007 © John Wiley & Sons, Inc. All rights reserved.

Differential Equations; First-Order Linear Equations
A differential equation of the form is called a first-order linear differential equation. Here p and q are given functions defined and continuous on some interval I. [In the simplest case, p(x) = 0 for all x, the equation reduces to y´ = q(x). The solutions of this equation are the antiderivatives of q.] Salas, Hille, Etgen Calculus: One and Several Variables Copyright 2007 © John Wiley & Sons, Inc. All rights reserved.

First-Order Linear Equations
Solving Equations y´ + p(x)y = q(x) Salas, Hille, Etgen Calculus: One and Several Variables Copyright 2007 © John Wiley & Sons, Inc. All rights reserved.

Example Find the general solution of the equation y´ + ay = b, a, b constants, a ≠ 0. Solution First we calculate an integrating factor: and therefore eH(x) = eax . Multiplying the differential equation by eax, we get eax y´ + a eax y = b eax . The left-hand side is the derivative of eax y. Thus, we have We integrate this equation and find that It follows that This is the general solution. Salas, Hille, Etgen Calculus: One and Several Variables Copyright 2007 © John Wiley & Sons, Inc. All rights reserved.

When a differential equation is used as a mathematical model in some application, there is usually an initial condition y(x0) = y0 that makes it possible to evaluate the arbitrary constant that appears in the general solution. The problem of finding a particular solution that satisfies a given condition is called an initial-value problem. Example Solve the initial-value problem: xy´ − 2y = 3x4, y(−1) = 2. Solution This differential equation does not have the form of (9.1.1), but we can put it in that form by dividing the equation by x: Solving we obtain the general solution: Applying the initial condition, we have This gives C = ½. The function is the solution of the initial-value problem. Salas, Hille, Etgen Calculus: One and Several Variables Copyright 2007 © John Wiley & Sons, Inc. All rights reserved.

Applications Newton’s Law of Cooling Newton’s law of cooling states that the rate of change of the temperature T of an object is proportional to the difference between T and the (assumed constant) temperature τ of the surrounding medium, called the ambient temperature. The mathematical formulation of Newton’s law takes the form where m is a constant. Salas, Hille, Etgen Calculus: One and Several Variables Copyright 2007 © John Wiley & Sons, Inc. All rights reserved.

A Mixing Problem Example A chemical manufacturing company has a 1000-gallon holding tank which it uses to control the release of pollutants into a sewage system. Initially the tank has 360 gallons of fluid containing 2 pounds of pollutant per gallon. Fluid containing 3 pounds of pollutant per gallon enters the tank at the rate of 80 gallons per hour and is uniformly mixed with the fluid already in the tank. Simultaneously, fluid is released from the tank at the rate of 40 gallons per hour. Determine the rate (lbs/gal) at which the pollutant is being released after 10 hours of this process. Solution Let P(t) be the amount of pollutant (in pounds) in the tank at time t. The rate of change of pollutant in the tank, dP/dt, satisfies the condition The pollutant is entering the tank at the rate of 3 × 80 = 240 pounds per hour (rate in). continued → Salas, Hille, Etgen Calculus: One and Several Variables Copyright 2007 © John Wiley & Sons, Inc. All rights reserved.

Example (continued) Fluid is entering the tank at the rate of 80 gallons per hour and is leaving at the rate of 40 gallons per hour. The amount of fluid in the tank is increasing at the rate of 40 gallons per hour, and so there are t gallons of fluid in the tank at time t. We can now conclude that the amount of pollutant per gallon in the tank at time t is given by the function and the rate at which pollutant is leaving the tank is (rate out). Therefore, our differential equation reads This is a first-order linear differential equation. Here we have continued → Salas, Hille, Etgen Calculus: One and Several Variables Copyright 2007 © John Wiley & Sons, Inc. All rights reserved.

Example (continued) As an integrating factor, we use eH(t) = e ln(9+t) = 9 + t. Multiplying the differential equation by 9 + t, we have Since the amount of pollutant in the tank is initially 2 × 360 = 720 (pounds), we see that which implies that C = −3240. Thus, the function (pounds). gives the amount of pollutant in the tank at any time t. After 10 hours there are (10) = 760 gallons of fluid in the tank, and there are pounds of pollutant. Therefore, the rate at which pollutant is being released into the sewage system after 10 hours is pounds per gallon. Salas, Hille, Etgen Calculus: One and Several Variables Copyright 2007 © John Wiley & Sons, Inc. All rights reserved.

