Lecture Notes in Mathematics. Arkansas Tech University Department of Mathematics

Lecture Notes in Mathematics Arkansas Tech University Department of Mathematics Introductory Notes in Ordinary Differential Equations for Physical Sciences and Engineering Marcel B. Finan c All Rights Reserved May 4, 2017

Preface Differential equations arise from the study of problems in virtually every area of physical sciences as well as in engineering. Many mathematical models of real world phenomena lead to a differential equation problem for which we seek its solution. In this course, we will explore analytical, numerical and graphical techniques for finding the solutions. To gain a solid grasp of differential equations, it is essential for the reader to solve ALL the exercises listed at the end of each section. Answers to most exercises are available at the end of the book. A solution guide is available by request: Email mfinan@atu.edu Marcel B. Finan Russellville, Arkansas May 2011 i

ii PREFACE

Contents Preface i 1 Basic Terminology 3 2 Existence and Uniqueness of Solutions to First Order Linear IVP 13 3 Analytical Solution: The Method of Integrating Factor 19 4 Existence and Uniqueness of Solutions to First Order Nonlinear IVP 29 5 Separable Differential Equations 35 6 Exact Differential Equations 43 7 Substitution Techniques: Bernoulli and Riccati Equations 51 8 Graphical Solution: Direction Field of y = f(t, y) 59 9 Numerical Solutions to ODEs: Euler s Method and its Variants 69 10 Second Order Linear Differential Equations: Existence and Uniqueness Results 75 11 The General Solution of 2 nd Order Linear Homogeneous Equations 81 1

2 CONTENTS 12 Second Order Linear Homogeneous Equations with Constant Coefficients: Distinct Characterisitc Roots 91 13 Characteristic Equations with Repeated Roots 97 14 Characteristic Equations with Complex Roots 103 15 Series Solutions of Differential Equations 111 16 The Structure of the General Solution of Linear Nonhomogeneous Equations 117 17 The Method of Variation of Parameters 123 18 The Laplace Transform: Basic Definitions and Results 129 19 Further Studies of Laplace Transform 141 20 The Laplace Transform and the Method of Partial Fractions151 21 Laplace Transforms of Periodic Functions 159 22 Convolution Integrals 167 Cheat Sheets 173 Answer Key 181 Index 223

1 Basic Terminology In many models, we will have equations involving the derivatives of a dependent variable y with respect to one or more independent variables and are interested in discovering this function y. Such equations are referred to as differential equations (abbreviated DE). They arise in many applications such as population growth, decay of radioactive substance, the motion of a falling object, electrical network, Newton s law of cooling and many more models. A First Source of Differential Equations: Vertical Motion of an Object We consider the model of uniform motion. Suppose that an object initially at height y 0 is moving straight up or down with initial velocity v 0. Let y(t) denote the distance of the object from the ground, v(t) the object s velocity, and a(t) the object s acceleration at time t. We assume y to be positive in the upward direction. If air resistance is neglected, then by Newton s second law, which states that the net force is equal to the product of mass and acceleration, we have ma(t) = mg. The negative sign on the right-hand of the equation is due to the fact that acceleration due to gravity is pointing downward. Using the fact that a(t) = y (t) and eliminating the mass, we obtain the equation y = g. To find the velocity v(t) we integrate for a first time and obtain v(t) = y (t) = gt + C 1. Since the initial velocity is v 0, we obtain C 1 = v(0) = v 0 so that v(t) = gt + v 0. 3

4 1 BASIC TERMINOLOGY Integrating for the second time we find the position function y(t) = 1 2 gt2 + v 0 t + C 2. Since y 0 is the initial height, we find C 2 = y 0 and so y(t) = 1 2 gt2 + v 0 t + y 0. Example 1.1 An object is dropped from the top of a cliff that is 144 feet above ground level. (a) When will the object reach ground level? (b) What is the velocity with which the object strikes the ground? Solution. (a) The motion of the object translates to the differential equation y = 32 with solution y(t) = 16t 2 + 144. The object reaches ground level when 144 16 y(t) = 0 or 16t 2 = 144. Solving for t we find t = = 3 sec. The object will reach the ground 3 seconds after it is dropped from the cliff. (b) The object strikes the ground with velocity v(3) = 32(3) = 96 ft/sec Basic Terms of Differential Equations We next discuss some basic notions of differential equations. There are two types of differential equations: ordinary and partial differential equations. By an ordinary differential equation (abbreviated ODE) we mean an equation that involves an unknown function (the dependent variable) of a single variable, its independent variable, and one or more of its derivatives. The highest order derivative that appears in the equation is known as the order of the equation. Thus, an n th order ordinary differential equation is an equation of the form or alternatively y (n) = f(t, y, y,, y (n 1) ) G(t, y, y, y,, y n ) = 0. A first-order ordinary differential equation, for example, takes the form f(t, y(t), y (t)) = 0, and may alternatively be written as y (t) = g(t, y(t))

for all t in the interval of existence of y. Similarly, a second-order ordinary differential equation takes the form f(t, y(t), y (t), y (t)) = 0 or y = h(t, y, y ). Example 1.2 Determine the order of each equation. (a) y + 2ty = e t2. (b) d2 y 5 dy + 6y(t) = 0. dt 2 dt (c) y + 3ty + 2y = sin (5t). Solution. (a) This is a first order differential equation because the highest derivative is the first derivative. (b) and (c) are second order differential equations since the highest derivative in each equation is the second order derivative When a dependent function is a function of two or more independent variables then the derivatives are known as partial derivatives. An equation that involves a function of more than two independent variables and its partial derivatives is called a partial differential equation (abbreviated PDE). For example, the wave equation is a partial differential equation of the form 2 u x 1 2 u 2 c 2 t = 0. 2 In this course, when we use the term differential equation, we ll mean an ordinary differential equation. A solution of a differential equation is a function that satisfies the equation: When you substitute this function and its derivatives into the differential equation, you get a true mathematical statement. Example 1.3 Show that the function y = 100+e t is a solution to the differential equation y = 100 y. Solution. Indeed, finding the first order derivative of y we have y = e t. Also, 100 y = 100 (100 + e t ) = e t. Thus, y = 100 y so that y = 100 + e t is a solution to the given DE 5

6 1 BASIC TERMINOLOGY Example 1.4 (A piecewise-defined solution) Consider the differential equation ty 4y = 0 on the interval (, ). Verify that the piecewise-defined function { t y = 4, t < 0 t 4, t 0 is a solution. Solution. For t < 0, we have ty 4y = t( t 4 ) 4( t 4 ) = 4t 4 + 4t 4 = 0. For t 0, we have ty 4y = t(t 4 ) 4t 4 = 4t 4 4t 4 = 0. Thus, the given function is a solution Solving a differential equation means finding all possible solutions of the equation. Example 1.5 Solve the differential equation: y = 2t. Solution. Integrating twice, all the solutions have the form y(t) = t3 3 + C 1t + C 2 Note that the function of the previous example defines all the solutions to the differential equation. Such a function will be referred to as the general solution. The constants C 1 and C 2 are called the parameters. Specific values of C 1 and C 2 determine what is called a particular solution. To find a particular solution additional conditions on the values of the function or its derivatives must be given. Such conditions are called initial conditions. A differential equation together with a set of initial conditions is called an initial value problem (abbreviated IVP). Example 1.6 Consider the differential equation y (t) 1 = 0. (a) Find the general solution of this equation. (b) Find the solution that satisfies the initial conditions y(1) = 1 and y (1) = 4.

7 Solution. (a) Integrating twice we find the general solution y(t) = t2 2 + C 1t + C 2. (b) Since y (t) = t + C 1 and y (1) = 4, we find 4 = 1 + C 1 so that C 1 = 3. Hence, y(t) = t2 + 3t + C 2 2. Now, since y(1) = 1, we have 1 = 1 + 3 + C 2 2. Solving for C 2 we find C 2 = 5. Hence, the solution to the IVP 2 is { y (t) 1 = 0 y (1) = 4, y(1) = 1 y(t) = t2 2 + 3t 5 2 The graph of a particular solution is called a solution curve. The function y(t) = Ce 3t + 2t + 1 is the general solution to the differential equation y + 3y = 6t + 5. A family of solution curves is shown in Figure 1.1. Notice for C 0 the solution curves have an oblique asymptote with equation y(t) = 2t + 1. Figure 1.1 Sometimes a differential equation possesses a solution that cannot be obtained by assigning values to the parameters in a family of solutions. Such a solution is called a singular solution.

8 1 BASIC TERMINOLOGY Example 1.7 Using the method of separation of variables (Chapter 5), the non-zero solutions to the differential equation y = ty 1 2 are given by y(t) = ( t2 + 4 C)2. Find the singular solution. Solution. The function y(t) 0 is a solution to the differential equation. This is a singular solution since it cannot be obtained from the family for any choice of the parameter C. The general solution consists of all the solutions of the form y(t) = ( t2 4 + C)2 together with the zero solution

Practice Problems Problem 1.1 A car starts from rest and accelerates in a straight line at 1.6 m/sec 2 for 10 seconds. (a) What is its final speed? (b) How far has it travelled in this time? Problem 1.2 An object is thrown upward at time t = 0. After t seconds, its height is y(t) = 4.9t 2 + 7t + 1.5 meters above the ground. (a) From what height was the object thrown? (b) What is the initial velocity of the object? (c) What is the acceleration due to gravity? Problem 1.3 An object thrown in the air on a planet in a distant galaxy is at height y(t) = 25t 2 + 72t + 40 feet at time t seconds after it is thrown. What is the acceleration due to gravity on that planet? With what velocity was the object thrown? From what height? Problem 1.4 An object is dropped from a 400-foot tower. When does it hit the ground and how fast is it going at the time of the impact? Problem 1.5 A ball that is dropped from a window hits the ground in five seconds. How high is the window? (Give your answer in feet.) Problem 1.6 A ball is thrown straight up from ground level and reaches its greatest height after 5 seconds. Find the initial velocity of the ball and the value of its maximum height above ground level. Problem 1.7 At time t = 0 an object having mass m is released from rest at a height y 0 above the ground. Let g represent the constant gravitational acceleration. Derive an expression for the impact time (the time at which the object strikes the ground). What is the velocity with which the object strikes the ground? 9

10 1 BASIC TERMINOLOGY Problem 1.8 At time t = 0, an object of mass m is released from rest at a height of 252 ft above the floor of an experimental chamber in which gravitational acceleration has been slightly modified. Assume (instead of the usual value of 32 ft/sec 2 ), that the acceleration has the form 32 ɛ sin ( ) πt 4 ft/sec 2, where ɛ is a positive constant. In addition, assume that the object strikes the ground exactly 4 sec after release. Can this information be used to determine the constant ɛ? If so, determine ɛ. Problem 1.9 Find the order of the following differential equations. (a) ty + y = t 3. (b) y + y 2 = 2. (c) sin (y ) + 3t 2 y = 6t. Problem 1.10 What is the order of the differential equation? (a) y (t) 1 = 0. (b) y (t) 1 = 0. (c) y (t) 2ty(t) = 0. (d) y (t)(y (t)) 1 2 t = 0. y(t) Problem 1.11 In the equation u x u y = x 2y, identify the independent variable(s) and the dependent variable. Problem 1.12 Classify the following equations as either ordinary or partial. (a) (y ) 4 + t2 = 0. (y ) 2 +4 (b) u x + y u y = y x y+x. (c) y 4y = 0. Problem 1.13 Solve the equation y (t) 2 = 0 by computing successive antiderivatives.

11 Problem 1.14 Solve the initial-value problem dy dt What is the domain of the solution? = 3y, y(0) = 50. Problem 1.15 For what real value(s) of λ is y = cos λt a solution of the equation y +9y = 0? Problem 1.16 For what value(s) of m is y = e mt a solution of the equation y +3y +2y = 0? Problem 1.17 Show that y(t) = e t is a solution to the differential equation ( y 2 + 2 ) ( y + 1 + 2 ) y = 0. t t Problem 1.18 Show that any function of the form y(t) = C 1 cos ωt + C 2 sin ωt satisfies the differential equation d 2 y dt 2 + ω2 y = 0. Problem 1.19 Suppose y(t) = 2e 4t is the solution to the initial value problem y + ky = 0, y(0) = y 0. Find the values of k and y 0. Problem 1.20 Consider t > 0. For what value(s) of the constant n, if any, is y(t) = t n a solution to the differential equation t 2 y 2ty + 2y = 0? Problem 1.21 (a) Show that y(t) = C 1 e 2t + C 2 e 2t is a solution of the differential equation y 4y = 0, where C 1 and C 2 are arbitrary constants. (b) Find the solution satisfying y(0) = 2 and y (0) = 0. (c) Find the solution satisfying y(0) = 2 and lim t y(t) = 0.

12 1 BASIC TERMINOLOGY Problem 1.22 Suppose that the graph below is the particular solution to the initial value problem y (t) = m + 1, y(1) = y 0. Determine the constants m and y 0 and then find the formula for y(t). Problem 1.23 Suppose that the graph below is the particular solution to the initial value problem y (t) = mt, y(t 0 ) = 1. Determine the constants m and t 0 and then find the formula for y(t). Problem 1.24 Show that y(t) = e 2t is not a solution to the differential equation y +4y = 0. Problem 1.25 Consider the initial-value problem y + 3y = 6t + 5, y(0) = 3. (a) Show that y = Ce 3t + 2t + 1 is a solution to the above differential equation. (b) Find the value of C.

2 Existence and Uniqueness of Solutions to First Order Linear IVP Solutions to differential equations can be given in one of the following forms: by an explicit formula: For example, the function y = t 3 + 1 is an explicit solution to the initial value problem 2yy = 3t 2, y(1) = 2; by an implicit equation: The solution y to the equation y = 1+yety is 1+te ty defined implicitly by the equation t + y + e ty = C; by a power series representation. For example the general solution to the equation (t 2 4)y + 3ty + y = 0 is given by ( y(t) = C 1 1 + 1 8 t2 + 3 128 t4 + 5 ( 1024 t6 + )+C 2 t + 1 6 t3 + 1 30 t5 + 1 ) 140 t7 + ; numerically (i.e., Euler s methods); graphically (i.e., direction fields). Before worrying about how to solve an initial value problem, either analytically, graphically, or numerically, one should first resolve the basic issues of existence and uniqueness. First, does a solution exist? If, not, it makes no sense trying to find one. Second, is the solution uniquely determined by the initial conditions? Otherwise, the differential equation probably has little relevance for physical applications since we cannot use it as a predictive tool. In this section we discuss the conditions needed to guarantee the existence of a unique solution to first order linear initial value problems. We start with the definition of a first order linear differential equation. Any differential equation that can be written in the form y + p(t)y = g(t) (2.1) 13

142 EXISTENCE AND UNIQUENESS OF SOLUTIONS TO FIRST ORDER LINEAR IVP where p(t) and g(t) are functions with common domain a < t < b, is called a first order linear differential equation. The term linear is used because L(y) = y + p(t)y is linear in y. That is, L(αy 1 + βy 2 ) = αl(y 1 ) + βl(y 2 ). A DE that is not linear is called non-linear. In mathematics and physics, linear generally means simple and non-linear means complicated. The theory for solving linear equations is very well developed because linear equations are simple enough to be solvable. Nonlinear equations can usually not be solved exactly and are the subject of much on-going research. Now, we say that Equation (2.1) is homogeneous if g(t) 0 for all a < t < b. If there is a a < t < b such that g(t) 0 then Equation (2.1) is called non-homogeneous. Example 2.1 Classify each of the following first order differential equations as linear or non-linear. If the equation is linear, decide whether it is homogeneous or non-homogeneous. (a) dy + y = ty. dt 10 (b) t 2 3y 2 + 2ty dy = 0. dt (c) t dy = dt t2 2y. (d) dy = t y. dt t+y Solution. (a) Notice that the given equation can be written as dy + ( 1 t)y = 0 which dt 10 is a homogeneous first order linear DE where p(t) = 1 t and g(t) = 0. 10 (b) This is non-linear because of the term y 2. (c) This is a non-homogeneous first order linear DE since the right-hand side is not identically zero on any interval. Here, we have p(t) = 2 and g(t) = t. t (d) This is non-linear because of the y in the denominator First order linear differential equations possess important linearity or superposition properties. Theorem 2.1 (a) If y 1 (t) and y 2 (t) are any two solutions of the homogeneous equation y +p(t)y = 0 then for any constants c 1 and c 2 the linear combination c 1 y 1 (t)+ c 2 y 2 (t) is also a solution of the homogeneous equation. (b) If y 1 (t) is a solution to the homogeneous equation y + p(t)y = 0 and

y 2 (t) is a solution to the non-homogeneous equation y + p(t)y = g(t) then Cy 1 (t) + y 2 (t) is also a solution to the non-homogeneous equation, where C is an arbitrary constant. Proof. (a) Since y 1 (t) and y 2 (t) are solutions to the homogeneous equation, we have (c 1 y 1 + c 2 y 2 ) + p(t)(c 1 y 1 + c 2 y 2 ) = c 1 (y 1 + p(t)y 1 ) + c 2 (y 2 + p(t)y 2 ) = 0 + 0 = 0. (b) We have (Cy 1 + y 2 ) + p(t)(cy 1 + y 2 ) = C(y 1 + p(t)y 1 ) + y 2 + p(t)y 2 = 0 + g(t) = g(t) Remark 2.1 Part (a) of the previous theorem is not true in general for non-homogeneous equations. For example, consider the equation y = 1. Then y 1 (1) = t and y 2 (t) = t + 1 are both solutions to the DE. However, y 1 (t)+y 2 (t) = 2t+1 is not a solution since (y 1 + y 2 ) = 2 1 Next, we look at the conditions that guarantee the existence of a unique solution to the IVP 15 y + p(t)y = g(t), y(t 0 ) = y 0. (2.2) Theorem 2.2 If p(t) and g(t) are continuous functions in the open interval I = (a, b) and t 0 a point inside I then the IVP (2.2) has a unique solution y(t) defined on I. Example 2.2 If p(t) is continuous on (a, b) and a < t 0 < b then what is the unique solution to the IVP y + p(t)y = 0, y(t 0 ) = 0? Solution. Since the trivial solution y(t) 0 satisfies the initial value problem, by Theorem 2.2, y(t) 0 is the unique solution Remark 2.2 Theorem 2.2 asserts that if the hypotheses are satisfied then a unique solution exists on an interval containing t 0. However, the solution might actually exist on a larger interval than what the theorem asserts. To be more precise, consider the IVP ty + 2y = 0, y(1) = 0. If we apply Theorem 2.2, then a unique solution exists say on the interval 0 < t < since this is the interval containing 1 and where p(t) = 1 t is defined and continuous. Actually, the solution is y(t) 0. But one can easily see that y(t) 0 is a solution for all < t <. So our theorem asserts the existence of a solution locally rather than globally

162 EXISTENCE AND UNIQUENESS OF SOLUTIONS TO FIRST ORDER LINEAR IVP Practice Problems Problem 2.1 Find p(t) and y 0 so that the function y(t) = 3e t2 y + p(t)y = 0, y(0) = y 0. is the solution to the IVP Problem 2.2 For each of the initial conditions, determine the largest interval a < t < b on which Theorem 2.2 guarantees the existence of a unique solution. (a) y(0) = π (b) y(π) = 0. y + 1 t 2 + 1 y = sin t Problem 2.3 For each of the initial conditions, determine the largest interval a < t < b on which Theorem 2.2 guarantees the existence of a unique solution y + t t 2 4 y = (a) y(5) = 2 (b) y( 3 2 ) = 1 (c) y( 6) = 2. et t 3 Problem 2.4 (a) For what values of the constant C and the exponent r is y = Ct r the solution of the IVP 2ty 6y = 0, y( 2) = 8? (b) Determine the largest interval of the form a < t < b on which Theorem 2.2 guarantees the existence of a unique solution. (c) What is the actual interval of existence for the solution found in part (a)? Problem 2.5 What information does Theorem 2.2 give about the initial value problem ty = y + t 3 cos t, y(1) = 1?y( 1) = 1? Problem 2.6 Consider the following differential equation (t 4)y + 3y = 1 t 2 + 5t. Without solving, find the interval over which a unique solution is guaranteed for each of the following initial conditions: (a) y( 3) = 4 (b) y(1.5) = 2 (c) y( 6) = 0 (d) y(4.1) = 3.