Integral Curves; Separable Equations
Introduction We begin with a pair of functions P and Q which have continuous derivatives P´ = p, Q´ = q and form the equation (1) P(x) + Q(y) = C, C constant. Note that in this equation the x’s and y’s are not intermingled; they are separated. Different values of the parameter C give different curves. If y = y(x) is a differentiable function which on its domain satisfies (1), then by implicit differentiation we find that P´(x) + Q´(y)y´ = 0. Since P´ = p and Q´ = q, we have (2) p(x) + q(y)y´ = 0. In this sense, curves (1) satisfy differential equation (2). Salas, Hille, Etgen Calculus: One and Several Variables Copyright 2007 © John Wiley & Sons, Inc. All rights reserved.

Our starting point is a differential equation of the form with p and q continuous. A differential equation which can be written in this form is called separable. This equation represents a one-parameter family of curves, and these curves are the integral curves (solution curves) of the differential equation. Salas, Hille, Etgen Calculus: One and Several Variables Copyright 2007 © John Wiley & Sons, Inc. All rights reserved.

Functions as Solutions Example Show that the differential equation y´ = xey−x is separable and find the integral curves. Show that these curves are the graphs of functions. Solution The equation is separable since it can be written as xe−x − e−y y´ = 0. Setting we have −xe−x − e−x + e−y = C and therefore e−y = xe−x + e−x + C. These equations give the integral curves. continued → Salas, Hille, Etgen Calculus: One and Several Variables Copyright 2007 © John Wiley & Sons, Inc. All rights reserved.

To show that these curves are the graphs of functions, we take the logarithm of both sides: The integral curves are the graphs of the functions Since we have y´ = xey−x This shows that the functions satisfy the differential equation. Salas, Hille, Etgen Calculus: One and Several Variables Copyright 2007 © John Wiley & Sons, Inc. All rights reserved.

Applications In the mid-nineteenth century the Belgian mathematician P. F. Verhulst used the differential equation where k and M are positive constants, to model population growth. This equation is now known as the logistic equation, and its solutions are functions called logistic functions. Life scientists have used this equation to model the spread of disease, and social scientists have used it to study the dissemination of information. In the case of disease, if M denotes the total number of people in the population under consideration and y(t) is the number of infected people at time t, then the differential equation states that the rate of growth of the disease is proportional to the product of the number of people who are infected and the number who are not infected. Salas, Hille, Etgen Calculus: One and Several Variables Copyright 2007 © John Wiley & Sons, Inc. All rights reserved.

Example Assume that a rumor spreads through a population of 5000 people at a rate proportional to the product of the number of people who have heard it and the number who have not heard it. Suppose that by a certain day, call it time t = 0, 100 people have heard the rumor and 2 days later the rumor is known to 500 people. How long will it take for the rumor to spread to half of the population? Solution Let y(t) denote the number of people aware of the rumor by time t. Then y satisfies the logistic equation with M = 5000 and the initial condition R = y(0) = 100. Thus, The constant of proportionality k can be determined from the condition y(2) = 500: continued → Salas, Hille, Etgen Calculus: One and Several Variables Copyright 2007 © John Wiley & Sons, Inc. All rights reserved.

Example (continued) Therefore −10, 000k = ln (9/49) and k ≈ which gives To determine how long it will take for the rumor to spread to half of the population, we seek the value of t for which Solving this equation for t: It will take slightly more than 4½ days for the rumor to spread to half of the population. Salas, Hille, Etgen Calculus: One and Several Variables Copyright 2007 © John Wiley & Sons, Inc. All rights reserved.

Differential Notation The equation (1) p(x) + q(y)y´ = 0 can obviously be written (2) In differential notation the equation reads (3) p(x) dx + q(y) dy = 0 These equations are equivalent and can all be solved by setting Salas, Hille, Etgen Calculus: One and Several Variables Copyright 2007 © John Wiley & Sons, Inc. All rights reserved.

The Equation y´´ +ay´ +by = 0
A differential equation of the form where a and b are real numbers, is called a homogeneous second-order linear differential equation with constant coefficients. By a solution of this equation, we mean a function y = y(x) that satisfies the equation for all real x. Salas, Hille, Etgen Calculus: One and Several Variables Copyright 2007 © John Wiley & Sons, Inc. All rights reserved.

The Characteristic Equation As you can readily verify, the function y = e−ax satisfies the differential equation y´ + ay = 0. This suggests that the differential equation y´´ + ay´ + by = 0 may have a solution of the form y = erx . If y = erx , then y´ = rerx and y´´ = r2erx . Substitution into the differential equation gives r2erx + arerx + berx = erx (r2 + ar + b) = 0, and since erx ≠ 0, r2 + ar + b = 0. This shows that the function y = erx satisfies the differential equation iff This quadratic equation in r is called the characteristic equation. Salas, Hille, Etgen Calculus: One and Several Variables Copyright 2007 © John Wiley & Sons, Inc. All rights reserved.

The nature of the solutions of the differential equation y´´ + ay´ + by = 0 depends on the nature of the characteristic equation. There are three cases to be considered: 1) the characteristic equation has two distinct real roots; 2) it has only one real root; 3) it has no real roots, i.e. it has two complex roots. Salas, Hille, Etgen Calculus: One and Several Variables Copyright 2007 © John Wiley & Sons, Inc. All rights reserved.

Case 1: The characteristic equation has two distinct real roots r1 and r2. In this case both y1 = er1x and y2 = er2x are solutions of the differential equation. Salas, Hille, Etgen Calculus: One and Several Variables Copyright 2007 © John Wiley & Sons, Inc. All rights reserved.

Case 2: The characteristic equation has only one real root r = α. In this case the characteristic equation takes the form (r − α)2 = 0. This can be written r2 − 2αr + α2 = 0. The differential equation then reads y´´ − 2αy´ + α2 y = 0. The substitution y = ueax gives u´´ = 0. This equation is satisfied by the constant function u1 = and the identity function u2 = x. Thus the original differential equation is satisfied by the products y1 = eαx and y2 = xeαx Salas, Hille, Etgen Calculus: One and Several Variables Copyright 2007 © John Wiley & Sons, Inc. All rights reserved.

Case 3: The characteristic equation has two complex roots r1 = α + iβ, r2 = α − iβ with β ≠ 0. In this case the characteristic equation takes the form (r − α − iβ)(r − α + iβ) = 0, which, multiplied out, becomes r2 − 2αr + (α2 + β2) = 0. The differential equation thus reads y´´ − 2αy´ + (α2 + β2)y = 0. The substitution y = ueαx eliminates α (see Exercise 33) and gives u´´ + β2u = 0. This equation, the equation of simple harmonic motion, is satisfied by the functions u1 = cos βx and u2 = sin βx. Thus, the original differential equation is satisfied by the products y1 = eαx cos βx and y2 = eαx sin βx. Salas, Hille, Etgen Calculus: One and Several Variables Copyright 2007 © John Wiley & Sons, Inc. All rights reserved.

Linear Combinations of Solutions; Existence and Uniqueness of Solutions; Wronskians Observe that if y1 and y2 are both solutions of the homogeneous equation, then every linear combination u(x) = C1 y1(x) + C2 y2(x) is also a solution. Salas, Hille, Etgen Calculus: One and Several Variables Copyright 2007 © John Wiley & Sons, Inc. All rights reserved.

The General Solution The arbitrary linear combination y = C1 y1 + C2 y2 of any two solutions with nonzero Wronskian is called the general solution. Salas, Hille, Etgen Calculus: One and Several Variables Copyright 2007 © John Wiley & Sons, Inc. All rights reserved.

Example Find the general solution of the equation Then find the particular solution that satisfies the initial conditions y(0) = 0, y´(0) = −1. Solution The characteristic equation is the quadratic r2 + 2r − 15 = 0. Factoring the left side, we have (r + 5)(r − 3) = 0. There are two real roots: −5 and 3. The general solution takes the form y = C1e−5x + C2e3x . Differentiating the general solution, we have y' = −5C1e−5x + 3C2e3x The conditions y(0) = 0, y´(0) = −1 are satisfied iff C1 + C2 = and − 5C1 + 3C2 = −1. Solving these two equations simultaneously, we find that C1 = 1/8, C2 = −1/8. The solution that satisfies the prescribed initial conditions is the function Salas, Hille, Etgen Calculus: One and Several Variables Copyright 2007 © John Wiley & Sons, Inc. All rights reserved.

Example Find the general solution of the equation y´´ + ω2 y = (ω ≠ 0) Solution The characteristic equation is r2 + ω2 = 0 and the roots are r1 = ωi, r2 = −ωi. Thus the general solution is y = C1 cos ωx + C2 sin ωx. Remark As you probably recall, the equation in Example 4 describes the oscillatory motion of an object suspended by a spring under the assumption that there are no forces acting on the spring-mass system other than the restoring force of the spring. Salas, Hille, Etgen Calculus: One and Several Variables Copyright 2007 © John Wiley & Sons, Inc. All rights reserved.

Chapter 9: Some Differential Equations

Similar presentations

Presentation on theme: "Chapter 9: Some Differential Equations"— Presentation transcript:

Similar presentations

About project

Feedback

Log in

Auth with social network:

Chapter 9: Some Differential Equations

Similar presentations

Presentation on theme: "Chapter 9: Some Differential Equations"— Presentation transcript:

Similar presentations

About project

Feedback