17 Problem 2.7 Without solving the initial value problem, (t 1)y + (ln t)y = 2 t 3, y(t 0) = y 0, state whether or not a unique solution is guaranteed to exist for the y(t 0 ) = y 0 listed below. If a unique solution is guaranteed, find the largest interval for which the solution satisfies the differential equation and the initial condition. (a) y(2) = 4 (b) y(0) = 0 (c) y(4) = 2. Problem 2.8 (a) State precisely the theorem (hypothesis and conclusion) for existence and uniqueness of a linear first order initial value problem. (b) Consider the equation y + t 2 y = e t3 with initial conditions y(t 0 ) = y 0. Briefly discuss if this has a solution, if it is unique and why. Problem 2.9 Is the differential equation linear or non-linear? If the equation is linear, decide whether it is homogeneous or non-homogeneous. (a) y = ty 2 (b) y = t 2 y (c) (cos t)y + e t y = sin t (d) y y + t3 = sin t, y > 0. Problem 2.10 Consider the initial value problem y + p(t)y = g(t), y(3) = 1. Suppose that p(t) and/or g(t) have discontinuities at t = 2, t = 0, and t = 5 but are continuous for all other values of t. What is the largest interval (a, b) on which Theorem 2.2 is applied. Problem 2.11 Determine α and y 0 so that the graph of the solution to the initial-value problem y + αy = 0, y(0) = y 0 passes through the points (1, 4) and (3, 1). Problem 2.12 Match the following equations with the correct description. Every equation matches exactly one description. (a) y = 3y 5t. (b) y t = 2 y t 2 + 2 y x 2.

182 EXISTENCE AND UNIQUENESS OF SOLUTIONS TO FIRST ORDER LINEAR IVP (c) y y 2 = sin t. (d) y + 3y = 0. (i) A partial differential equation. (ii) A homogeneous first order linear differential equation. (iii) A non-linear first order differential equation. (iv) A non-homogenous first order linear differential equation. Problem 2.13 Consider the differential equation y = t 2 y + sin y. What is the order of this equation? Is it linear or non-linear? Problem 2.14 Show that y = t+y t is a linear first order non-homogeneous equation.

3 Analytical Solution: The Method of Integrating Factor If a unique solution to the initial value problem (2.2) exists then this solution is found by using the initial condition in the general solution to the differential equation to determine the value of the parameter. The purpose of this section is to find analytically the general solution to the first order linear non-homogeneous equation y + p(t)y = g(t) (3.1) where p(t) and g(t) are continuous on the open interval a < t < b. The Method of Integrating Factor We discuss a technique for solving equation (3.1). Since p(t) is continuous, it has an antiderivative namely p(t)dt. Let µ(t) = e p(t)dt. Multiply Equation (3.1) by µ(t) and notice that the left hand side of the resulting equation is the derivative of a product. Indeed, d dt (µ(t)y) = µ (t)y(t) + µ(t)y (t) = µ(t)(y (t) + p(t)y(t)) = µ(t)g(t). Integrate both sides of the last equation with respect to t to obtain µ(t)y = µ(t)g(t)dt + C. Now, divide through by µ(t), we find y(t) = 1 µ(t) or y(t) = e p(t)dt µ(t)g(t)dt + C µ(t) e p(t)dt g(t)dt + Ce p(t)dt. 19

203 ANALYTICAL SOLUTION: THE METHOD OF INTEGRATING FACTOR One can write the above function in the form y(t) = Cy 1 (t) + y 2 (t) where y 1 (t) = e p(t)dt and y 2 (t) = e p(t)dt e p(t)dt g(t)dt. Notice that y 1 is a solution for the homogeneous equation y + p(t)y = 0 and y 2 is a solution to the non-homogeneous equation as shown in the next example. Thus, the general solution to Equation (3.1) is the sum of a particular solution of the non-homogeneous equation and the general solution of the homogeneous equation. This solution structure will appear again when discussing second order linear equations. Remark 3.1 Notice that multiplying Equation (3.1) by µ(t) was a key factor in the integration step discussed above. That s why µ(t) is known as the integrating factor. Example 3.1 Show that y p = e p(t)dt e p(t)dt g(t)dt satisfies Equation (3.1). Solution. We have y p + p(t)y p = p(t)e p(t)dt +p(t)e p(t)dt =g(t) e p(t)dt g(t)dt + e p(t)dt e p(t)dt g(t) e p(t)dt g(t)dt Example 3.2 Solve the initial-value problem y y t = 4t, y(1) = 5. Solution. By Theorem 2.2, the solution exists and is unique on the interval (0, ) since 1 belongs to that interval. We have p(t) = 1 t so that µ(t) = 1 t. Multiplying the given equation by the integrating factor and using the product rule we notice that ( ) 1 t y = 4.

21 Integrating with respect to t and then solving for y we find that the general solution is given by y(t) = t 4dt + Ct = 4t 2 + Ct. The condition y(1) = 5 implies C = 1 and hence the unique solution to the IVP is y(t) = 4t 2 + t, 0 < t < Example 3.3 Find the general solution to the equation y + 2 y = ln t, t > 0. t Solution. The integrating factor is µ(t) = e 2 t dt = t 2. Multiplying the given equation by t 2 to obtain (t 2 y) = t 2 ln t. Integrating with respect to t we find t 2 y = t 2 ln tdt + C. The integral on the right-hand side is evaluated using integration by parts with u = ln t, dv = t 2 dt, du = dt t, v = t3 3 obtaining Thus, t 2 y = t3 3 ln t t3 9 + C. y = t 3 ln t t 9 + C t 2 Remark 3.2 Instead of using indefinite integrals in the above discussion one can use definite integrals. For example, replace p(t)dt by t t 0 p(s)ds for some fixed t 0. Using definite integral is proven to be useful when p(t) does not have an elementary function as an antiderivative. For example, when p(t) = sin t t or p(t) = sin (t 2 ). We illustrate this idea in the next example. Example 3.4 Solve y 1 t y = sin t, y(1) = 3. Express your answer in terms of the sine integral, Si(t) = t sin s 0 s ds.

223 ANALYTICAL SOLUTION: THE METHOD OF INTEGRATING FACTOR Solution. Since p(t) = 1 t we find µ(t) = 1 t. Thus, ( ) 1 t y = sin t ( t = t 0 1 y(t) =Si(t) + C t y(t) =tsi(t) + Ct. ) sin s s ds Since y(1) = 3, we find C = 3 Si(1). Hence, y(t) = tsi(t) + (3 Si(1))t Case when either p(t) or g(t) has a jump discontinuity Consider the IVP y + p(t)y = g(t), y(a) = y 0, a t b (3.2) where p(t) and g(t) are continuous in a t b except at t = c where either p(t) or g(t) has a jump discontinuity at a < c < b. Although y is not continuous at c, we seek a solution y(t) that is continuous at t = c. To solve this problem, we first solve the initial value problem on the interval a t < c where both p(t) and g(t) are continuous. Let y 1 (t) be the unique solution. Since we are seeking a continuous solution to (3.2), we expect y 1 (t) to have a one-sided limit at c, i.e., lim y 1(t) = y 1 (c ). t c Next, we find the unique solution y 2 (t) to the IVP y + p(t)y = g(t), y(c) = y 1 (c ) where c t b. The unique solution to the original IVP is then given by { y1 (t), if a t < c y(t) = y 2 (t) if c t b. We illustrate this process in the next example. Example 3.5 Find the solution to the IVP where y + 1 y = g(t), y(1) = 1 t g(t) = The graph of g(t) is given in Figure 3.1(a). { 3t, if 1 t 2 0 if 2 < t 3.

23 Solution. First, we solve the IVP y + 1 y = 3t, y(1) = 1, 1 t 2. t The integrating factor is µ(t) = t and the general solution is y 1 (t) = t 2 + C t. Since y(1) = 1, we have C = 0. Hence, y 1 (t) = t 2 and y 1 (2) = 4. Next, we solve the IVP y + 1 y = 0, y(2) = 4, 2 < t 3. t The integrating factor is µ(t) = t and the general solution is y 2 (t) = C t. Since y 2 (2) = 4 we find C = 8. Thus, { t y(t) = 2, if 1 t 2 8 t if 2 < t 3. The graph of y(t) is given in Figure 3.1(b). As you can see from the graph, y(t) is continuous on [1, 3] but not differentiable at t = 2 Figure 3.1

243 ANALYTICAL SOLUTION: THE METHOD OF INTEGRATING FACTOR Practice Problems Problem 3.1 Solve the IVP Problem 3.2 Solve the IVP y = 2ty, y(1) = 1. y + e t y = 0, y(0) = 2. Problem 3.3 Consider the first order linear non-homogeneous IVP y + p(t)y = αp(t), y(t 0 ) = y 0. (a) Show that the IVP can be reduced to a first order linear homogeneous IVP by the change of variable z(t) = y(t) α. (b) Solve this initial value problem for z(t) and use the solution to determine y(t). Problem 3.4 Apply the results of the previous problem to solve the IVP Problem 3.5 The unique solution to the IVP y + 2ty = 6t, y(0) = 4. ty αy = 0, y(1) = y 0 goes through the points (2, 1) and (4, 4). Find the values of α and y 0. Problem 3.6 The table below lists values of t and ln [y(t)] where y(t) is the unique solution to the IVP y + t n y = 0, y(0) = y 0. (a) Determine the values of n and y 0. (b) What is y( 1)? t 1 2 3 4 ln [y(t)] 0.25 4.00 20.25 64.00

25 Problem 3.7 Consider the differential equation y + p(t)y = 0. Find p(t) so that y = c t general solution. is the Problem 3.8 Consider the differential equation y + p(t)y = 0. Find p(t) so that y = ct 3 is the general solution. Problem 3.9 Solve the initial-value problem: y 3 t y = 0, y(2) = 8. Problem 3.10 Solve the differential equation y 2ty = 0. Problem 3.11 Find the function f(t) that crosses the point (0, 4) and whose slope satisfies f (t) = 2f(t). Problem 3.12 Find the general solution to the differential equation y 2y = 0. Hint: Let y = z. Problem 3.13 Solve the IVP: y + 2ty = t, y(0) = 0. Problem 3.14 Find the general solution: y + 3y = t + e 2t. Problem 3.15 Find the general solution: y + 1 t y = 3 cos t, t > 0. Problem 3.16 Find the general solution: y + 2y = cos (3t). Problem 3.17 Given that the solution to the IVP ty +4y = αt 2, y(1) = 1 3 < t <. What is the value of the constant α? exists on the interval Problem 3.18 Suppose that y(t) = Ce 2t +t+1 is the general solution to the equation y +p(t)y = g(t). Determine the functions p(t) and g(t).

263 ANALYTICAL SOLUTION: THE METHOD OF INTEGRATING FACTOR Problem 3.19 Suppose that y(t) = 2e t + e t + sin t is the unique solution to the IVP y + y = g(t), y(0) = y 0. Determine the constant y 0 and the function g(t). Problem 3.20 Find the value (if any) of the unique solution to the IVP y + (1 + cos t)y = 1 + cos t, y(0) = 3 in the long run, i.e., lim t y(t)? Problem 3.21 Find the solution to the IVP where Problem 3.22 Find the solution to the IVP y + p(t)y = 2, y(0) = 1 { 0, if 0 t 1 p(t) = 1 t if 1 < t 2. y + (sin t)y = g(t), y(0) = 3 where g(t) = { sin t, if 0 t π sin t if π < t 2π. Problem 3.23 Find the solution to the IVP y + y = g(t), t > 0, y(0) = 3 where g(t) = { 1, if 0 t 1 0 if t > 1. Problem 3.24 Find the solution to the IVP where y + p(t)y = 0, y(0) = 3 p(t) = 2t 1, if 0 t 1 0, if 1 < t 3 1 t if 3 < t 4. Sketch an accurate graph of the solution, and discuss the long-run behavior of the solution. Is the solution differentiable on the interval (0, )? Explain your answer.

Problem 3.25 Solve the initial-value problem y + y = e t y 2, y(0) = 1 using the substitution u(t) = 1 y(t). 27

283 ANALYTICAL SOLUTION: THE METHOD OF INTEGRATING FACTOR

4 Existence and Uniqueness of Solutions to First Order Nonlinear IVP When a mathematical model is constructed for physical systems, two reasonable demands are made. First, solutions should exist if the model is to be useful at all. Second, to work effectively in predicting the future behavior of the physical system, the model should produce only one solution for a particular set of initial conditions. Existence and uniqueness theorems help to meet these demands. In this section we discuss the conditions that guarantee the existence of a unique solution to the initial value problem y = f(t, y), y(t 0 ) = y 0 (4.1) where f(t, y) needs not be linear in y. Before we proceed to the major result of this section, we remind the reader of the definition of a partial derivative. Partial Derivatives If f(t, y) is a function of two variables t and y then the partial derivative f t of f(t, y) is the derivative of f(t, y) with respect to t, while pretending y is a constant. The partial derivative f y is the derivative of f(t, y) with respect to y, while pretending t is constant. The precise definitions are f t (t, y) = lim h 0 f(t+h,y) f(t,y) h and f y (t, y) = lim h 0 f(t,y+h) f(t,y) h. Example 4.1 Find f t Solution. We have and f y if f(t, y) = t4 y 3 + t 5. f t (t, y) = 4t3 y 3 + 5t 4 and f y (t, y) = 3t4 y 2 29

304 EXISTENCE AND UNIQUENESS OF SOLUTIONS TO FIRST ORDER NONLINEAR IV The major existence and uniqueness theorem is stated next. Theorem 4.1 Suppose that f(t, y) and f y (t, y) are continuous on the open rectangle Then for any (t 0, y 0 ) in R the IVP R = {(t, y) : a < t < b, c < y < d}. y = f(t, y), y(t 0 ) = y 0 has a unique solution defined on an interval of the form [t 0 h, t 0 + h] [a, b] for some positive h. Remark 4.1 Although existence can be proved with no hypotheses on f beyond continuity, some assumption such as the continuity of f y is necessary for uniqueness. For example, the IVP y = y 1 3, y(0) = 0 has as solutions the functions y(t) 0 and y(t) = { 0 if t 0 ( 2 3 t) 3 2 if t > 0. Note that f y (t, y) = 1 3 y 2 3 is not continuous at y = 0 Remark 4.2 If the conditions of Theorem 4.1 are not satisfied then this does not mean that the problem does not have a unique solution. Example 4.2 Consider the differential equation y = y 1 3 t(y 2). Does the existence theorem guarantee the existence of a unique solution to the following IVPs: (a) y(3) = 4 (b) y(0) = 7 (c) y(0) = 2 (d) y(1) = 2?

31 Solution. The function f(t, y) = y 1 3 t(y 2) is continuous for t 0 and y 2. The function f (t, y) = y 2 2y 3t(y 2) 2 y 2 3 is continuous for t 0 and y 0, 2. Thus, f and f y are continuous for t 0 and y 0, 2. Theorem 4.1 guarantees the existence of a unique solution for the initial value problem in (a) whereas there is no guarantee that there is a unique solution or any solution for the remaining IVPs Example 4.3 On what rectangle do we expect a unique solution to the IVP Solution. We have and y = y2, y(0) = 0? 1 t2 f(t, y) = f (t, y) = 2y y y2 1 t 2 1 t 2. These are both continuous functions as long as we avoid the lines t = ±1. Theorem 4.1 tells us that we can expect one and only one solution of in the strip Example 4.4 Consider the IVP y = y2 1 t 2, y(0) = 0 R = {(t, y) : 1 < t < 1, < y < } y = 1 2 ( t + t 2 + 4y), y(2) = 1. (a) Show that y(t) = 1 t and y(t) = t2 4 (b) Does this contradict Theorem 4.1? are two solutions to the above IVP. Solution. (a) You can verify that the two functions are solutions by substitution. (b) Since f(t, y) = 1 2 ( t+ t 2 1 + 4y) and f y (t, y) =, these two functions are t 2 +4y not continuous at (2, 1). Thus, we can not apply Theorem 4.1 for this problem

324 EXISTENCE AND UNIQUENESS OF SOLUTIONS TO FIRST ORDER NONLINEAR IV Example 4.5 Consider the initial value problem: t 2 y y 2 = 0, y(1) = 1. (a) Determine the largest open rectangle in the ty plane, containing the point (t 0, y 0 ) = (1, 1), in which the hypotheses of Theorem 4.1 are satisfied. (b) A solution of the initial value problem is y(t) = t. This solution exists on < t <. Does this fact contradicts Theorem 4.1? Explain your answer. Solution. (a) We have f(t, y) = y2 t 2, f y (t, y) = 2y t 2. So R = {(t, y) : 0 < t <, < y < }. (b) No. Theorem 4.1 is a local existence theorem and not a global one Example 4.6 Is it possible to find a function f(t, y) that is continuous and has continuous partial derivatives in the entire ty plane such that the functions y 1 (t) = cos t and y 2 (t) = 1 sin t are both solutions to the initial value problem y = f(t, y), y ( π 2 ) = 0? Solution. Since f is continuous and has continuous partial derivatives in the entire ty-plane, the equation y = f(t, y) satisfies the conditions of Theorem 4.1. Notice that y 1 ( π 2 ) = y 2( π 2 ) = 0, so the curves y 1(t) = cos t and y 2 (t) = 1 sin t have a common point ( π 2, 0), so if they were both solutions of our equation, by the uniqueness theorem they would have to agree on any rectangle containing ( π 2, 0). Since they do not, they cannot both be solutions of the equation y = f(t, y)

33 Practice Problems For the given initial value problem in Problems 4.1-4.5, (a) Rewrite the differential equation, if necessary, to obtain the form y = f(t, y), y(t 0 ) = y 0. Identify the function f(t, y). (b) Compute f f y. Determine where in the ty plane both f(t, y) and y are continuous. (c) Determine the largest open rectangle in the ty plane that contains the point (t 0, y 0 ) and in which the hypotheses of Theorem 4.1 are satisfied. Problem 4.1 3y + 2t cos y = 1, y( π 2 ) = 1. Problem 4.2 3ty + 2 cos y = 1, y( π 2 ) = 1. Problem 4.3 Problem 4.4 Problem 4.5 2t + (1 + y 3 )y = 0, y(1) = 1. (y 2 9)y + e y = t 2, y(2) = 2. cos yy = 2 + tan t, y(0) = 0. Problem 4.6 Give an example of an initial value problem for which the open rectangle R = {(t, y) : 0 < t < 4, 1 < y < 2} represents the largest region in the ty plane where the hypotheses of Theorem 4.1 are satisfied.

344 EXISTENCE AND UNIQUENESS OF SOLUTIONS TO FIRST ORDER NONLINEAR IV Problem 4.7 Can we apply Theorem 4.1 to the following problem? Explain what (if anything) we can conclude, and why (or why not): y = y t, y(0) = 2. Problem 4.8 Consider the differential equation y = t y t+y. For which of the following initial value conditions does Theorem 4.1 apply? (a) y(0) = 0 (b) y(1) = 1 (c) y( 1) = 1. Problem 4.9 Does the initial value problem y = y t Theorem 4.1? + 2, y(0) = 1 satisfy the conditions of Problem 4.10 Does the initial value problem y = y sin y + t, y(0) = 1 satisfy the conditions of Theorem 4.1?

5 Separable Differential Equations In this section, we discuss an analytical method for solving a type of nonlinear first order differential equations, the separable differential equations. A first order differential equation is separable if it can be written with one variable only on the left and the other variable only on the right: f(y)y = g(t). To solve this equation, we proceed as follows. Let F (t) be an antiderivative of f(t) and G(t) be an antiderivative of g(t). Then by the Chain Rule Thus, It follows that which is equivalent to d df dy F (y) = dt dy dt = f(y)y. f(y)y g(t) = d dt F (y) d dt G(t) = d [F (y) G(t)] = 0. dt F (y) G(t) = C f(y)y dt = g(t)dt + C. As you can see, the result is generally an implicit equation involving a function of y and a function of t. It may or may not be possible to solve this to get y explicitly as a function of t. For an initial value problem, substitute the values of t and y by t 0 and y 0 to get the value of C. Remark 5.1 If F is a differentiable function of y and y is a differentiable function of t and both F and y are given then the chain rule allows us to find df dt given by df dt = df dy dy dt. 35

36 5 SEPARABLE DIFFERENTIAL EQUATIONS For separable equations, we are given f(y)y = df dt and we are asked to find F (y). This process is referred to as reversing the chain rule. Example 5.1 Solve the initial value problem y = 6ty 2, y(1) = 1 25. Solution. Since f(t, y) = 6ty 2 and f y (t, y) = 12ty are continuous in the rectangle R = {(t, y) : < t <, < y < } and R contains the point ( 1, 1 25), by Theorem 4.1, the IVP has a unique solution on some interval containing t = 1. Separating the variables and integrating both sides we obtain or Thus, y y 2 dt = ( ) 1 dt = y 6tdt 1 y(t) = 3t2 + C. 6tdt. Since y(1) = 1 25, we find C = 28. The unique solution to the IVP is then given explicitly by 1 y(t) = 28 3t 2. The next question is the question of the interval of existence of this solution. Recall that there are two conditions that define an interval of validity. First, it must be a continuous interval with no breaks or holes in it. Second it must contain the value of the independent variable in the initial condition, t = 1 in this case. There are three possible intervals where y(t) is continuous: 28 < t < 3, 28 28 28 3 < t < 3, t > 3. Only one of these will contain the value of t from the initial condition and so we can see that 28 28 3 < t < 3

37 must be the interval of existence for this solution. Figure 5.1 shows the graph of the solution Figure 5.1 Example 5.2 Solve the IVP yy = 4 sin (2t), y(0) = 1. Solution. This is a separable differential equation. Integrating both sides we find ( ) y 2 dt = 4 sin (2t)dt. 2 Thus, y 2 = 4 cos (2t) + C. Since y(0) = 1, we find C = 5. Now, solving explicitly for y(t) we find y(t) = ± 4 cos t + 5. Since y(0) = 1, we find y(t) = 4 cos t + 5. The interval of existence of the solution is the interval < t < Example 5.3 Solve the initial value problem y = 1 y 2, y(0) = 0. Solution. Separating the variables and then integrating we find y dt = dt 1 y 2

38 5 SEPARABLE DIFFERENTIAL EQUATIONS or arcsin y = t + C. Since y(0) = 0, we find C = 0 and consequently y(t) = sin t where π 2 t π 2. Now, notice that y 1 (t) = 1 and y 1 (t) = 1 are solutions to the differential equations. Moreover, y( π 2 ) = 1 and y( π 2 ) = 1 so that the graph of y = arcsin t is connected to the solution y 1 at the point t = π 2 and to the solution y 2 at the point t = π 2. Thus, the solution to the given IVP is 1, < t < π 2 y(t) = sin t, π 2 t π 2 π 1 2 < t <. The graph of this function is shown in Figure 5.2 Figure 5.2 Linear Versus Nonlinear Differential Equations Next, we turn our attention to some general questions about differential equations and explore in more detail some important ways in which nonlinear differential equations differ from linear ones. 1. Existence and Uniqueness of Solutions If the functions p(t) and g(t) for a linear differential equation are continuous on an interval a < t < b containing the initial condition t 0 then Theorem 2.2 asserts the existence of a unique solution throughout the interval a < t < b. This interval can be identified by glancing at the functions p(t) and g(t). Turning now to nonlinear differential equations, the function f(t, y) = 6ty 2 in Example 5.1 is continuous everywhere however the interval of existence of solution to the initial value problem y = 6ty 2 and y(1) = 1 25 is 28 3 < t < 28 3 and one cannot predict this by simply looking at the equation y = 6ty 2. 2. General Solutions: There may not be a single formula that gives all solutions

39 For a first order linear differential equation the general solution involves a constant from which all particular solutions are obtained by specifying values for this constant. In contrast, this is not the case for nonlinear differential equations. If we consider the initial value problem y = 6ty 2, y(0) = 0 then we see that the 1 3t 2 +C family of solutions y(t) = does not include the zero solution. Indeed, there is no choice of C that yields the unique solution y 0 guaranteed by Theorem 4.1. Such a solution is called a singular solution. 3. Implicit Solutions Recall again that, for an initial value problem of the form y +p(t)y = g(t), y(t 0 ) = y 0, the unique solution is always given explicitly. In contrast, the situation for nonlinear differential equations is much less satisfactory. For example, the solutions in Examples 5.1-5.3, are defined explicitly. However, the unique solution to the initial value problem (2y sin y)y = sin t t, y(0) = 0 is defined by the implicit equation y 2 + cos y + cos t + t2 2 = 2 (See Problem 5.11) where y can not be written explicitly as a function of t.

40 5 SEPARABLE DIFFERENTIAL EQUATIONS Practice Problems Problem 5.1 Solve the (separable) differential equation y = te t2 ln y 2. Problem 5.2 Solve the (separable) differential equation y = t2 y 4y t + 2. Problem 5.3 Solve the (separable) differential equation ty = 2(y 4). Problem 5.4 Solve the (separable) differential equation y = 2y(2 y). Problem 5.5 Solve the IVP: y = 4 sin (2t), y(0) = 1. y Problem 5.6 Solve the IVP: Problem 5.7 Solve the IVP: Problem 5.8 Solve the IVP: Problem 5.9 Solve the IVP: yy = sin t, y( π 2 ) = 2. y + 1 = 0, y(1) = 0. y + 1 y ty 3 = 0, y(0) = 2. y = 1 + y 2, y( π 4 ) = 1.

41 Problem 5.10 Solve the IVP: Problem 5.11 Solve the IVP: y = t ty 2, y(0) = 1 2. (2y sin y)y = sin t t, y(0) = 0. Problem 5.12 For what values of the constants α, y 0, and integer n is the function y(t) = (4+t) 1 2 a solution of the initial value problem y + αy n = 0, y(0) = y 0? Problem 5.13 State an initial value problem, with initial condition imposed at t 0 = 2, having implicit solution y 3 + t 2 + sin y = 4. Problem 5.14 Consider the initial value problem y = 2y 2, y(0) = y 0. For what value(s) of y 0 will the solution have a vertical asymptote at t = 4, where the t interval of existence is < t < 4? Problem 5.15 Assume that y sin y 3t + 3 = 0 is an implicit solution of the initial value problem y = f(y), y(1) = 0. What is f(y)? What is an implicit solution to the initial value problem y = t 2 f(y), y(1) = 0? Problem 5.16 Find all the solutions to the differential equation y = 2ty 1+t. Problem 5.17 Solve the initial-value problem y = cos 2 y cos 2 t, y(0) = π 4. Problem 5.18 Solve the initial-value problem y = e t+y, y(0) = 0 and determine the interval on which the solution y(t) is defined.

42 5 SEPARABLE DIFFERENTIAL EQUATIONS Problem 5.19 (a) Solve the initial-value problem y = t2 ey e y t 2. State the name of the method you are using. (b) Find the solution which satisfies the condition y(1) = 1.

6 Exact Differential Equations We shall now present another technique for solving first order, non-linear, ordinary differential equations. This technique is a generalization of the one we used for separable equations. We have seen that the solution procedure of separable equations consists of reversing the chain rule. This same procedure works for exact equations but this time the chain rule is for functions of two variables. The Extended Chain Rule You recall the chain rule for functions of one variable: If u is differentiable at t and f is differentiable at u(t) then the composite function y = f(u(t)) is also differentiable at t with derivative given by dy dt = dy du du dt. Example 6.1 Find the derivative of the function y = e t. Solution. Let u(t) = t and f(t) = e t. Then du dt = 1 2 t dy dt = 1 eu 2 t = e t 2 t and dy du = eu. Hence, The above chain rule can be extended to functions of two variables. Suppose that u and v are differentiable at t and f is a differentiable function of u and v. Then the function z(t) = f(u(t), v(t)) is differentiable at t with derivative dz dt = f du u dt + f dv v dt. Example 6.2 Let z = f(u, v) = u 2 + 2u uv + v 2 where u(t) = t 2 + 1 and v(t) = t 3 t 2. Find dz t=2 in two different ways. dt 43

44 6 EXACT DIFFERENTIAL EQUATIONS Solution. First notice that u(2) = 5 and v(2) = 4. By using the extended chain rule we have Thus, dz dt = f du u dt + f dv v dt =(2u + 2 v)(2t) + (2v u)(3t 2 2t). dz dt = (10 + 2 4)(4) + (8 5)(8) = 56. t=2 A different way for finding the derivative is to write z as only a function of t obtaining z(t) = t 6 3t 5 + 3t 4 t 3 + 5t 2 + 3. Finding the derivative of z(t) Finally, z (2) = 56 z (t) = 6t 5 15t 4 + 12t 3 3t 2 + 10t. Exact Differential Equations The basic idea underlying separable equations is to reverse the chain rule for functions of one variable. The basic idea underlying exact equations is to reverse the extended chain rule. To this end, consider the differential equation M(t, y) + N(t, y) dy dt Let H(t, y) be a function satisfying the two conditions H (t, y) = M(t, y) and H t Then Equation (6.1) can be written as H t + H dy y dt By the extended chain rule, Equation (6.3) is the same as d H(t, y) = 0. dt Therefore, we obtain an implicitly defined solution given by H(t, y) = C. = 0. (6.1) (t, y) = N(t, y). (6.2) y = 0. (6.3)

45 An equation like (6.1) is called exact if there is a function H(t, y) satisfying the conditions in (6.2). Testing a Differentiable Equation for Exactness The next question is the question of telling whether or not Equation (6.1) is exact. This is answered by the following theorem. Theorem 6.1 Suppose that the functions M(t, y) and N(t, y) in (6.1) are continuous and have continuous first partial derivatives M N y and t in an open rectangle Then (6.1) is exact in R if and only if for all (t, y) in R. R = {(t, y) : a < t < b, c < y < d}. M y = N t Remark 6.1 Every separable differential equation is exact. Indeed, since g(t) + f(y)y = 0 we have g f y = 0 and t = 0. However, not every exact equation is separable. For example, the differential equation (2t + y) + (2y + t)y = 0 is exact since M y = 1 = N t = 1. This equation is clearly not separable Example 6.3 Determine whether or not the equation is exact. (a) ty 2 + t + t 2 yy = 0. (b) y 2 + 1 + tyy = 0. (c) cos y + (y 2 + t sin y)y = 0. (d) cos y + (y 2 t sin y)y = 0. Solution. (a) Since y (ty2 + t) = 2ty and t (t2 y) = 2ty, the given equation is exact. (ty) = y, the given equation is not exact. (b) Since y (y2 + 1) = 2y and t (c) Since y (cos y) = sin y and t (y2 + t sin y) = sin y, the given equation is not exact. (d) Since y (cos y) = sin y and t (y2 t sin y) = sin y, the equation is exact

46 6 EXACT DIFFERENTIAL EQUATIONS Example 6.4 Consider the initial value problem t + y + (t + 2y)y = 0, y(0) = 1. Show that the differential equation is exact and solve the IVP. Solution. We have M(t, y) = t + y and N(t, y) = t + 2y. Since M(t, y) y = N(t, y) t = 1 we have by Theorem 6.1 that the differential equation is exact. Thus, H(t, y) = (t + y)dt = ty + t2 2 + c 1(y). Hence, It follows that t + 2y = H(t, y) y = t + c 1(y). c 1 (y) = (2y)dy = y 2 + C. Hence, ty + y 2 + t2 2 = C. Since y(0) = 1, we find C = 1. Thus, y satisfies the implicit equation 2ty + 2y 2 + t 2 = 2

47 Practice Problems Problem 6.1 Find f t Problem 6.2 Find f t and f y if f(t, y) = y ln y e ty. and f y if f(t, y) = ln (ty) + t2 +1 y 5. Problem 6.3 Let f(u, v) = 2u 3uv where u(t) = 2 cos t and v(t) = 2 sin t. Find df dt. In Problems 6.4-6.8, determine whether the given differential equation is exact. If the equation is exact, find an implicit solution and (where possible) an explicit solution. Problem 6.4 Problem 6.5 Problem 6.6 Problem 6.7 Problem 6.8 yy + 3t 2 2 = 0, y( 1) = 2. y = (3t 2 + 1)(y 2 + 1), y(0) = 1. (6t + y 3 )y + 3t 2 y = 0, y(1) = 2. (e t+y + 2y)y + (e t+y + 3t 2 ) = 0, y(0) = 0. (sin (t + y) + y cos (t + y) + t + y)y + (y cos (t + y) + y + t) = 0, y(1) = 1. Problem 6.9 For what values of the constants m, n, and α (if any) is the following differential equation exact? t m y 2 y + αt 3 y n = 0.

48 6 EXACT DIFFERENTIAL EQUATIONS Problem 6.10 Assume that N(t, y)y +t 2 +y 2 sin t = 0 is an exact differential equation. Determine the general form of N(t, y). Problem 6.11 Assume that t 3 y + e t + y 2 = 5 is an implicit solution to the differential equation N(t, y)y + M(t, y) = 0, y(0) = y 0. Determine possible functions M(t, y), N(t, y), and the possible value(s) for y 0. Problem 6.12 Assume that y = t 4 t 2 is an explicit solution of the following initial value problem (y + at)y + (ay + bt) = 0, y(0) = y 0. Determine values for the constants a, b and y 0. Problem 6.13 Let k be a positive constant. Use the exactness criterion to determine whether or not the population equation dp dt = kp is exact. Do NOT try to solve the equation or carry out any further calculation. Problem 6.14 Consider the differential equation (2t + 3) + (2y 2)y = 0. Determine whether this equation is exact or not. If it is, solve it. Problem 6.15 Consider the differential equation (ye 2ty + t) + bte 2ty y = 0. Determine for which value of b this equation is exact, and then solve it with this value of b. Problem 6.16 Consider the differential equation y + (2t ye y )y = 0. Check that this equation is not exact. Now multiply the equation by y. Check that the new equation is exact, and solve it. Problem 6.17 (a) Consider the differential equation y + p(t)y = g(t) with p(t) 0. Show that this equation is not exact. (b) Let µ(t) = e p(t)dt. Show that the equation is exact and solve it. µ(t)(y + p(t)y) = µ(t)g(t)

49 Problem 6.18 Use the method of the previous problem to solve the linear, first-order equation y y t = 1, with initial condition y(1) = 7. First, check that this equation is not exact. Next, find µ(t). Multiply the equation by µ(t) and check that the new equation is exact. Solve it, using the method of exact equations. Problem 6.19 Put the following differential equation in the Exact Differential Equation form and find the general solution y = y3 2ty t 2 3ty 2. Problem 6.20 The following differential equations are exact. Solve them by that method. (a) (4t 3 y + 4t + 4)y = 8 4y 6t 2 y 2, y( 1) = 1. (b) (6 4y + 16t) + (10y 4t + 2)y = 0, y(1) = 2.

50 6 EXACT DIFFERENTIAL EQUATIONS

7 Substitution Techniques: Bernoulli and Riccati Equations A well-known non-linear equation that reduces to a linear one with an appropriate substitution is the Bernoulli equation given by y + p(t)y = g(t)y n (7.1) where n is a real number different from 0 and 1. Notice that for n = 0 or n = 1 the equation becomes linear. To solve (7.1), first notice that y(t) 0 is the trivial solution. If y(t) 0 then (7.1) can be written as y y n + p(t)y1 n = g(t). (7.2) Let z = y 1 n. Then by the chain rule of differentiation we have z = (1 n)y n y and therefore y y = 1 n 1 n z. Thus, (7.1) reduces to 1 1 n z + p(t)z = g(t) which is a first order linear differential equation that can be solved by the technique of integrating factor. Once z is found then the desired solution is y(t) = z 1 1 n. Example 7.1 Solve the Bernoulli equation y 1 t y = ty2, t > 0. Solution. Divide by y 2 and then let z = y 1 to obtain z + 1 t z = t. 51

527 SUBSTITUTION TECHNIQUES: BERNOULLI AND RICCATI EQUATIONS The integrating factor is µ(t) = t and the general solution is z(t) = 1 t t 2 dt + Ct 1 = t2 3 + Ct 1. Thus, the solutions to the differential equation are y = 1 z = 1 and y(t) = t2 +Ct 1 3 0 Example 7.2 Solve the Bernoulli equation y + 3y = e 3t y 2. Solution. Dividing by y 2 to obtain Let z = y 1. Then, y 2 y + 3y 1 = e 3t. z 3z = e 3t. Solving this equation using the integrating factor method with µ(t) = e 3t we find z(t) = e 3t e 3t ( e 3t )dt + Ce 3t = te 3t + Ce 3t = e 3t (C t). The solutions are y(t) = e 3t (C t) 1 and y(t) = 0 Example 7.3 Solve: y + 2 t y = t2 y 2 cos t. Solution. Dividing through by y 2 to obtain y 2 y + 2 t y 1 = t 2 cos t. Letting z = y 1 to obtain z 2 t z = t2 cos t. Solving this equation by the method of integrating factor with µ(t) = 1 t 2 z(t) = t 2 cos tdt + Ct 2 = t 2 sin t + Ct 2. we find The solutions are y(t) = (t 2 sin t + Ct 2 ) 1 and y(t) = 0

53 Riccati Equation A first order differential equation is called a Riccati equation if it can be written in the form: y = a(t)y 2 + b(t)y + c(t) (7.3) where a, b and c are functions of t. Clearly, this is a first order nonlinear and nonseparable differential equation. The solution of a Riccati equation requires knowledge of a particular solution to (7.3). To solve (7.3), we first find a particular solution y 1 to (7.3). Then we use the substitution 1 z = y y 1. Thus, y = 1 z + y 1. Now, (7.3) reduces to z z 2 + y 1 =a(t)( 1 z 2 + 2y 1 z + y2 1) + b(t)( 1 z + y 1) + c(t) z z 2 + a(t)y2 1 + b(t)y 1 + c(t) = a(t) z 2 + 2a(t)y 1 + a(t)y1 2 + b(t) z z + b(t)y 1 + c(t) z = a(t) [2a(t)y 1 + b(t)]z. Thus, a Riccati equation can be reduced to a linear equation z + [2a(t)y 1 + b(t)]z = a(t) (7.4) that can be solved by the method of integrating factor. Once z(t) is found then the solution to the original equation is y(t) = 1 z(t) + y 1(t). As the next example illustrates, in many cases a solution of a Riccati equation cannot be expressed in terms of elementary functions. Example 7.4 Solve : y = 2 2ty + y 2 given that y 1 (t) = 2t. Solution. We have a(t) = 1, b(t) = 2t, and c(t) = 2. Substituting in (7.4) to obtain The integrating factor is µ(t) = e t2 z + 2tz = 1. so that ( e t2 z) = e t 2.

547 SUBSTITUTION TECHNIQUES: BERNOULLI AND RICCATI EQUATIONS Now, the integral t t 0 e s2 ds cannot be expressed in terms of elementary functions. Thus, we write t e t2 z(t) = e s2 ds + e t2 0 z(t0 ). t 0 Solving for z, we find t z(t) = e t2 (e t2 0 z(t0 ) e s2 ds). t 0 Finally, y(t) = 1 e t2 (e t2 0z(t 0 ) t t 0 e s2 ds) + 2t Example 7.5 Solve the IVP: y = 2 + 2y + y 2, variables. y(0) = 0 using the method of separation of Solution. Notice first that 2 + 2y + y 2 = 1 + (1 + y) 2. Separating the variables we find y 1 + (1 + y) 2 = 1. Integrating both sides with respect to t to obtain But y(0) = 0 so that C = π 4. Thus, arctan (1 + y) = t + C. arctan (1 + y) = t + π 4. For the inverse arc tangent we must have π 2 < t + π 4 < π 2 or 3π 4 < t < π 4. Hence, y(t) = tan (t + π 4 ) 1, 3π 4 < t < π 4 Example 7.6 Solve the differential equation y = 5 t 2 + 2ty y 2 given that y 1 (t) = t 2 is a particular solution.

55 Solution. Let 1 z = y t + 2. Then the given equation reduces to z + 4z = 1. Solving this equation by the method of integrating factor with µ(t) = e 4t to obtain z(t) = e 4t e 4t dt + Ce 4t = 1 4 + Ce 4t. Thus, y(t) = ( 1 4 + Ce 4t ) 1 + t 2

567 SUBSTITUTION TECHNIQUES: BERNOULLI AND RICCATI EQUATIONS Practice Problems Problem 7.1 Solve the Bernoulli equation y = t2 + 3y 2, t > 0. 2ty Problem 7.2 Find the general solution of y + ty = te t2 y 3. Problem 7.3 Solve the IVP ty + y = t 2 y 2, y(0.5) = 0.5. Problem 7.4 Solve the IVP y 1 t y = y2, y(1) = 1. Problem 7.5 Solve the IVP y = y(1 y), y(0) = 1 2. Problem 7.6 Solve y + y = ty 4. Problem 7.7 Solve the differential equation y = 1 + t 2 y 2 given that y 1 (t) = t is a particular solution. Problem 7.8 Perform a change of variable that changes the Bernoulli equation y + y + y 2 = 0 into a linear equation in the new variable. Do NOT try to solve the equation or proceed further than with any calculations. Problem 7.9 Consider the equation y = ɛy σy 3, ɛ > 0, σ > 0. (a) Use the Bernoulli transformation to change this nonlinear equation into a linear equation. (b) Solve the resulting linear equation in part (a) and use the solution to find the solution of the given differential equation above.

57 Problem 7.10 Consider the differential equation y = f ( y t ). (a) Show that the substitution z = y t leads to a separable differential equation in z. (b) Use the above method to solve the initial-value problem Problem 7.11 Solve: y + y 3 = et y 4. Problem 7.12 Solve: ty + y = ty 3. Problem 7.13 Solve: ty + y = t 2 y 2 ln t. y = t + y, y(1) = 0. t y Problem 7.14 Verify that y 1 (t) = 2 is a particular solution to the Riccati equation and then find the general solution. Problem 7.15 Verify that y 1 (t) = 2 t and then find the general solution. y = 2 y + y 2, is a particular solution to the Riccati equation y = 4 t 2 1 t y + y2,

587 SUBSTITUTION TECHNIQUES: BERNOULLI AND RICCATI EQUATIONS

8 Graphical Solution: Direction Field of y = f(t, y) Since explicit solutions of differential equations are often unobtainable, we explore methods of finding properties of solutions from the differential equation itself; the principal tool is the geometry of direction field. A direction field (also known as slope field) consists of an array of short line segments in the ty-plane having the property that the line plotted at a point (t, y) has slope f(t, y). Direction fields are basically used to visualize the family of solutions of a given differential equation without the need of solving the equation. Direction fields give qualitative information about solutions of ODEs. In this section we use direction fields for solving initial value problems of the form dy dt = f(t, y), y(t 0) = y 0. In the special case where f(t, y) = f(y), i.e. the independent variable t does not appear on the right side, the first order DE dy dt = f(y) is called autonomous. Producing slope fields by hand is a daunting task. For that reason, slope fields are usually created by means of an electronic generator. One such a generator can be found at http://slopefield.nathangrigg.net/ Example 8.1 Find the direction field of the differential equation dy dt = 2t. What is the form of the general solution? through (0, 1). Graph the particular solution going Solution. Figure 8.1 shows the slope field and the graph of the particular solution to the given 59

60 8 GRAPHICAL SOLUTION: DIRECTION FIELD OF Y = F (T, Y ) DE passing through the point (0, 1). The solution curves look like parabolas. Thus, the general solution is given by the equation y = t 2 + C Figure 8.1 Example 8.2 Using direction field, guess the form of the solution curves of the differential equation dy dt = t y. Solution. The direction field is given in Figure 8.2. The solution curves look like circles centered at the origin. Thus, the general solution is given implicitly by the equation t 2 + y 2 = C where C is a positive constant Figure 8.2

61 Remark 8.1 We point out here that even though one can draw solution curves, some do not have simple formulas. For instance, the equation dy dt = 1+yety 1+te does not have ty explicit solutions. Equilibrium Solutions and Stability for Autonomous Equations A physical system is often said to be in equilibrium if it doesn t change in time. We adopt this idea and say that a solution to a differential equation is an equilibrium solution if it is a constant function. Thus, in a direction field of an autonomous equation, equilibrium solutions are solution curves represented by horizontal lines. It follows that the equations of such solutions have the form y(t) c where c is a constant. The following result tells us where to look for equilibrium solutions. Theorem 8.1 The function y(t) c, where c is a constant, is an equilibrium solution to y = f(y) if and only if c is a root of f(y) = 0. Example 8.3 Find the equilibrium solutions to the DE dy dt = 2y(1 y). Solution. The roots of f(y) = 2y(1 y) = 0 are y = 0 and y = 1. According to the previous theorem, the equilibrium solutions are y(t) 0 and y(t) 1. The direction field of the DE is shown in Figure 8.3 Figure 8.3

62 8 GRAPHICAL SOLUTION: DIRECTION FIELD OF Y = F (T, Y ) Remark 8.2 Equilibrium solutions can be defined for nonautonomous differential equations. For example, the function y(t) 1 is an equilibrium solution to the DE y = (1 y)t 2. The direction field of a given differential equation indicates that as t increases without bound, every solution either moves towards or moves away from an equilibrium solution. If all nearby solutions move towards a certain equilibrium solution, then that equilibrium solution is called asymptotically stable, stable, or attracting. The solution y = 1 in Figure 8.3 is attracting. An equilibrium solution is called unstable or repelling when all nearby solutions move away from it. The solution y = 0 in Figure 8.3 is repelling. If solutions on one side of an equilibrium solution move towards the equilibrium solution and on the other side of the equilibrium solution move away from it then we call the equilibrium solution semi-stable. An equilibrium solution does not necessarily have to be either attracting or repelling. The next example illustrates this situation. Example 8.4 Sketch the field direction of the differential equation y = 4y(1 y) 2. Show that y = 1 is neither stable or unstable. Solution. The direction field is shown in Figure 8.4. Note that the equilibrium solution y(t) 1 is neither stable or unstable. Nearby solutions that start below it are attracted upward towards it but nearby solutions that start above it are repelled upward and away from it Figure 8.4

63 A very suitable qualitative representation of a differential equation for the study of stability is the so-called phase line. A phase line consists of solid dots and arrows. The solid dots represent the equilibrium points and the arrows indicate the directions that solutions move as t increases. Figure 8.5(a) shows an example of a phase line. We see that the equilibrium b is stable, whereas the equilibria a and c are unstable. Example 8.5 Consider the autonomous differential equation dy dt = f(y) where the graph of f(y) is given in Figure 8.5(b). (a) Sketch the phase line. (b) Sketch the direction field of this differential equation. (c) Sketch the graph of the solution to the IVP y = f(y), y(0) = 3 2. Find lim t y(t). (d) Sketch the graph of the solution to the IVP y = f(y), y(0) = 1 2. Find lim t y(t). (e) Sketch the graph of the solution to the IVP y = f(y), y(0) = 3 2. Find lim t y(t). Figure 8.5 Solution. (a) Note that for y < 1, f(y) < 0 so that the solution y(t) is decreasing. For 1 < y(t) < 0, we have f(y) > 0 so that y(t) is increasing. For 0 < y(t) < 1, we have that f(y) < 0 so that y(t) is decreasing. Finally, for y(t) > 1 we have that f(y) > 0 so that y(t) is increasing. Hence, the phase line is given by Figure 8.5(c). (b) The direction field is given in Figure 8.6. (c) We notice from Figure 8.6 that lim t y(t) =. (d) We notice from Figure 8.6 that lim t y(t) = 0. (e) We notice from Figure 8.6 that lim t y(t) =

64 8 GRAPHICAL SOLUTION: DIRECTION FIELD OF Y = F (T, Y ) Figure 8.6

65 Practice Problems Problem 8.1 Sketch the direction field for the differential equation in the window 3 t 3, 3 y 3. (a) y = y (b) y = t y. Problem 8.2 Match each direction field with the equation that the slope field could represent. Each direction field is drawn in the portion of the ty-plane defined by 6 t 6, 4 y 4. (a) y = t (b) y = sin t (c) y = 1 y (d) y = y(2 y). Problem 8.3 State whether or not the equation is autonomous. (a) y = t (b) y = sin t (c) y = 1 y (d) y = y(2 y). Problem 8.4 Find all the equilibrium solutions of each of the autonomous differential equations below (a) y = (y 1)(y 2). (b) y = (y 1)(y 2) 2. (c) y = (y 1)(y 2)(y 3). Problem 8.5 Find an autonomous differential equation with an equilibrium solution at y = 1 and satisfying y < 0 for < y < 1 and 1 < y <.

66 8 GRAPHICAL SOLUTION: DIRECTION FIELD OF Y = F (T, Y ) Problem 8.6 Find an autonomous differential equation with no equilibrium solutions and satisfying y > 0. Problem 8.7 Find an autonomous differential equation with equilibrium solutions y = n 2, where n is an integer. Problem 8.8 Find an autonomous differential equation with equilibrium solutions y = 0 and y = 2 and satisfying the properties y > 0 for 0 < y < 2; y < 0 for y < 0 or y > 2. Problem 8.9 Classify whether the equilibrium solutions are stable, unstable, or neither. (a) y = 1 y 2. (b) y = (y + 1) 2. Problem 8.10 Consider the direction field below. Classify the equilibrium points, as asymptotically stable, semi-stable, or unstable. Problem 8.11 Sketch the direction field of the equation y = y 3. Sketch the solution satisfying the condition y(1) = 1. What is the domain of this solution? Problem 8.12 Find the equilibrium solutions and determine their stability y = y 2 (y 2 1).

67 Problem 8.13 Find the equilibrium solutions of the equation y = y 2 4y then decide whether they are stable or unstable. What is the long-run behavior if y(0) = 5?y(0) = 4?y(0) = 3? Problem 8.14 What is lim t y(t) for the initial-value problem y = sin (y(t)), y(0) = π 2?

68 8 GRAPHICAL SOLUTION: DIRECTION FIELD OF Y = F (T, Y )

9 Numerical Solutions to ODEs: Euler s Method and its Variants Whenever a mathematical problem is encountered in science or engineering, which cannot readily or rapidly be solved by a traditional mathematical method, then a numerical method is usually sought and carried out. In this section, we study Euler s method and its variants for approximating the solution to the initial value problem y = f(t, y), y(t 0 ) = y 0, a < t < b. (9.1) Denote the (unique) solution by y(t). Then y(t) satisfies y (t) = f(t, y(t)). A numerical method for solving the initial value problem provides an algorithm for calculating approximate values y 0, y 1, y 2,, y n, of the solution y(t) at a set of points t 0 < t 1 < t 2 < < t n <. Usually this is a step-by-step process. Euler s Method Euler s method is a very simple numerical method for solving the first-order initial value problem (9.1). Suppose we want to find approximate values y 0, y 1, y 2,, y n, of the solution y(t) at equally spaced points t 0 < t 1 < t 2 < < t n < with step size h. We begin by integrating Equation (9.1) between t n and t n+1 obtaining tn+1 t n y (s)ds = tn+1 t n f(s, y(s))ds. By the fundamental theorem of calculus we can write tn+1 t n y (s)ds = y(t n+1 ) y(t n ). 69

709 NUMERICAL SOLUTIONS TO ODES: EULER S METHOD AND ITS VARIANTS Thus, tn+1 y(t n+1 ) = y(t n ) + f(s, y(s))ds. (9.2) t n Computationally, this last equation can not be used since we do not know f(s, y(s)) for t n s t n+1. If we estimate the integral by a left Riemann sum we find tn+1 t n f(s, y(s))ds hf(t n, y(t n )). Hence, or y(t n+1 ) y(t n ) + hf(t n, y(t n )). y n+1 = y n + hf(t n, y n ) (9.3) where y n y(t n ). Equation (9.3) is known as Euler s method. Therefore, we can view Euler s method as a left Riemann sum approximation of the integral equation (9.2). Example 9.1 Suppose that y(0) = 1 and dy dt = y. Estimate y(0.5) in 5 steps using Euler s method. Solution. The step size is h = 0.5 0 5 = 0.1. The following chart lists the steps needed: k t k y k f(t k, y k )h 0 0 1 0.1 1 0.1 1.1 0.11 2 0.2 1.21 0.121 3 0.3 1.331 0.1331 4 0.4 1.4641 0.14641 5 0.5 1.61051 Thus, y(0.5) 1.61051. Note that the exact value is y(0.5) = e 0.5 = 1.6487213 The Improved Euler s Method: Heun s Method Euler s method is derived from estimating the integral equation by a left Riemann sum. A better estimate to the integral is obtained by using the trapezoid rule obtaining tn+1 t n f(s, y(s))ds h 2 [f(t n, y(t n )) + f(t n+1, y(t n+1 ))].

71 This leads to the estimate y n+1 = y n + h 2 [f(t n, y n ) + f(t n+1, y n+1 )]. Note that y n+1 is defined implicitly so solving for y n+1 requires solving a nonlinear equation say by the Newton s method for root-finding. Despite the inconvenience of working with an implicit method, the trapezoid method has two advantages over the ordinary Euler method: First, it is more accurate and second it is stable (this means that it is not sensitive to small changes in the initial values). Heun s method is based on the evaluation of an estimate of f(t n+1, y n+1 ) by replacing the value of y n+1 with an estimate derived from the original Euler method, with the resulting iteration scheme: ỹ n+1 = y n + hf(t n, y n ), y n+1 = y n + h 2 [f(t n, y n ) + f(t n+1, ỹ n+1 )] or y n+1 = y n + h 2 [f(t n, y n ) + f(t n+1, y n + hf(t n, y n ))]. This equation is known as Heun s method or the improved Euler s method. Example 9.2 Suppose that y(0) = 1 and dy dt Euler s method. = y. Estimate y(0.5) in 5 steps using the improved Solution. The step size is h = 0.5 0 5 = 0.1 The following chart lists the steps needed: y 0 =1 y 1 =y 0 + h 2 [f(t 0, y 0 ) + f(t 1, y 0 + hf(t 0, y 0 ))] = 1.105 y 2 =y 1 + h 2 [f(t 1, y 1 ) + f(t 2, y 1 + hf(t 1, y 1 ))] = 1.221025 y 3 =y 2 + h 2 [f(t 2, y 2 ) + f(t 3, y 2 + hf(t 2, y 2 ))] = 1.349232625 y 4 =y 3 + h 2 [f(t 3, y 3 ) + f(t 4, y 3 + hf(t 3, y 3 ))] = 1.490902050625 y 5 =y 4 + h 2 [f(t 4, y 4 ) + f(t 5, y 4 + hf(t 4, y 4 ))] = 1.647446765940625 Thus, y(0.5) 1.647446765940625. Note that the exact value is y(0.5) = e 0.5 = 1.6487213

729 NUMERICAL SOLUTIONS TO ODES: EULER S METHOD AND ITS VARIANTS The Modified Euler s Method Another numerical integration scheme is the midpoint rule. In this case, we have tn+1 f(s, y(s))ds hf (t n + h2 (, y t n + h )). 2 t n This implies that y(t n+1 ) y(t n ) + hf (t n + h2 (, y t n + h )). 2 Using Euler s method we can write ( y t n + h ) y(t n ) + h 2 2 f(t n, y(t n )). Hence, y n+1 = y n + hf ( t n + h 2, y n + h ) 2 f(t n, y n ). This last equation is known as the modified Euler s method. Example 9.3 Suppose that y(0) = 1 and dy dt Euler s method. = y. Estimate y(0.5) in 5 steps using the modified Solution. The step size is h = 0.5 0 5 = 0.1. The following chart lists the steps needed: y 0 =1 y 1 =y 0 + hf(t 0 + h 2, y 0 + h 2 f(t 0, y 0 )) = 1.105 y 2 =y 1 + hf(t 1 + h 2, y 1 + h 2 f(t 1, y 1 )) = 1.221025 y 3 =y 2 + hf(t 2 + h 2, y 2 + h 2 f(t 2, y 2 )) = 1.349232625 y 4 =y 3 + hf(t 3 + h 2, y 3 + h 2 f(t 3, y 3 )) = 1.490902050625 y 5 =y 4 + hf(t 4 + h 2, y 4 + h 2 f(t 4, y 4 )) = 1.647446765940625 Thus, y(0.5) 1.647446765940625. Note that the exact value is y(0.5) = e 0.5 = 1.6487213

73 Practice Problems In Problems 9.1-9.3, answer the following questions: (a) Solve the initial value problem analytically, using an appropriate solution technique. (b) For the given initial value problem express y n+1 in terms of y n using Heun s method. (c) For the given initial value problem express y n+1 in terms of y n using the Modified Euler s method. (d) Use a step size h = 0.1. Compute the first three approximations y 1, y 2, y 3 using the method in part (b). (e) Use a step size h = 0.1. Compute the first three approximations y 1, y 2, y 3 using the method in part (c). (f) For comparison, calculate and list the exact solution values y(t 1 ), y(t 2 ), y(t 3 ). Problem 9.1 Problem 9.2 Problem 9.3 Problem 9.4 Consider the initial value problem y = 2t 1, y(1) = 0. y = y, y(0) = 1. y 2 y + t = 0, y(0) = 1. y = 1 + y 2, y(0) = 1. (a) Find the exact solution of the given initial value problem. (b) Use step size h = 0.05. Compute 4 steps of Euler s method, Heun s method, and modified Euler s method. Compare the numerical values obtained at t = 1 by calculating the error y(1) y 4. Problem 9.5 Consider the initial value problem y + 2y = 4, y(0) = 3.

749 NUMERICAL SOLUTIONS TO ODES: EULER S METHOD AND ITS VARIANTS (a) Find the exact solution of the given initial value problem. (b) Use step size h = 0.05. Compute 4 steps of Euler s method, Heun s method, and modified Euler s method. Compare the numerical values obtained at t = 1 by calculating the error y(1) y 4. In Problems 9.6-9.8, the given iteration is the result of applying N steps of Euler s method, Heun s method, or the modified Euler s method to an initial value problem of the form y = f(t, y), y(t 0 ) = y 0, t 0 t t 0 + T. Identify the numerical method and determine t 0, N, T, and f(t, y). Problem 9.6 Problem 9.7 Problem 9.8 y n+1 = y n + h(y n + t 2 ny 3 n), y 0 = 1 t n = 2 + nh, h = 0.02, n = 0, 1, 2,, 49. y n+1 = y n + h ( t n + h ) sin (y 2 n + h2 ) 2 t n sin 2 y n, y 0 = 1 t n = nh, h = 0.01, n = 0, 1, 2,, 199. y n+1 = y n + h 2 [t ny 2 n + 2 + (t n + h)(y n + h(t n y 2 n + 1)) 2 ], y 0 = 2 t n = 1 + nh, h = 0.05, n = 0, 1, 2,, 99.

10 Second Order Linear Differential Equations: Existence and Uniqueness Results To this point we have considered only first order differential equations. However, many of the most interesting differential equations involve second derivatives. Indeed, since acceleration is the second derivative of position, Newton s second law of motion, F = ma, is a second order differential equation. In this and the coming sections we turn our attention to linear second-order differential equations. By a second-order linear differential equation we mean an equation of the form y + p(t)y + q(t)y = g(t). If g(t) 0 we say that the equation is homogeneous. Otherwise the equation is non-homogeneous. Initial-value problems for second-order linear differential equations require two initial-conditions. In this section we will consider the question of existence and uniqueness of solutions to the initial value problem y + p(t)y + q(t)y = g(t), y(t 0 ) = y 0, y (t 0 ) = y 0. (10.1) Existence and uniqueness results similar to first-order equations exist for secondorder equations as well. The following theorem tells us the conditions for the existence and uniqueness of solution to (10.1). Note how this theorem is analogous to the corresponding theorem for first order linear ODE s with initial conditions. Theorem 10.1 If p(t), q(t), and g(t) are continuous functions over an interval a < t < b containing t 0 then the initial value problem (10.1) has a unique solution in the interval a < t < b. 75

7610 SECOND ORDER LINEAR DIFFERENTIAL EQUATIONS: EXISTENCE AND UNIQU Example 10.1 Find the largest interval where (t 2 1)y + 3ty + (cos t)y = e t, y(0) = 4, y (0) = 5 is guaranteed to have a unique solution. Solution. We first put it into the standard form y + 3t t 2 1 y + cos t t 2 1 y = et t 2 1, y(0) = 4, y (0) = 5. We see that p, q, and g are all continuous except at t = 1 and t = 1. Theorem 10.1 tells us that there is a unique solution on 1 < t < 1 since this interval contains 0 Remark 10.1 The above theorem does not give the largest t interval of existence as shown in the next example. Thus, Theorem 10.1 is a local existence theorem. Remark 10.2 If the conditions of the above theorem are not guaranteed, this does not mean the the problem does not have a solution. Example 10.2 Consider the initial value problem t 2 y ty + y = 0, y(1) = 1, y (1) = 1. (a) What is the largest interval on which Theorem 10.1 guarantees the existence of a unique solution? (b) Show by direct substitution that the function y(t) = t is the unique solution to this initial value problem. What is the interval on which this solution actually exists? (c) Does this example contradict the assertion of Theorem 10.1? Explain. Solution. (a) Writing the equation in standard form to obtain y 1 t y + 1 t 2 y = 0. From this, we see that the functions p(t) = 1 t and q(t) = 1 are continuous for all t 2 t 0. Since t 0 = 1, the largest t interval according to Theorem 10.1 is 0 < t <. (b) If y(t) = t then y (t) = 1 and y (t) = 0 so that y 1 t y + 1 y = 0 1 t 2 t + 1 t = 0, y(1) = y (1) = 1. So y(t) = t is a solution so that by Theoren 10.1 it is the only solution. The t interval for this solution is < t <. (c) No, because the theorem is a local existence theorem and not a global one

77 Example 10.3 Consider the graphs shown. Each graph displays a portion of the solution of one of the four initial value problems given. Match each graph with the appropriate initial value problem. (a) y + y = 2 sin t, y(0) = 1, y (0) = 1; (b) y + y = 2t, y(0) = 1, y (0) = 1; (c) y y = t 2, y(0) = y (0) = 1; (d) y y = 2 cos t, y(0) = y (0) = 1. Solution. (a) B since y (0) = 1 < 0 and y (0) = 2 sin 0 y(0) = 1 > 0. (b) D since y (0) = 1 < 0 and y (0) = 2(0) y(0) = 1 < 0. (c) A since y (0) = 1 > 0 and y (0) = 0 2 + y(0) = 1 > 0. (d) C since y (0) = 1 > 0 and y (0) = 2 cos 0 + y(0) = 1 < 0 Example 10.4 Give an example of an initial value problem of the form y + p(t)y + q(t)y = 0, y(t 0 ) = y 0, y (t 0 ) = y 0 for which the given t interval 1 < t < 5 is the largest on which Theorem 10.1 guarantees a unique solution. Solution. One such an answer is

7810 SECOND ORDER LINEAR DIFFERENTIAL EQUATIONS: EXISTENCE AND UNIQU y + 1 t+1 y + y = 1 t 5, y(0) = 1, y (0) = 2 Example 10.5 Is there a solution y(t) to the initial value problem y + 2y + 1 t 3 y = 0, y(1) = 1, y (1) = 2 such that lim t 0 + y(t) =? That is, y(t) has an infinite discontinuity at t = 0. Solution. Since p(t) = 2, q(t) = 1 t 3, and t 0 = 1, according to Theorem 10.1 the largest interval for which the solution y(t) is defined is < t < 3. Since 0 is in that interval and y(t) is continuous, the limit cannot hold

Practice Problems In Problems 10.1-10.6, determine the largest t interval on which Theorem 10.1 guarantees the existence of a unique solution. Problem 10.1 79 Problem 10.2 y + y + 3ty = tan t, y(π) = 1, y (π) = 1. Problem 10.3 e t y + 1 t 2 1 y = 4 t, y( 2) = 1, y ( 2) = 2. Problem 10.4 ty + sin 2t t 2 9 y + 2y = 0, y(1) = 0, y (1) = 1. Problem 10.5 ty (1 + t)y + y = t 2 e 2t, y( 1) = 0, y ( 1) = 1. (sin 2 t)y (2 sin t cos t)y + (cos 2 t + 1)y = sin 3 t, y( π 4 ) = 0, y ( π 4 ) = 2. Problem 10.6 t 2 y + ty + y = sec (ln t), y( π 3 ) = 0, y ( π 3 ) = 1. In Problems 10.7-10.8, give an example of an initial value problem of the form y + p(t)y + q(t)y = 0, y(t 0 ) = y 0, y (t 0 ) = y 0 for which the given t interval is the largest on which Theorem 10.1 guarantees a unique solution. Problem 10.7 < t <.

8010 SECOND ORDER LINEAR DIFFERENTIAL EQUATIONS: EXISTENCE AND UNIQU Problem 10.8 3 < t <. Problem 10.9 Determine the largest interval in which the initial-value problem (t 3)y + ty + (ln t )y = 0, y(1) = 0, y (1) = 1 is certain to have a unique solution. Problem 10.10 Theorem 10.1 tells us that the initial-value problem y + t 2 y = 0, y(0) = y (0) = 0 defines exactly one function y(t). Using only Theorem 10.1, show that this function has the additional property y( t) = y(t). That is, y(t) is an even function. Problem 10.11 State the existence and uniqueness theorem for second order linear differential equations with initial consitions. Problem 10.12 Consider the homogeneous second order linear differential equation ty y + 4t 3 y = 0. Show that y = cos (t 2 ) is a solution to this differential equation. Problem 10.13 Find the largest interval where 16 t 2 y + ln (t + 1)y + (cos t)y = 0, y(0) = 2, y (0) = 0 is guaranteed to have a unique solution. Problem 10.14 Find the largest interval where (t + 2)y + ty + cot ty = t 2 + 1, y(2) = 11, y (2) = 2 is guaranteed to have a unique solution. Problem 10.15 Define L(y) = y +p(t)y +q(t)y. Show that L(αy 1 +βy 2 ) = αl(y 1 )+βl(y 2 ), where y 1 and y 2 are twice differentiable functions. That is, L is a linear transformation.

11 The General Solution of 2 nd Order Linear Homogeneous Equations In this section we discuss the structure of the general solution to the homogeneous second order linear differential equation y + p(t)y + q(t)y = 0 (11.1) where p(t) and q(t) are continuous functions for a < t < b. The first key property of (11.1) is its linear property also known as the principle of superposition. Theorem 11.1 (Principle of Superposition) If y 1 and y 2 are respective solutions of (11.1) then for any constants c 1 and c 2, the function y = c 1 y 1 + c 2 y 2 is also a solution to (11.1). The function c 1 y 1 + c 2 y 2 is called a linear combination of the functions y 1 and y 2. Example 11.1 Write y = 3 cos ( 2t + π 4 ) as a linear combination of y1 = cos 2t and y 2 = sin 2t. Solution. Using the identity cos (α + β) = cos α cos β sin α sin β 81

8211 THE GENERAL SOLUTION OF 2 ND ORDER LINEAR HOMOGENEOUS EQUATIONS we arrive at ( 3 cos 2t + π ) =3 cos 2t cos π 4 4 3 sin 2t sin π [ 4 = 3 cos π ] [ cos 2t + 3 sin π ] sin 2t ( 4 4 3 ) ( ) 2 = cos 2t + 3 2 sin 2t 2 2 Theorem 11.1 states that if y 1 and y 2 are two given solutions of (11.1) then one can build many new solutions by taking a linear combination y = c 1 y 1 + c 2 y 2. However, this theorem does not say if every solution to (11.1) has to be written as a linear combination of y 1 and y 2. So our next interest is to find out if one can express every solution of (11.1) as a linear combination of two solutions of (11.1). If there are such solutions y 1 and y 2, we shall say that the set {y 1, y 2 } forms a fundamental set of solutions to (11.1). It follows that if we know a fundamental set of solutions {y 1, y 2 } then we know the general solution to (11.1) which is given by y(t) = c 1 y 1 (t) + c 2 y 2 (t). Identifying Fundamental Sets Given a particular homogeneous differential equation and two solutions of that differential equation, is there a convenient way for checking whether or not these two solutions form a fundemental set of solutions? The answer is in the affirmative according to the following theorem. Theorem 11.2 Let y 1 (t) and y 2 (t) be two solutions to the homogeneous second order linear differential equation y + p(t)y + q(t)y = 0, a < t < b where p(t) and q(t) are continuous in a < t < b. If there is a a < t 0 < b such that W (y 1 (t 0 ), y 2 (t 0 )) = y 1(t 0 ) y 2 (t 0 ) y 1 (t 0) y 2 (t 0) = y 1(t 0 )y 2(t 0 ) y 1(t 0 )y 2 (t 0 ) 0 then {y 1, y 2 } is a fundamental set of solutions. We call the function W the Wronskian function. Proof. We need to show that if y(t) is a solution to (11.1) then we can write y(t) as a linear combination of y 1 and y 2. That is y(t) = c 1 y 1 (t) + c 2 y 2 (t).

83 So the problem reduces to finding the constants c 1 and c 2. These are found by solving the following linear system of two equations in the unknowns c 1 and c 2 : By the method of elimination we find and c 1 y 1 (t 0 ) + c 2 y 2 (t 0 ) = y(t 0 ) c 1 y 1(t 0 ) + c 2 y 2(t 0 ) = y (t 0 ). c 1 = y(t 0)y 2 (t 0) y (t 0 )y 2 (t 0 ) W (y 1 (t 0 ), y 2 (t 0 )) c 2 = y (t 0 )y 1 (t 0 ) y(t 0 )y 1 (t 0). W (y 1 (t 0 ), y 2 (t 0 )) Note that c 1 and c 2 exist since W (y 1 (t 0 ), y 2 (t 0 )) 0 Example 11.2 Consider the differential equation y + 4y = 0. (11.2) (a) Show that y 1 (t) = cos 2t and y 2 (t) = sin 2t are solutions to (11.2). (b) Show that {cos 2t, sin 2t} is a fundamental set of solutions. (c) Write the solution y(t) = 3 cos (2t + π 4 ) as a linear combination of y 1 and y 2. Solution. (a) A simple calculation shows (b) For any t we have W (y 1 (t), y 2 (t)) = y 1 + 4y 1 = 4 cos 2t + 4 cos 2t = 0 y 2 + 4y 2 = 4 sin 2t + 4 sin 2t = 0. cos 2t 2 sin 2t sin 2t 2 cos 2t = 2 cos2 2t + 2 sin 2 2t = 2 0. Thus, {y 1, y 2 } is a fundamental set of solutions. (c) Using the formulas for c 1 and c 2 with t 0 = 0 we find c 1 = y(0)y 2 (0) y (0)y 2 (0) W (y 1 (0), y 2 (0)) = 6 cos π 4 cos 0 + 6 sin π 4 sin 0 2 = 3 2 2

8411 THE GENERAL SOLUTION OF 2 ND ORDER LINEAR HOMOGENEOUS EQUATIONS and c 2 = y (0)y 1 (0) y(0)y 1 (0) W (y 1 (0), y 2 (0)) = 6 sin π 4 cos 0 + 6 cos π 4 sin 0 2 = 3 2 2 Theorem 11.2 says that if one can find a < t 0 < b such that W (y 1 (t 0 ), y 2 (t 0 )) 0 then the set {y 1, y 2 } is a fundamental set of solutions. The following theorem shows that the condition W (y 1 (t 0 ), y 2 (t 0 )) 0 implies that W (y 1 (t), y 2 (t)) 0 for all t in the interval (a, b). That is, the theorem tells us that we can choose our test point t 0 on the basis of convenience any test point t 0 will do. That s why we chose t 0 = 0 in the previous example. Theorem 11.3 (Abel s) Let y 1 (t) and y 2 (t) be two solutions to the homogeneous second order linear differential equation y + p(t)y + q(t)y = 0, a < t < b where p(t) and q(t) are continuous in a < t < b. If t 0 is any point in (a, b) then W (y 1 (t), y 2 (t)) = W (y 1 (t 0 ), y 2 (t 0 ))e t t 0 p(s)ds. Thus, if W (y 1 (t 0 ), y 2 (t 0 )) 0 then W (y 1 (t), y 2 (t)) 0 for all a < t < b. Remark 11.1 It is easy to check that W satisfies the differential equation W + pw = 0. Thus, Abel s formula is just the solution to the initial value problem W + pw = 0, W (y 1 (t 0 ), y 2 (t 0 )) = α 0. In the case y 1 and y 2 form a fundamental set of solutions then W (y 1 (t), y 2 (t)) is never zero in the interval a < t < b as shown in the following theorem. Theorem 11.4 Suppose that {y 1, y 2 } is a fundamental set of solutions to (11.1). Then W (y 1 (t), y 2 (t)) 0 for all a < t < b. Combining Theorem 11.2 and Theorem 11.4 we obtain the following corollary characterizing a fundamental set of solutions. Corollary 11.1 Let y 1 (t) and y 2 (t) be two solutions of (11.1). Let W (y 1 (t), y 2 (t)) denote the Wronskian of y 1 and y 2. Then {y 1, y 2 } is a fundamental set of solution if and only if W (y 1 (t), y 2 (t)) 0 for all a < t < b.

85 Example 11.3 Consider the initial value problem y 1 t y 3 t 2 y = 0, y(1) = 4, y (1) = 8, 0 < t <. (a) Show that y 1 (t) = t 3 and y 2 (t) = t 1 are solutions to the differential equation. (b) Show that {y 1, y 2 } is a fundamental set of solutions to the differential equation. (c) Solve the given initial value problem. Solution. (a) By substitution and simple calculation we find y 1 1 t y 1 3 t 2 y 1 = 6t 1 t 3t2 3 t 2 t3 = 0 y 2 1 t y 2 3 t 2 y 2 = 2t 3 1 t ( t 2 ) 3 t 2 t 1 = 0. (b) Finding the Wronskian at t 0 = 1 we see W (y 1 (1), y 2 (1)) = 1 1 3 1 = 4 0. Thus, {y 1, y 2 } is a fundamental set of solution. (c) The general solution to the differential equation has the form y(t) = c 1 y 1 (t) + c 2 y 2 (t). The initial conditions yield the following linear system in the unknowns c 1 and c 2. or c 1 y 1 (1) + c 2 y 2 (1) =4 c 1 y 1(1) + c 2 y 2(1) =8 c 1 + c 2 =4 3c 1 c 2 =8. Solving this system by the method of elimination we find c 1 = 3 and c 2 = 1. Thus, y(t) = 3t 3 + 1 t, 0 < t < Example 11.4 Suppose that y 1 (t) = t is a solution to the differential equation Find a second solution y 2. t 2 y (t + 2)ty + (t + 2)y = 0.

8611 THE GENERAL SOLUTION OF 2 ND ORDER LINEAR HOMOGENEOUS EQUATIONS Solution. Rewriting the given equation in the form y ( 2 t + 1)y + ( 2 t 2 + 1 )y = 0. t Thus, p(t) = ( 2 t + 1). But W + pw = 0 so that W ( 2 t + 1)W = 0. Using the method of integrating factor we find W (t) = Ct 2 e t. To find y 2, note that ( y 2 ) t = W (t) = e t. Integrating both sides, we find t 2 y 2 (t) = t e t dt = te t where we chose the constant of integration to be 0

87 Practice Problems In Problems 11.1-11.7, the t interval of solution is < t < unless indicated otherwise. (a) Determine whether the given functions are solutions to the differential equation. (b) If both functions are solutions, calculate the Wronskian. Does this calculation show that the two functions form a fundamental set of solutions? (c) If the two functions have been shown in (b) to form a fundamental set, construct the general solution and determine the unique solution satisfying the initial value problem. Problem 11.1 Problem 11.2 y 4y = 0, y 1 (t) = e 2t, y 2 (t) = 2e 2t, y(0) = 1, y (0) = 2. Problem 11.3 y + y = 0, y 1 (t) = sin t cos t, y 2 (t) = sin t, y( π 2 ) = 1, y ( π 2 ) = 1. Problem 11.4 y 4y + 4y = 0, y 1 (t) = e 2t, y 2 (t) = te 2t, y(0) = 2, y (0) = 0. ty + y = 0, y 1 (t) = ln t, y 2 (t) = ln 3t, y(3) = 0, y (3) = 3, 0 < t <. Problem 11.5 t 2 y ty 3y = 0, y 1 (t) = t 3, y 2 (t) = t 1, y( 1) = 0, y ( 1) = 2, t < 0. Problem 11.6 y = 0, y 1 (t) = t + 1, y 2 (t) = t + 2, y(1) = 4, y (1) = 1.

8811 THE GENERAL SOLUTION OF 2 ND ORDER LINEAR HOMOGENEOUS EQUATIONS Problem 11.7 4y + 4y + y = 0, y 1 (t) = e t 2, y 2 (t) = te t 2, y(1) = 1, y (1) = 0. Problem 11.8 The functions y 1 (t) = t and y 2 (t) = t ln t form a fundamental set of solutions to the differential equation t 2 y ty + y = 0, 0 < t <. (a) Show that y(t) = 2t + t ln 3t is a solution to the differential equation. (b) Find c 1 and c 2 such that y(t) = c 1 y 1 (t) + c 2 y 2 (t). Problem 11.9 The functions y 1 (t) = e 3t and y 2 (t) = e 3t are known to be solutions of y + αy + βy = 0, where α and β are constants. Determine α and β. Problem 11.10 The functions y 1 (t) = t and y 2 (t) = e t are known to be solutions of y + p(t)y + q(t)y = 0. (a) Determine the functions p(t) and q(t). (b) On what t intervals are the functions p(t) and q(t) continuous? (c) Compute the Wronskian of these two functions. On what t intervals is the Wronskian nonzero? (d) Are the observations in (b) and (c) consistent with Abel s Theorem? Problem 11.11 It is known that two solutions of y + ty + 2y = 0 has a Wronskian W (y 1 (t), y 2 (t)) that satisfies W (y 1 (1), y 2 (1)) = 4. What is W (y 1 (2), y 2 (2))? Problem 11.12 The pair of functions {y 1, y 2 } is known to form a fundamental set of solutions of y + αy + βy = 0, where α and β are constants. One solution is y 1 (t) = e 2t, and the Wronskian formed by these two solutions is W (y 1 (t), y 2 (t)) = e t. Determine the constants α and β. Problem 11.13 The Wronskian of a pair of solutions of y + p(t)y + 3y = 0 is W (t) = e t2. What is the coefficient function p(t)?

89 Problem 11.14 Prove that if y 1 and y 2 have maxima or minima at the same point in an interval I, then they cannot be a fundamental set of solutions on that interval. Problem 11.15 Without solving the equation, find the Wronskian of two solutions of Bessel s equation t 2 y + ty + (t 2 µ 2 )y = 0. Problem 11.16 If W (y 1, y 2 ) = t 2 e t and y 1 (t) = t then find y 2 (t). Problem 11.17 The functions t 2 and 1/t are solutions to a 2nd order, linear homogeneous ODE on t > 0. Verify whether or not the two solutions form a fundamental solution set. Problem 11.18 Show that t 3 and t 4 can t both be solutions to a differential equation of the form y + p(t)y + q(t)y = 0 where p and q are continuous functions defined on the real numbers. Problem 11.19 Suppose that t 2 + 1 is the Wronskian of two solutions to the differential equation y + p(t)y + q(t)y = 0. Find p(t).

9011 THE GENERAL SOLUTION OF 2 ND ORDER LINEAR HOMOGENEOUS EQUATIONS

12 Second Order Linear Homogeneous Equations with Constant Coefficients: Distinct Characterisitc Roots In the previous section we discussed the structure of the general solution of a second order linear homogeneous differential equation. As we saw, the general solution is a linear combination of two solutions that form a fundamental set of solutions. In this and the next two sections we discuss methods for finding the fundamental set of solutions for second order homogeneous equations with constant coefficients, i.e., equations of the form ay + by + cy = 0 (12.1) where a, b and c are constants with a 0. Notice first that for b = 0 and c 0 the function y is a constant multiple of y. So it makes sense to look for a function with such property. One such function is y(t) = e rt. Substituting this function into (12.1) leads to ay + by + cy = ar 2 e rt + bre rt + ce rt = (ar 2 + br + c)e rt = 0. Since e rt > 0 for all t, the previous equation leads to ar 2 + br + c = 0. (12.2) Thus, a function y(t) = e rt is a solution to (12.1) when r satisfies equation (12.2). We call (12.2) the characteristic equation for (12.1) and the polynomial C(r) = ar 2 + br + c is called the characteristic polynomial. Example 12.1 Solve: y 5y 6y = 0. 91

9212 SECOND ORDER LINEAR HOMOGENEOUS EQUATIONS WITH CONSTANT COEF Solution. The characteristic polynomial for this equation is C(r) = r 2 5r 6 = (r 2)(r 3). Thus, the roots of the characteristic equation are r = 2 and r = 3. Since W (t) = e 2t e 3t = e5t 0 2e 2t the functions y 1 (t) = e 2t and y 2 (t) = e 3t form a fundamental set of solutions. Hence, the general solution is given by y(t) = c 1 e 2t + c 2 e 3t where c 1 and c 2 are arbitrary constants We conclude from the previous example that the two distinct real solutions to the characteristic equation lead to the general solution. Does this result apply to any equation (12.1) whose characteristic equation has distinct solutions? The answer is in the affirmative. To see this, let r 1 and r 2 be the two distinct solutions to (12.2). Then W (t) = e r 1t r 1 e r 1t e r 2t r 2 e r 2t 3e 3t = r 2e (r 1+r 2 )t r 1 e (r 1+r 2 )t = (r 2 r 1 )e (r 1+r 2 )t 0 since both r 1 r 2 and e (r 1+r 2 )t are not equal to 0. Hence, e r 1t and e r 2t form a fundamental set of solutions. As a result, the general solution of (12.1) is given by y(t) = c 1 e r 1t + c 2 e r 2t where c 1 and c 2 are arbitrary constants. Example 12.2 Solve the initial value problem y y 6y = 0, y(0) = 1, y (0) = 2. Describe the behavior of the solution y(t) as t and t. Solution. The characteristic polynomial is C(r) = r 2 r 6 = (r 3)(r + 2) so that the characteristic equation r 2 r 6 = 0 has the solutions r 1 = 3 and r 2 = 2. The general solution is then given by Taking the derivative to obtain y(t) = c 1 e 3t + c 2 e 2t. y (t) = 3c 1 e 3t 2c 2 e 2t.

93 The conditions y(0) = 1 and y (0) = 2 lead to the system c 1 + c 2 = 1 3c 1 2c 2 = 2. Solving this system by the method of elimination we find c 1 = 4 5 and c 2 = 1 5. Hence, the unique solution to the initial value problem is y(t) = 1 5 (4e3t + e 2t ). As t, e 3t 0 and e 2t. Thus, y(t). Similarly, y(t) as t Remark 12.1 In this section, we have only considered the case when (12.2) has two distinct real solutions. In Section 13, we discuss the case of (12.2) having repeated solutions and in Section 14 we look at the complex solutions Example 12.3 Consider the initial value problem y + αy + βy = 0, y(0) = 1, y (0) = y 0, where α, β, and y 0 are constants. It is known that one solution of the differential equation is y 1 (t) = e 3t and that the solution of the initial value problem satisfies lim t y(t) = 2. Determine the constants α, β, and y 0. Solution. Since r = 3 is a solution to the characteristic equation, we have ( 3) 2 + α( 3) + β = 0 or 3α + β = 9. Also, since lim t y(t) = 2, the second root for the characteristic equation must be r = 0. In this case, β = 0 and solving for α we find α = 3. Hence, y(t) = c 1 e 3t + c 2. Since lim t y(t) = 2, we find c 2 = 2. Since y(0) = 1, we find c 1 + 2 = 1 so that c 1 = 1. Thus, y(t) = e 3t + 2 and y (t) = 3e 3t. Therefore, y 0 = y (0) = 3 Example 12.4 Consider the initial value problem y + αy + βy = 0, y(0) = 3, y (0) = 5. The differential equation has a fundamental set of solutions {y 1, y 2 }. It is known that y 1 (t) = e t and that the Wronskian formed by the two members of the fundamental set is W (t) = 4e 2t. (a) Determine y 2 (t). (b) Determine the constants α and β. (c) Solve the initial value problem.

9412 SECOND ORDER LINEAR HOMOGENEOUS EQUATIONS WITH CONSTANT COEF Solution. (a) The second solution is of the form y 2 (t) = e rt. In this case, W (t) = e t e rt = (r + 1)e(r 1)t. e t re rt But W (t) = 4e 2t and this leads to r = 3. Hence, y 2 (t) = e 3t. (b) Since r = 1 and r = 3 are the roots for the characteristic equation, we have (r + 1)(r 3) = 0 or r 2 2r 3 = 0. This implies that y 2y 3y = 0 so that α = 2 and β = 3 (c) The initial conditions and y (t) = c 1 e t +3c 2 e 3t lead to the system c 1 +c 2 = 3 and c 1 + 3c 2 = 5. Solving this system we find c 1 = 1 and c 2 = 2. Thus, y(t) = e t + 2e 3t Example 12.5 Solve the initial-value problem 4y y = 0, y(0) = 2, y (0) = β. Then find β so that the solution approaches zero as t. Solution. The characteristic equation 4r 2 1 = 0 has roots r = 1 2 and r = 1 2. Thus, y(t) = c 1 e t 2 + c2 e t 2. The initial conditions and y (t) = c 1 2 e t 2 + c 2 2 e t 2 lead to the system c 1 + c 2 = 2 and c 1 c 2 = 2β. Solving this system we find c 1 = 1 β and c 2 = 1 + β. Thus, y(t) = (1 β)e t 2 + (1 + β)e t 2. Since lim t y(t) = 0 and lim t e t =, we must have 1+β = 0. Thus, β = 1 Example 12.6 Show that if λ is a root of aλ 3 + bλ 2 + cλ + d = 0, then e λt is a solution of ay + by + cy + dy = 0. Solution. We have ay + by + cy + dy = aλ 3 e λt + bλ 2 e λt + cλe λt + de λt = (aλ 3 + bλ 2 + cλ + d)e λt = 0

95 Practice Problems Problem 12.1 Solve the initial value problem y + y 2y = 0, y(0) = 3, y (0) = 3. Describe the behavior of the solution y(t) as t and t. Problem 12.2 Solve the initial value problem y 4y + 3y = 0, y(0) = 1, y (0) = 1. Describe the behavior of the solution y(t) as t and t. Problem 12.3 Solve the initial value problem y y = 0, y(0) = 1, y (0) = 1. Describe the behavior of the solution y(t) as t and t. Problem 12.4 Solve the initial value problem y + 5y + 6y = 0, y(0) = 1, y (0) = 1. Describe the behavior of the solution y(t) as t and t. Problem 12.5 Solve the initial value problem y 4y = 0, y(3) = 0, y (3) = 0. Describe the behavior of the solution y(t) as t and t. Problem 12.6 Solve the initial value problem 2y 3y = 0, y( 2) = 3, y ( 2) = 0. Describe the behavior of the solution y(t) as t and t.

9612 SECOND ORDER LINEAR HOMOGENEOUS EQUATIONS WITH CONSTANT COEF Problem 12.7 Solve the initial value problem y + 4y + 2y = 0, y(0) = 0, y (0) = 4. Describe the behavior of the solution y(t) as t and t. Problem 12.8 Solve the initial value problem 2y y = 0, y(0) = 2, y (0) = 2. Describe the behavior of the solution y(t) as t and t. Problem 12.9 Obtain the general solution to the differential equation y 5y + 6y = 0. Problem 12.10 A particle of mass m moves along the x axis and is acted upon by a drag force proportional to its velocity. The drag constant is denoted by k. If x(t) represents the particle position at time t, Newton s law of motion leads to the differential equation mx (t) = kx (t). (a) Obtain the general solution to this second order linear differential equation. (b) Solve the initial value problem if x(0) = x 0 and x (0) = v 0. (c) What is lim t x(t)? Problem 12.11 Find a homogeneous second-order linear ordinary differential equation whose general solution is y(t) = c 1 e 2t + c 2 e t. Problem 12.12 Find the general solution of the differential equation y 3y 4y = 0. Problem 12.13 Find the general solution of the differential equation y + 4y 5y = 0. Problem 12.14 Find the general solution of the differential equation 3y + 2y + y = 0. Problem 12.15 Solve the initial-value problem: y + 3y 4y = 0, y(0) = 1, y (0) = 1. Problem 12.16 Solve the initial-value problem: 2y + 5y 3y = 0, y(0) = 2, y (0) = 1.

13 Characteristic Equations with Repeated Roots In this section we consider the question of the characteristic equation having a repeated real solution. This occurs when b 2 4ac = 0. The two equal roots are given by r 1 = r 2 = b 2a. The computation based on the trial form y(t) = e rt yields only one solution, namely y 1 (t) = e b 2a t. Since a fundamental set of solutions consists of two functions having a nonzero Wronskian, there must be another solution having a different functional form. The second solution follows from the following theorem. Theorem 13.1 Suppose that y 1 (t) is a nontrivial solution to the differential equation y + p(t)y + q(t)y = 0. (13.1) Then any solution y 2 (t) can be written in the form ( e ) p(t)dt y 2 (t) = C y1 2(t) dt y 1 (t) + C y 1 (t) (13.2) where C and C are arbitrary constants. Proof. First, recall that the Wronskian W (t) of any two solutions to (13.1) satisfies the differential equation W +p(t)w = 0 so that W (t) = Ce p(t)dt. If y 2 is a solution to (13.1) then ( ) y2 = W (t) e p(t)dt = C (t) y1 2(t). y 1 y 2 1 97

98 13 CHARACTERISTIC EQUATIONS WITH REPEATED ROOTS Integrating this last equation we find ( e ) p(t)dt y 2 (t) = C y1 2(t) dt y 1 (t) + C y 1 (t) where C and C are arbitrary constants Notice that the term C y 1 (t) is simply a constant multiple of y 1 (t). Since the general solution of the differential equation (13.1) contains y 1 (t) multiplied by an arbitrary constant, we lose no generality by setting C = 0. We can likewise take C = 1 since y 2 (t) will also be multiplied by an arbitrary constant in the general solution. With these simplification the second solution is Now, for the equation we have p(t) = b a. If y 1(t) = e b 2a t then ( e ) p(t)dt y 2 (t) = y1 2(t) dt y 1 (t). ay + by + cy = 0 (13.3) ( ) e b a t y 2 (t) = e dt e b b a t 2a t = te b 2a t. Hence, the general solution to (13.3) is given by y(t) = c 1 e b 2a t + c 2 te b 2a t. Example 13.1 Solve the initial value problem: y + 2y + y = 0, y(0) = 1, y (0) = 1. Solution. The characteristic equation r 2 + 2r + 1 = 0 has a repeated root: r 1 = r 2 = 1. Thus, the general solution is given by y(t) = c 1 e t + c 2 te t. The two conditions y(0) = 1 and y (0) = 1 lead to c 2 = 0 and c 1 = 1. Hence, the unique solution is y(t) = e t

99 Example 13.2 Consider the differential equation t 2 y + 2ty 2y = 0, 0 < t <. Find the general solution given that y 1 (t) = t is a solution to the differential equation. Solution. Since t > 0, we can rewrite the given equation in the form In this case, p(t) = 2 t and y 2 (t) = ( e 2 y + 2 t y 2 t 2 y = 0. t 2 t dt dt ) t = Hence, the general solution is y(t) = c 1 t + c 2 t 2 ( ) 1 t 4 dt t = 1 3t 2. Example 13.3 The graph of a solution y(t) of the differential equation 4y + 4y + y = 0 passes through the points (1, e 1 2 ) and (2, 0). Determine y(0) and y (0). Solution. The characteristic equation 4r 2 + 4r + 1 = 0 has the roots r 1 = r 2 = 1 2 the general solution is y(t) = c 1 e t 2 + c2 te t 2. so that Since y(2) = 0, we find c 1 + 2c 2 = 0. Since y(1) = e 1 2, we have c 1 + c 2 = 1. Solving the system of two equations, we find c 1 = 2 and c 2 = 1. Hence, y(t) = 2e t 2 te t 2. Now, y(0) = 2. Also, replacing t = 0 in y (t) = 2e t 2 + t 2 e t 2 to obtain y (0) = 2 Example 13.4 Find a homogeneous second order linear differential equation whose general solution is given by y(t) = c 1 e 3t + c 2 te 3t.

100 13 CHARACTERISTIC EQUATIONS WITH REPEATED ROOTS Solution. The characteristic equation has the double roots r 1 = r 2 = 3 so that r 2 +6r+9 = 0. Hence, the differential equation is y + 6y + 9y = 0 Example 13.5 Show that if λ is a double root of at 3 + bt 2 + ct + d = 0, then te λt is also a solution of ay + by + cy + dy = 0. Solution. Since λ is a double root, we have aλ 3 + bλ 2 + cλ + d = 0 and 3aλ 2 + 2bλ + c = 0. Let y(t) = te λt. Then y = e λt + λte λt, y = 2λe λt + λ 2 te λt, y = 3λ 2 e λt + λ 3 te λt. Substituting into the equation we find ay + by + cy + dy =[(aλ 3 + bλ 2 + cλ + d)t + (3aλ 2 + 2bλ + c)]e λt =0

101 Practice Problems In Problems 13.1-13.5 answer the following questions. (a) Obtain the general solution of the differential equation. (b) Impose the initial conditions to obtain the unique solution of the initial value problem. (c) Describe the behavior of the solution as t and t. Problem 13.1 Problem 13.2 Problem 13.3 Problem 13.4 Problem 13.5 9y 6y + y = 0, y(3) = 2, y (3) = 5 3. 25y + 20y + 4y = 0, y(5) = 4e 2, y (5) = 3 5 e 2. y 4y + 4y = 0, y(1) = 4, y (1) = 0. y + 2 2y + 2y = 0, y(0) = 1, y (0) = 0. 3y + 2 3y + y = 0, y(0) = 2 3, y (0) = 3. Problem 13.6 Find the general solution of y 6y + 9y = 0. Problem 13.7 Find the general solution of 4y 4y + y = 0. Problem 13.8 Solve the initial-value problem: y + y + y 4 = 0, y(0) = 2, y (0) = 0.

102 13 CHARACTERISTIC EQUATIONS WITH REPEATED ROOTS

14 Characteristic Equations with Complex Roots In this section, we solve the linear second order homogeneous differential equation with constant coefficients when the roots of the characteristic equation ay + by + cy = 0, a 0 (14.1) ar 2 + br + c = 0 (14.2) are complex numbers. This occurs when b 2 4ac < 0. In this case, the complex roots of equation (14.2) are given by where i = 1. We will write r 1,2 = b ± i 4ac b 2 2a r 1,2 = α ± iβ where α = b 2a and β = 4ac b 2 2a. Like before, we would like to conclude that the functions e (α+iβ)x and e (α iβ)x are solutions to (14.1). These are complex solutions, we would like to have real solutions to the original real differential equation. This requires the use of the so-called the complex exponential function which we introduce and discuss some of its properties. For any complex number z = α + iβ we define the Euler s function e z = e α (cos β + i sin β). 103

104 14 CHARACTERISTIC EQUATIONS WITH COMPLEX ROOTS The exponential function satisfies the usual laws of exponentials such as e z e w = e z+w. To see this, we let z = α 1 + iβ 1 and w = α 2 + iβ 2. Then e z e w =e α 1 (cos β 1 + i sin β 1 )e α 2 (cos β 2 + i sin β 2 ) =e α 1+α 2 [(cos β 1 cos β 2 sin β 1 sin β 2 ) + i(sin β 1 cos β 2 + cos β 1 sin β 2 )] =e α 1+α 2 [cos (β 1 + β 2 ) + i sin (β 1 + β 2 )] =e z+w. From the above rule we can write where n is a positive integer. (e z ) n = e z e z e z = e z+z+ +z = e nz It follows from the above discussion that the complex solutions to the differential equation are linear combinations of e αt cos βt and e αt sin βt. Now letting y 1 (t) = e αt cos βt and y 2 (t) = e αt sin βt we find ay 1 + by 1 + cy 1 =a(α 2 e αt cos βt β 2 e αt cos βt 2αβe αt sin βt) +b(αe αt cos βt βe αt sin βt) + ce αt cos βt =e αt cos βt(a(α 2 β 2 ) + bα + c) e αt sin βt(2aαβ + bβ) ( ( ) ( ) ) b =e αt 2 4ac b2 b cos βt a 4a2 4a 2 + b + c 2a ( ( e αt sin βt 2a b ) ) 4ac b 2 + b 4ac b 2 = 0. 2a 2a 2a Thus, y 1 (t) = e αt cos βt is a solution to equation (14.1). Similarly, we show that y 2 (t) = e αt sin βt is a solution to equation (14.1). Moreover, W (t) = e αt cos βt e αt sin βt αe αt cos βt βe αt sin βt αe αt sin βt + βe αt cos βt = βe2αt 0. Hence, {y 1, y 2 } is a fundamental set of solutions to equation (14.1) so that the general solution is given by where c 1 and c 2 are real numbers. y(t) = e αt (c 1 cos βt + c 2 sin βt)

105 Example 14.1 Solve: y + 2y + 5y = 0. Solution. The characteristic equation r 2 + 2r + 5 = 0 has complex roots r 1,2 = 1 ± 2i. The general solution is y(t) = e t (c 1 cos 2t + c 2 sin 2t) Example 14.2 Solve the initial value problem y 10y + 29y = 0, y(0) = 1, y (0) = 3. Solution. The characteristic equation r 2 10r + 29 = 0 has the complex roots r 1,2 = 5 ± 2i. Thus, the general solution is given by the expression Finding y we obtain y(t) = e 5t (c 1 cos 2t + c 2 sin 2t). y (t) = e 5t [(5c 1 + 2c 2 ) cos 2t + (5c 2 2c 1 ) sin 2t]. The initial conditions yield c 1 = 1 and c 2 = 1. Thus, the unique solution to the initial value problem is y(t) = e 5t (cos 2t sin 2t) Next, we consider the question of representing the general solution y(t) = e αt (c 1 cos βt+ c 2 sin βt) in the form y(t) = Ke αt cos (βt δ), where 0 δ < 2π. For this, we let P (c 1, c 2 ) be a coordinate point in the plane and let δ be the angle between the t-axis and ray OP. See Figure 14.1. Then cos δ = c 1 K and sin δ = c 2 K where K = c 2 1 + c2 2. Then in terms of K and δ we can write ( c1 c 1 cos βt + c 2 sin βt =K K cos βt + c ) 2 K sin βt =K(cos δ cos βt + sin δ sin βt) = K cos (βt δ). It follows that y(t) = Ke αt cos (βt δ).

106 14 CHARACTERISTIC EQUATIONS WITH COMPLEX ROOTS Figure 14.1 We call Ke αt the amplitude of the oscillations. This means that the graph of y(t) is bounded by the graphs of ±Ke αt. The angle δ β is the phase shift. The term ( phase) shift reflects the fact that we obtain the graph of cos (βt δ) = cos β t δ β by shifting the graph of cos βt to the right by an amount t = δ β. Example 14.3 Put the solution of the initial value problem in the form y(t) = Ke αt cos (βt δ). y 2y + 17y = 0, y(0) = 4, y (0) = 8 Solution. The characteristic equation r 2 2r + 17 = 0 has the complex roots r 1,3 = 1 ± 4i. Thus, the general solution to the differential equation is y(t) = e t (c 1 cos 4t + c 2 sin 4t). Since y(0) = 4, we find c 1 = 4. Also y (t) = e t (c 1 cos 4t+c 2 sin 4t)+e t (4c 2 cos 4t 4c 1 sin 4t) and y (0) = 8 imply 8 = 4 + 4c 2 so that c 2 = 3. Thus, the unique solution to the initial value problem is y(t) = e t (3 sin 4t 4 cos 4t). Now, K = 9 + 16 = 25 = 5. Thus, tan δ = 3 4 so that δ = arctan ( 3 4 ). Hence, ( ( y(t) = 5e t cos 4t + arctan 3 )). 4

107 The graph of y(t) together with the envelope containing it is shown in Figure 14.2 Figure 14.2

108 14 CHARACTERISTIC EQUATIONS WITH COMPLEX ROOTS Practice Problems Problem 14.1 For any z = α + iβ we define the conjugate of z to be the complex number z = α iβ. Show that α = 1 2 (z + z) and β = 1 2i (z z). Problem 14.2 Write each of the complex numbers in the form α + iβ, where α and β are real numbers. 1. 2e i π 3. 2. (2 i)e i 3π 2. 3. ( 2e i π 6 ) 4. Problem 14.3 Write each function in the form e αt (A cos βt + ib sin βt), where α, β, A, and B are real numbers. 1. 2e i 2t. 2. 1 2 e2t+i(t+π). 3. ( 3e (1+i)t ) 3. In Problems 14.4-14.8 (a) Determine the roots of the characteristic equation. (b) Obtain the general solution as a linear combination of real-valued solutions. (c) Impose the initial conditions and solve the initial value problem. Problem 14.4 Problem 14.5 Problem 14.6 Problem 14.7 y + 2y + 2y = 0, y(0) = 3, y (0) = 1. 2y 2y + y = 0, y( π) = 1, y ( π) = 1. y + 4y + 5y = 0, y( π 2 ) = 1 2, y ( π 2 ) = 2. y + 4π 2 y = 0, y(1) = 2, y (1) = 1.

109 Problem 14.8 9y + π 2 y = 0, y(3) = 2, y (3) = π. In Problems 14.9-14.10, the function y(t) is a solution of the initial value problem y +ay +by = 0, y(t 0 ) = y 0, y (t 0 ) = y 0, where the point t 0 is specified. Determine the constants a, b, y 0, and y 0. Problem 14.9 y(t) = 2 sin 2t + cos 2t, t 0 = π 4. Problem 14.10 y(t) = e t π 6 cos 2t e t π 6 sin 2t, t0 = π 6. In Problems 14.11-14.13, rewrite the function y(t) in the form y(t) = Ke αt cos (βt δ), where 0 δ < 2π. Use this representation to sketch a graph of the given function, on a domain sufficiently large to display its main features. Problem 14.11 y(t) = sin t + cos t. Problem 14.12 y(t) = e t cos t + 3e t sin t. Problem 14.13 y(t) = e 2t cos 2t e 2t sin 2t. Problem 14.14 Consider the differential equation y + ay + 9y = 0, where a is a real number. Suppose that we know the Wronskian of a fundamental set of solutions of this differential equation is constant: W (t) = 1 for all real numbers t. Find the general solution of this differential equation. Problem 14.15 Rewrite y(t) = 2 cos 7t 11 sin 7t in phase-angle form. Give the exact function (so your answer will involve the inverse tangent function.)

110 14 CHARACTERISTIC EQUATIONS WITH COMPLEX ROOTS Problem 14.16 Find a homogeneous linear ordinary differential equation whose general solution is y(t) = c 1 e 2t cos (3t) + c 2 e 2t sin (3t). Problem 14.17 Rewrite y(t) = 5e (5 2i)t 3e (5+2i)t, without complex exponents, using sines and cosines. What ODE of the form ay + by + cy = 0, has y as a solution? Problem 14.18 Consider the function y(t) = 3 cos 2t 4 sin 2t. Find a second order linear IVP that y satisfies. Problem 14.19 An equation of the form t 2 y + αty + βy = 0, t > 0 where α and β are real constants is called an Euler equation. Show that the substitution x = ln t transforms Euler equation into an equation with constant coefficients. Hint: dy dt = dy dx dx dt. Problem 14.20 Use the result of the previous problem to solve the differential equation t 2 y + ty + y = 0.

15 Series Solutions of Differential Equations We know how to find the general solution of a second order linear differential equation with constant coefficients. In this section, we discuss finding the general solution to the homogeneous equation y + p(t)y + q(t)y = 0 (15.1) where p(t) and q(t) are not necessarily constants. Let t = t 0. We are interested in answering the question: When is it possible to represent the general solution of (15.1) in terms of a power series that converge in some open interval centered at t 0? In order to answer this question, we need to introduce the following definitions: We say that a function f(t) is analytic at t = t 0 if f(t) can be expressed as a Taylor series f (n) (t 0 ) f(t) = (t t 0 ) n. n! n=0 Examples of analytic functions are polynomial functions and rational functions. We say that t = t 0 is an ordinary point when both p(t) and q(t) are analytic at t 0. If p(t) and/or q(t) is not analytic at t 0, we say that t 0 is a singular point. Example 15.1 Consider the differential equation (1 t 2 )y + (tan t)y + t 5 3 y = 0 in the open interval 2 < t < 2. Classify each point in this interval as an ordinary point or a singular point. Solution. We first rewrite the equation in the form (15.1), y + tan t 1 t 2 y + t 5 3 1 t 2 y = 0. 111

112 15 SERIES SOLUTIONS OF DIFFERENTIAL EQUATIONS Therefore, p(t) = tan t 1 t 2 and q(t) = t 5 3 1 t 2. The function p(t) is not analytic at t = ±1 (where the denominator vanishes) and at t = ± π 2 (where the tangent has a vertical asymptotes). The function q(t) is not analytic at t = ±1 and at t = 0 (q(t) does not have a second derivative at 0.) Thus, in the interval 2 < t < 2, the five points t = 0, ±1, ± π 2 are singular points and all other points are ordinary points The following theorem answers our earlier question about a series solution to the equation (15.1). Theorem 15.1 Let p(t) and q(t) be analytic at t 0 and let R be the smaller of the two radii of convergence 1 of their respective Taylor series representations. The general solution to (15.1) can be expressed as y(t) = a n (t t 0 ) n = c 1 y 1 (t) + c 2 y 2 (t), n=0 where c 1 and c 2 are arbitrary constants. The functions y 1 (t) and y 2 (t) form a fundamental set of solutions, analytic in the interval t t 0 < R. Example 15.2 The point t 0 = 0 is an ordinary point of the differential equation y + (t 3 + 1)y ty = 0. Determine the value of R as stated in the previous theorem. Solution. Since p(t) and q(t) are polynomial functions, we have R p = R q =. Hence, R = The Method of Solution We assume that t 0 is an ordinary point of the equation (15.1), so it has a series solution near t 0 of the form y(t) = n=0 a n(t t 0 ) n that converges for t t 0 < R. 1 The radius of convergence of a function f(t), about t 0, is the distance between t 0 and the nearest singularity of f(t).

113 We want to determine the coefficients a 0, a 1, a 2, in the series solution. Let s differentiate the series termwise twice: y(t) =a 0 + a 1 (t t 0 ) + a 2 (t t 0 ) 2 + y (t) =a 1 + 2a 2 (t t 0 ) + 3a 3 (t t 0 ) 2 + y (t) =2a 2 + 6a 3 (t t 0 ) + 12a 4 (t t 0 ) 2 + Substituting in the differential equation (15.1), we get K 0 + K 1 (t t 0 ) + K 2 (t t 0 ) 2 + = 0 (15.2) where K n is function of certain coefficients a i. In order that series (15.2) be valid for all t in the interval t t 0 < R, we must have K n = 0, n = 0, 1, 2,. This leads to a set of conditions that the coefficients a n must satisfy. Example 15.3 Find the series solution of the differential equation near the ordinary point 0. (t 2 4)y + 3ty + y = 0. Solution. The singular points of the given differential equation are 2 and 2. Thus, we will find a series solution of the differential equation in the interval t < 2. The solution is of the form y(t) = a n t n. Differentiating this series twice we find n=2 n=0 y (t) = n=1 na nt n 1 and y (t) = n=2 n(n 1)a nt n 2. Substituting these series in the given equation we find t 2 n(n 1)a n t n 2 4 n(n 1)a n t n 2 + 3t na n t n 1 + a n t n =0 n=0 n=2 n=2 n(n 1)a n t n 4n(n 1)a n t n 2 + 3na n t n + a n t n =0 n=0 n=2 n(n 1)a n t n 4(n + 2)(n + 1)a n+2 t n + 3na n t n + a n t n =0 n=1 n=1 n=0 n=0 n=0 n=0 [n(n 1)a n 4(n + 1)(n + 2)a n+2 + 3na n + a n ] t n =0. n=0

114 15 SERIES SOLUTIONS OF DIFFERENTIAL EQUATIONS Equating all the coefficients to zero we find and obtain the recurrence formula, From this formula, we find (n + 1) 2 a n 4(n + 1)(n + 2)a n+2 = 0 a n+2 = (n + 1)a n 4(n + 2), n 0. a 2 = a 0 8 a 3 = a 1 6 a 4 = 3a 0 128 a 5 = a 1 30 a 6 = 5a 0 1024 a 7 = a 1 140.. Notice that each even coefficient is expressed in terms of a 0 and each odd coefficient is expressed in terms of a 1. The general solution is: ( y(t) = a 0 1 + 1 8 t2 + 3 128 t4 + 5 ( 1024 t6 + )+a 1 t + 1 6 t3 + 1 30 t5 + 1 ) 140 t7 +

115 Practice Problems Problem 15.1 Identify the singular and ordinary points of the differential equation y + ty + (t 2 + 2)y = 0. Problem 15.2 Identify the singular and ordinary points of the differential equation (t 2 1)y + ty + 1 t y = 0. Problem 15.3 Identify the singular and ordinary points of the differential equation (t 2 4)y + 3ty + y = 0. Problem 15.4 The point t 0 = 0 is an ordinary point of the differential equation (t 2 4)y + 3ty + y = 0. Determine a value of R as stated in Theorem 15.1. Problem 15.5 Find the series solution near the ordinary point 0 to the differential equation y + ty + (t 2 + 2)y = 0. Problem 15.6 Find the series solution near the ordinary point 0 to the differential equation y ty t 2 y = 0. Problem 15.7 Find the series solution near the ordinary point 0 to the differential equation (t 2 + 1)y y + y = 0. Problem 15.8 Write the following as a single power series: n=0 na n (t 2) n+1 + n=2 n 2 a n (t 2) n.

116 15 SERIES SOLUTIONS OF DIFFERENTIAL EQUATIONS Problem 15.9 Find the series solution near the ordinary point 0 to the differential equation y + ty + y = 0. Problem 15.10 Find the series solution near the ordinary point 0 to the differential equation (1 t) 2 y 2y = 0.

16 The Structure of the General Solution of Linear Nonhomogeneous Equations In this section, we consider the question of finding the general solution to the differential equation y + p(t)y + q(t)y = g(t) (16.1) where p(t), q(t) and g(t) are continuous functions for a < t < b. The following theorem provides the structure of the general solution to equation (16.1). Theorem 16.1 Let {y 1 (t), y 2 (t)} be a fundamental set of solutions to the homogeneous equation y + p(t)y + q(t)y = 0 and y p (t) be a particular solution of the nonhomogeneous equation y + p(t)y + q(t)y = g(t), a < t < b. The general solution of the nonhomogeneous equation is given by for constants c 1 and c 2. y(t) = y p (t) + c 1 y 1 (t) + c 2 y 2 (t) Proof. Let y(t) be any solution to equation (16.1). Because y p (t) is also a solution then (y y p ) + p(t)(y y p ) + q(t)(y y p ) =(y + p(t)y + q(t)y) (y p + p(t)y p + q(t)y p ) =g(t) g(t) = 0. 117

11816 THE STRUCTURE OF THE GENERAL SOLUTION OF LINEAR NONHOMOGENEO Therefore, y y p is a solution to the homogeneous equation. But {y 1, y 2 } is a fundamental set of solutions to the homogeneous equation so that there exist unique constants c 1 and c 2 such that y(t) y p (t) = c 1 y 1 (t) + c 2 y 2 (t). Hence, y(t) = y p (t) + c 1 y 1 (t) + c 2 y 2 (t) It follows from the above theorem that finding the general solution to nonhomogeneous equations consists of three steps: 1. Find the general solution of the associated homogeneous equation y + p(t)y + q(t)y = 0. 2. Find a single solution of the original equation y + p(t)y + q(t)y = g(t). 3. Add together the solutions found in steps 1 and 2. Example 16.1 Use the fact that y p (t) = 3t 1 to find the unique solution to the initial value problem y 2y 3y = 9t 3, y(0) = 1, y (0) = 3. Solution. Because we are given y p then we need to find the general solution of the homogeneous equation y 2y 3y = 0. The associated characteristic equation r 2 2r 3 = 0 has roots r 1 = 1 and r 2 = 3. Hence, the general solution to the differential equation is y(t) = c 1 e t + c 2 e 3t + 3t 1. The derivative of this function is given by y (t) = c 1 e t +3c 2 e 3t +3. The condition y(0) = 1 leads to c 1 + c 2 = 2. The condition y (0) = 3 leads to c 1 + 3c 2 = 0. Solving for c 1 and c 2 we find c 1 = 3 2 and c 2 = 1 2. The unique solution is given by y(t) = 3 2 e t + 1 2 e3t + 3t 1 Note that y 1 (t) = 3t 1 and y 2 (t) = e 3t + 3t 1 both are particular solutions to the given differential equation. However, the sum u(t) = y 1 (t)+y 2 (t) = e 3t +6t 2 is not a solution since u 2u 3u = 18t 6 9t 3. This shows that the superposition of solutions is valid only for homogeneous equations and not true in general for nonhomogeneous equations. However, we can have a property of superposition of nonhomogeneous if one is adding two solutions of two different nonhomogeneous equations. More precisely, we have

119 Theorem 16.2 Let y 1 (t) be a solution of y + p(t)y + q(t)y = g 1 (t) and y 2 (t) a solution of y + p(t)y + q(t)y = g 2 (t). Then for any constants c 1 and c 2 the function Y (t) = c 1 y 1 (t) + c 2 y 2 (t) is a solution of the equation Proof. We have y + p(t)y + q(t)y = c 1 g 1 (t) + c 2 g 2 (t). Y + p(t)y + q(t)y =c 1 y 1 + c 2 y 2 + p(t)c 1 y 1 + p(t)c 2 y 2 + q(t)c 1 y 1 + q(t)c 2 y 2 =c 1 (y 1 + p(t)y 1 + q(t)y 1 ) + c 2 (y 2 + p(t)y 2 + q(t)y 2 ) =c 1 g 1 (t) + c 2 g 2 (t) Example 16.2 The functions u 1 (t) and u 2 (t) are solutions to the following differential equations u 1 + p(t)u 1 + q(t)u 1 =2e t t 1 u 2 + p(t)u 2 + q(t)u 2 =3t Use the functions u 1 and u 2 to construct a particular solution of the differential equation u + p(t)u + q(t)u = 4e t 2. Solution. The right-hand side of the given equation can be written as 4e t 2 = 2(2e t t 1) + 2 3 (3t) so that by the previous theorem, the function u(t) = 2u 1(t) + 2 3 u 2(t) is the required particular solution Example 16.3 Consider the IVP y y 2y = 4e t, y(0) = 0, y (0) = 0, y p (t) = 4 3 te t. (a) Verify that the given function, y p (t), is a particular solution of the differential equation. (b) Determine the general solution,y h, of the homogeneous equation. (c) Find the general solution to the differential equation and impose the initial conditions to obtain the unique solution of the initial value problem.

12016 THE STRUCTURE OF THE GENERAL SOLUTION OF LINEAR NONHOMOGENEO Solution. (a) y p = 4 3 e t + 4 3 te t, y p = 8 3 e t 4 3 te t. y p y p 2y p = 8 3 e t 4 3 te t + 4 3 e t 4 3 te t + 8 3 te t =4e t (b) The associated characteristic equation r 2 r 2 = 0 has roots r 1 = 1 and r 2 = 2. Hence, the general solution to the homogeneous differential equation is y h (t) = c 1 e t + c 2 e 2t (c) The general solution to the differential equation is y(t) = c 1 e t + c 2 e 2t 4 3 te t. The derivative of this function is given by y (t) = c 1 e t + 2c 2 e 2t 4 3 e t + 4 3 te t. The condition y(0) = 0 leads to c 1 + c 2 = 0. The condition y (0) = 0 leads to c 1 + 2c 2 = 4 3. Solving for c 1 and c 2 we find c 1 = 4 9 and c 2 = 4 9. The unique solution is given by Example 16.4 Determine the function g(t) if y(t) = 4 9 (e2t e t 3te t ) y 2y 3y = g(t), y p (t) = 3e 5t. Solution. We have y p = 15e 5t and y p = 75e 5t. Thus, g(t) =y p 2y p 3y p = 75e 5t 30e 5t 9e 5t = 36e 5t Example 16.5 The general solution of the nonhomogeneous differential equation y + αy + βy = g(t) is given below, where c 1 and c 2 are arbitrary constants. Determine the constants α and β and the function g(t). y(t) = c 1 e t + c 2 e 2t + 2e 2t. Solution. From the given general solution we see that the roots of the characteristic equation are r 1 = 1 and r 2 = 2. Thus, the characteristic equation is (r 1)(r 2) = r 2 3r + 2 = 0. The associated differential equation is y 3y + 2y = 0. Hence, α = 3 and β = 2. Now, letting y p (t) = 2e t, we have g(t) =y p 3y p + 2y p =8e 2t + 12e 2t + 4e 2t = 24e 2t

121 Practice Problems In Problems 16.1-16.6, answer the following three question. (a) Verify that the given function, y p (t), is a particular solution of the differential equations. (b) Determine the general solution,y h, of the homogeneous equation. (c) Find the general solution to the differential equation and impose the initial conditions to obtain the unique solution of the initial value problem. Problem 16.1 Problem 16.2 Problem 16.3 Problem 16.4 Problem 16.5 y 2y 3y = e 2t, y(0) = 1, y (0) = 0, y p (t) = 1 3 e2t. y y 2y = 10, y( 1) = 0, y ( 1) = 1, y p (t) = 5. y + y = 2e t, y(0) = 2, y (0) = 2, y p (t) = 2te t. y + 4y = 10e t π, y(π) = 2, y (π) = 0, y p (t) = 2e t π. y 2y + 2y = 5 sin t, y( π 2 ) = 1, y ( π 2 ) = 0, y p(t) = 2 cos t + sin t. Problem 16.6 y 2y + y = t 2 + 4 + 2 sin t, y(0) = 1, y (0) = 3, y p (t) = t 2 + 4t + 10 + cos t. The functions u 1, u 2, and u 3 are solutions to the following differential equations u + p(t)u + q(t)u =2e t + 1 u + p(t)u + q(t)u =2e t t 1 u + p(t)u + q(t)u =3t. In Problems 16.7-16.8, use the functions u 1, u 2 and u 3 to construct a particular solution of the differential equation.

12216 THE STRUCTURE OF THE GENERAL SOLUTION OF LINEAR NONHOMOGENEO Problem 16.7 Problem 16.8 u + p(t)u + q(t)u = e t + 2t + 1 2. u + p(t)u + q(t)u = et + e t. 2 In Problems 15.9-15.11, determine the function g(t). Problem 16.9 Problem 16.10 Problem 16.11 y 2y = g(t), y p (t) = 3t + t, t > 0. y + y = g(t), y p (t) = ln (1 + t), t > 1. y + 2y + y = g(t), y p (t) = t 2. In Problems 16.12-16.13, the general solution of the nonhomogeneous differential equation y + αy + βy = g(t) is given, where c 1 and c 2 are arbitrary constants. Determine the constants α and β and the function g(t). Problem 16.12 Problem 16.13 y(t) = c 1 e t + c 2 te t + t 2 e t. y(t) = c 1 sin 2t + c 2 cos 2t 1 + sin t. Problem 16.14 Given that the function et 5 satisfies the differential equation y + 4y = e t, write a general solution of the differential equation y + 4y = e t. Problem 16.15 Find the general solution to the differential equation y (4) + 9y = 24 + 108t 2 given a particular solution y p (t) = cos 3t + sin 3t + t 4.

17 The Method of Variation of Parameters In this section, we discuss a method for finding a particular solution to a nonhomogeneous differential equation y + p(t)y + q(t)y = g(t), a < t < b. (17.1) This method has no prior conditions to be satisfied by either p(t), q(t), or g(t). To use this method, we first find the general solution to the homogeneous equation y(t) = c 1 y 1 (t) + c 2 y 2 (t). Then we replace the parameters c 1 and c 2 by two functions u 1 (t) and u 2 (t) to be determined. From this the method got its name. Thus, obtaining y p (t) = u 1 (t)y 1 (t) + u 2 (t)y 2 (t). Observe that if u 1 and u 2 are constant functions then the above y is just the homogeneous solution to the differential equation. In order to determine the two functions one has to impose two constraints. Finding the derivative of y p we obtain Finding the second derivative to obtain y p = (y 1u 1 + y 2u 2 ) + (y 1 u 1 + y 2 u 2). y p = y 1u 1 + y 1u 1 + y 2u 2 + y 2u 2 + (y 1 u 1 + y 2 u 2). Since it is up to us to choose u 1 and u 2 we decide to do that in such a way to make our computation simple. One way to achieving that is to impose the condition y 1 u 1 + y 2 u 2 = 0. (17.2) 123

124 17 THE METHOD OF VARIATION OF PARAMETERS Under such a constraint y p and y p are simplified to and y p = y 1u 1 + y 2u 2 y p = y 1u 1 + y 1u 1 + y 2u 2 + y 2u 2. In particular, y p does not involve u 1 and u 2. Inserting y p, y p, and y p into equation (17.1) to obtain [y 1 u 1 + y 1 u 1 + y 2 u 2 + y 2 u 2 ] + p(t)(y 1 u 1 + y 2 u 2) + q(t)(u 1 y 1 + u 2 y 2 ) = g(t). Rearranging terms, [y 1 + p(t)y 1 + q(t)y 1]u 1 + [y 2 + p(t)y 2 + q(t)y 2]u 2 + [u 1 y 1 + u 2 y 2 ] = g(t). Since y 1 and y 2 are solutions to the homogeneous equation, the previous equation yields our second constraint u 1y 1 + u 2y 2 = g(t). (17.3) Combining equation (17.2) and (17.3) we find the system of two equations in the unknowns u 1 and u 2 y 1 u 1 + y 2 u 2 =0 u 1y 1 + u 2y 2 =g(t). Since {y 1, y 2 } is a fundamental set, the expression W (t) = y 1 y 2 y 1 y 2 is nonzero so that one can find unique u 1 and u 2. Using the method of elimination, these functions are given by u 1 (t) = y 2(t)g(t) W (t) Computing antiderivatives to obtain and u 2 (t) = y 1(t)g(t) W (t). u 1 (t) = y 2(t)g(t) W (t) Example 17.1 Find the general solution of dt and u 2 (t) = y 1 (t)g(t) W (t) dt. y y 2y = 2e t using the method of variation of parameters.

125 Solution. The characteristic equation r 2 r 2 = 0 has roots r 1 = 1 and r 2 = 2. Thus, y 1 (t) = e t, y 2 (t) = e 2t and W (t) = 3e t. Hence, e 2t 2e t u 1 (t) = 3e t dt = 2 3 t and The particular solution is The general solution is then given by e t 2e t u 2 (t) = 3e t dt = 2 9 e 3t. y p (t) = 2 3 te t 2 9 e t. y(t) = c 1 e t + c 2 e 2t 2 3 te t 2 9 e t Example 17.2 Find the general solution to (2t 1)y 4ty + 4y = (2t 1) 2 e t if y 1 (t) = t and y 2 (t) = e 2t form a fundamental set of solutions to the equation. Solution. First we rewrite the equation in standard form y 4t 2t 1 y + 4 2t 1 y = (2t 1)e t. Since W (t) = (2t 1)e 2t we find e 2t (2t 1)e t u 1 (t) = (2t 1)e 2t dt = e t and Thus, t (2t 1)e t u 2 (t) = (2t 1)e 2t dt = 1 3 te 3t 1 9 e 3t. y p (t) = te t 1 3 te t 1 9 e t = 2 3 te t 1 9 e t. The general solution is y(t) = c 1 t + c 2 e 2t + 2 3 te t 1 9 e t

126 17 THE METHOD OF VARIATION OF PARAMETERS Example 17.3 Find the general solution to the differential equation y + y = ln t, t > 0. Solution. The characterisitc equation r 2 + r = 0 has roots r 1 = 0 and r 2 = 1 so that y 1 (t) = 1, y 2 (t) = e t, and W (t) = e t. Hence, e t ln t u 1 (t) = dt = ln tdt = t ln t t e t ln t e u 2 (t) = dt = e t ln tdt = e t t ln t + e t t dt Thus, and Example 17.4 Find the general solution of y p (t) = t ln t t ln t + e t e t t dt y(t) = c 1 + c 2 e t + t ln t t ln t + e t e t y + y = 1 2 + sin t. Solution. Since the characteristic equation r 2 + 1 = 0 has roots r = ±i, the general solution of the corresponding homogeneous equation y + y = 0 is given by Since W (t) = 1 we find u 1 (t) = u 2 (t) = y h (t) = c 1 cos t + c 2 sin t dt = t + sin t 2 + sin t cos t dt = ln (2 + sin t) 2 + sin t Hence, the particular solution is y p (t) = sin t ln (2 + sin t) + cos t( and the general solution is y(t) = c 1 cos t + c 2 sin t + y p (t) t dt 2 2 + sin t dt 2 dt t) 2 + sin t

127 Practice Problems Problem 17.1 Solve y + y = sec t by variation of parameters. Problem 17.2 Solve y y = e t by variation of parameters. Problem 17.3 Solve the following 2nd order equation using the variation of parameter method: y + 4y = t 2 + 8 cos 2t. Problem 17.4 Find a particular solution by the variation of parameters to the equation y + 2y + y = e t ln t. Problem 17.5 Solve the following initial value problem by using variation of parameters: y + 2y 3y = te t, y(0) = 1 64, y (0) = 59 64. Problem 17.6 (a) Verify that {e t, e t } is a fundamental set for the equation 4ty + 2y y = 0 on the interval (0, ). You may assume that the given functions are solutions to the equation. (b) Use the method of variation of parameters to find one solution to the equation 4ty + 2y y = 4 te t. Problem 17.7 Use the method of variation of parameters to find the general solution to the equation y + y = sin t.

128 17 THE METHOD OF VARIATION OF PARAMETERS Problem 17.8 Consider the differential equation t 2 y + 3ty 3y = 0, t > 0. (a) Determine r so that y = t r is a solution. (b) Use (a) to find a fundamental set of solutions. (c) Use the method of variation of parameters for finding a particular solution to t 2 y + 3ty 3y = 1 t 3, t > 0. Problem 17.9 Use the method of variation of parameters to find the general solution to the differential equation y + y = sin 2 t.

18 The Laplace Transform: Basic Definitions and Results Laplace transform is yet another operational tool for solving constant coefficients linear differential equations. The process of solution consists of three main steps: The given hard problem is transformed into a simple equation. This simple equation is solved by purely algebraic manipulations. The solution of the simple equation is transformed back to obtain the solution of the given problem. In this way the Laplace transformation reduces the problem of solving a differential equation to an algebraic problem. The third step is made easier by tables, whose role is similar to that of integral tables in integration. The above procedure can be summarized by Figure 18.1. Figure 18.1 In this section we introduce the concept of Laplace transform and discuss some of its properties. The Laplace transform is defined in the following way. Let f(t) be defined for t 0. Then the Laplace transform of f, which is denoted by L[f(t)] or by F (s), is defined by the following equation T L[f(t)] = F (s) = lim f(t)e st dt = T 0 0 f(t)e st dt. The integral which defines a Laplace transform is an improper integral. An improper integral may converge or diverge, depending on the integrand. When the improper integral is convergent then we say that the function f(t) possesses 129

13018 THE LAPLACE TRANSFORM: BASIC DEFINITIONS AND RESULTS a Laplace transform. So what types of functions possess Laplace transforms, that is, what type of functions guarantees a convergent improper integral? Example 18.1 Find the Laplace transform, if it exists, of each of the following functions. (a) f(t) = e at (b) f(t) = 1 (c) f(t) = t (d) f(t) = e t2. Solution. (a) Using the definition of Laplace transform we see that But L[e at ] = T 0 0 e (s a)t dt = T e (s a)t dt = lim e (s a)t dt. T 0 { T if s = a 1 e (s a)t s a if s a. For the improper integral to converge we need s > a. In this case, L[e at ] = F (s) = 1 s a, s > a. (b) In a similar way to what was done in part (a), we find (c) We have L[1] = 0 T e st dt = lim e st dt = 1 T 0 s, s > 0. L[t] = 0 te st dt = [ te st s ] e st s 2 = 1 0 s 2, s > 0. (d) Again using the definition of Laplace transform we find L[e t2 ] = 0 e t2 st dt. If s 0 then t 2 st 0 so that e t2 st 1 and this implies that 0 e t2 st dt 0 dt. Since the integral on the right is divergent, by the comparison theorem of improper integrals the integral on the left is also divergent. Now, if s > 0 then 0 e t(t s) dt s dt. By the same reasoning the integral on the left is divergent. This shows that the function f(t) = e t2 does not possess a Laplace transform

131 The above example raises the question of what class or classes of functions possess a Laplace transform. Looking closely at Example 18.1(a), we notice that for s > a the integral 0 e (s a)t dt is convergent and a critical component for this convergence is the type of the function f(t). To be more specific, if f(t) is a continuous function such that f(t) Me at, t C (18.1) where M 0 and a and C are constants, then this condition yields 0 f(t)e st dt C 0 f(t)e st dt + M C e (s a)t dt. Since f(t) is continuous in 0 t C, by letting A = max{ f(t) : 0 t C} we have C C ( ) 1 f(t)e st dt A e st dt = A 0 s e sc <. s 0 Also, by Example 18.1(a), the integral C e (s a)t dt is convergent for s > a. By the comparison theorem of improper integrals the integral on the left is also convergent. That is, f(t) possesses a Laplace transform. We call a function that satisfies condition (18.1) a function with an exponential order a. Graphically, this means that the graph of f(t) is contained in the region bounded by the graphs of y = Me at and y = Me at for t C. Note also that this type of functions controls the negative exponential in the transform integral so that to keep the integral from blowing up. If C = 0 then we say that the function is exponentially bounded. Note that f has exponential order a if lim t e at f(t) = 0. Example 18.2 (a) Show that f(t) = t n where n is a positive integer, has an exponential order. (b) Show that any bounded function f(t) for t 0 is exponentially bounded. (c) Show that f(t) = e t2 is not of exponential order. Solution. (a) Let a > 0. Using L Hôpital s rule n times we can see that lim t e at t n = 0. Thus, there is a positive constant C such that e at t n 1 for all t C. That is, t n e at for all t C. (b) Since f(t) is bounded for t 0, there is a positive constant M such that f(t) M for all t 0. But this is the same as (18.1) with a = 0 and C = 0. Thus, f(t) is exponentially bounded. (c) Suppose not. That is, let M and C be non-negative constants such that

13218 THE LAPLACE TRANSFORM: BASIC DEFINITIONS AND RESULTS e t2 Me at for some constant a and for all t C. Hence, e t2 at M for all t C. Letting t we find M which is impossible since M <. It follows that f(t) = e t2 is not of exponential order Another question that comes to mind is whether it is possible to relax the condition of continuity on the function f(t). Let s look at the following situation. Example 18.3 Show that the square wave function whose graph is given in Figure 18.2 possesses a Laplace transform. Figure 18.2 Note that the function is periodic of period 2. Solution. Since f(t)e st e st, 0 f(t)e st dt 0 e st dt. But the integral on the right is convergent for s > 0 so that the integral on the left is convergent as well. That is, L[f(t)] exists for s > 0 The function of the above example belongs to a class of functions that we define next. A function is called piecewise continuous on an interval if it consists of a finite number of continuous pieces with possibly either removable or jump discontinuities (but no inifinite discontinuities). Below is a sketch of a piecewise continuous function. Figure 18.3 An example of a function that is not piecewise continuous is the function f(t) = 1 t 1 on the interval [0, ) since at t = 1 the continuity is infinite.

133 Example 18.4 Show that the following function is piecewise continuous and exponentially bounded. { e t sin (t 1) if t 1 f(t) = 1 2 if 0 t < 1. Solution. Since e t sin (t 1) = (e sin (t 1) ) t (e 1 ) t = e t for all t 0, f(t) is piecewise continuous and exponentially bounded The following theorem guarantees the existence of the Laplace transform for all functions that are piecewise continuous and have exponential order. Theorem 18.1 (Existence) Suppose that f(t) is piecewise continuous on t 0 and has an exponential order with f(t) Me at for t C. Then the Laplace transform F (s) = 0 f(t)e st dt exists as long as s > a. Note that the two conditions above are sufficient, but not necessary, for F (s) to exist. Example 18.5 Consider the floor function f(t) = t, where for any integer n we have t = n for all n t < n + 1. (a) Is f(t) continuous on 0 t <, discontinuous but piecewise continuous on 0 t <, or neither? (b) Are there fixed numbers a and M such that f(t) Me at for 0 t <? Solution. (a) The floor function is a piecewise continuous function on 0 t <. (b) Since t t < e t for 0 t < we find M = 1 and a = 1 Example 18.6 Consider the function f(t) = t 2 e t. (a) Is f(t) continuous on 0 t <, discontinuous but piecewise continuous on 0 t <, or neither? (b) Are there fixed numbers a and M such that f(t) Me at for 0 t <?

13418 THE LAPLACE TRANSFORM: BASIC DEFINITIONS AND RESULTS Solution. (a) Since t 2 and e t are continuous everywhere, f(t) = t 2 e t is continuous on 0 t <. (b) By L Hôpital s rule one has t 2 lim t e t = 0. Since f(0) = 0, f(t) is bounded. Since f (t) = (2t t 2 )e t, f(t) has a maximum when t = 2. The value of this maximum is f(2) = 4e 2. Hence, M = 4e 2 and a = 0 In what follows, we will denote the class of all piecewise continuous functions with exponential order by PE. The next theorem shows that any linear combination of functions in PE is also in PE. The same is true for the product of two functions in PE. Theorem 18.2 Suppose that f(t) and g(t) are two elements of PE with f(t) M 1 e a 1t, t C 1 and g(t) M 2 e a 1t, t C 2. (i) For any constants α and β the function αf(t) + βg(t) is also a member of PE. Moreover L[αf(t) + βg(t)] = αl[f(t)] + βl[g(t)]. (ii) The function h(t) = f(t)g(t) is an element of PE. Example 18.7 Use the linearity property of Laplace transform to find L[5e 7t + t + 2e 2t ]. Find the domain of F (s). Solution. We have L[e 7t ] = 1 s+7, s > 7, L[t] = 1, s > 0, and L[e 2t ] = 1 s 2 s 2, s > 2. Hence, L[5e 7t + t + 2e 2t ] = 5L[e 7t ] + L[t] + 2L[e 2t ] = 5 s + 7 + 1 s 2 + 2 s 2, s > 2 We next discuss the problem of how to determine the function f(t) if F (s) is given. That is, how do we invert the transform. The following result on uniqueness provides a possible answer. This result establishes a one-to-one correspondence between the set PE and its Laplace transforms. Alternatively, the following theorem asserts that the Laplace transform of a member in PE is unique.

135 Theorem 18.3 Let f(t) and g(t) be two elements in PE with Laplace transforms F (s) and G(s) such that F (s) = G(s) for some s > a. Then f(t) = g(t) for all t 0 where both functions are continuous. With the above theorem, we can now officially define the inverse Laplace transform as follows: For a piecewise continuous function f with exponential order whose Laplace transform is F, we call f the inverse Laplace transform of F and write f = L 1 [F (s)]. Symbolically Example( 18.8) Find L 1 1 s 1, s > 1. f(t) = L 1 [F (s)] F (s) = L[f(t)]. Solution. From Example 18.1(a), we have that L[e at ] = 1 s a we find that L[e t ] = 1 s 1, s > 1. Hence, L 1 ( 1 s 1, s > a. In particular, for a = 1 ) = e t, t 0 The above theorem states that if f(t) is continuous and has a Laplace transform F (s), then there is no other function that has the same Laplace transform. To find L 1 [F (s)], we can inspect tables of Laplace transforms of known functions to find a particular f(t) that yields the given F (s). When the function f(t) is not continuous, the uniqueness of the inverse Laplace transform is not assured. The following example addresses the uniqueness issue. Example 18.9 Consider the two functions f(t) = h(t)h(3 t) and g(t) = h(t) h(t 3) where { 1 if t 0 h(t) = 0 otherwise. (a) Are the two functions identical? (b) Show that L[f(t)] = L[g(t). Solution. (a) We have and f(t) = g(t) = { 1, 0 t 3 0, t > 3 { 1, 0 t < 3 0, t 3.

13618 THE LAPLACE TRANSFORM: BASIC DEFINITIONS AND RESULTS Since f(3) = 1 and g(3) = 0, f and g are not identical. (b) We have L[f(t)] = L[g(t)] = 3 0 e st dt = 1 e 3s, s > 0. s Thus, both functions f(t) and g(t) have the same Laplace transform even though they are not identical. However, they are equal on the interval(s) where they are both continuous The inverse Laplace transform possesses a linear property as indicated in the following result. Theorem 18.4 Given two Laplace transforms F (s) and G(s) then for any constants a and b. Example( 18.10 ) Find L 1 2 s+2 + 2 s 2. Solution. We have L 1 ( 2 s + 2 + 2 s 2 L 1 [af (s) + bg(s)] = al 1 [F (s)] + bl 1 [G(s)] ) ( ) ( ) 1 1 = 2L 1 + 2L 1 = 2(e 2t + e 2t ), t 0 s + 2 s 2

137 Practice Problems Problem 18.1 Determine whether the integral 1 0 dt converges. If the integral converges, 1+t 2 give its value. Problem 18.2 Determine whether the integral t 0 dt converges. If the integral converges, 1+t 2 give its value. Problem 18.3 Determine whether the integral 0 e t cos (e t )dt converges. If the integral converges, give its value. Problem 18.4 Using the definition, find L[e 3t ], if it exists. If the Laplace transform exists then find the domain of F (s). Problem 18.5 Using the definition, find L[t 5], if it exists. If the Laplace transform exists then find the domain of F (s). Problem 18.6 Using the definition, find L[e (t 1)2 ], if it exists. If the Laplace transform exists then find the domain of F (s). Problem 18.7 Using the definition, find L[(t 2) 2 ], if it exists. If the Laplace transform exists then find the domain of F (s). Problem 18.8 Using the definition, find L[f(t)], if it exists. If the Laplace transform exists then find the domain of F (s). { 0, 0 t < 1 f(t) = t 1, t 1. Problem 18.9 Using the definition, find L[f(t)], if it exists. If the Laplace transform exists then find the domain of F (s). f(t) = 0, 0 t < 1 t 1, 1 t < 2 0, t 2.

13818 THE LAPLACE TRANSFORM: BASIC DEFINITIONS AND RESULTS Problem 18.10 Let n be a positive integer. Using integration by parts establish the reduction formula t n e st dt = tn e st + n t n 1 e st dt, s > 0. s s Problem 18.11 For s > 0 and n a positive integer evaluate the limits (a) lim t 0 t n e st (b) lim t t n e st. Problem 18.12 (a) Use the previous two problems to derive the reduction formula for the Laplace transform of f(t) = t n, L[t n ] = n s L[tn 1 ], s > 0. (b) Calculate L[t k ], for k = 1, 2, 3, 4, 5. (c) Formulate a conjecture as to the Laplace transform of f(t), t n with n a positive integer. From a table of integrals, e αu αu α sin βu β cos βu sin βudu =e α 2 + β 2 e αu αu α cos βu + β sin βu cos βudu =e α 2 + β 2 Problem 18.13 Use the above integrals to find the Laplace transform of f(t) = cos ωt, if it exists. If the Laplace transform exists, give the domain of F (s). Problem 18.14 Use the above integrals to find the Laplace transform of f(t) = sin ωt, if it exists. If the Laplace transform exists, give the domain of F (s). Problem 18.15 Use the above integrals to find the Laplace transform of f(t) = cos ω(t 2), if it exists. If the Laplace transform exists, give the domain of F (s). Problem 18.16 Use the above integrals to find the Laplace transform of f(t) = e 3t sin t, if it exists. If the Laplace transform exists, give the domain of F (s).

139 Problem 18.17 Consider the function f(t) = tan t. (a) Is f(t) continuous on 0 t <, discontinuous but piecewise continuous on 0 t <, or neither? (b) Are there fixed numbers a and M such that f(t) Me at for 0 t <? Problem 18.18 Consider the function f(t) = et2 e 2t +1. (a) Is f(t) continuous on 0 t <, discontinuous but piecewise continuous on 0 t <, or neither? (b) Are there fixed numbers a and M such that f(t) Me at for 0 t <? Problem( 18.19 ) Find L 1 3 s 2. Problem 18.20 ) Find L ( 1 2 + 1 s 2 s+1.

14018 THE LAPLACE TRANSFORM: BASIC DEFINITIONS AND RESULTS

19 Further Studies of Laplace Transform Properties of the Laplace transform enable us to find Laplace transforms without having to compute them directly from the definition. In this section, we establish properties of Laplace transform that will be useful for solving ODEs. Laplace Transform of the Heaviside Step Function The Heaviside step function is a piecewise continuous function defined by h(t) = { 1, t 0 0, t < 0. Figure 19.1 displays the graph of h(t). Figure 19.1 Taking the Laplace transform of h(t) we find L[h(t)] = 0 h(t)e st dt = 0 e st dt = 141 ] [ e st s 0 = 1 s, s > 0.

142 19 FURTHER STUDIES OF LAPLACE TRANSFORM A Heaviside function at α 0 is the shifted function h(t α) (α units to the right). For this function, the Laplace transform is L[h(t α)] = 0 h(t α)e st dt = α e st dt = Laplace Tranform of e at The Laplace transform for the function f(t) = e at is [ L[e at ] = e (s a)t dt = e (s a)t s a Laplace Tranforms of sin at and cos at Using integration by parts twice we find 0 ] L[sin at] = e st sin atdt 0 [ = e st sin at ae st cos at s s 2 = a s 2 a2 L[sin at] s2 ( s 2 + a 2 ) s 2 L[sin at] = a s 2 L[sin at] = a s 2 + a 2, s > 0. A similar argument shows that L[cos at] = Laplace Transforms of cosh at and sinh at Using the linear property of L we can write A similar argument shows that 0 ] [ e st s α = 1 s a, s > a. ] 0 s s 2 + a 2, s > 0. a2 s 2 L[cosh at] = 1 ( L[e at ] + L[e at ] ) 2 = 1 ( 1 2 s a + 1 ), s > a s + a = s s 2 a 2, s > a. L[sinh at] = a s 2 a 2, s > a. = e sα s, s > 0. 0 e st sin atdt

143 Laplace Transform of a Polynomial Let n be a positive integer. Using integration by parts we can write 0 [ t t n e st n e st dt = s ] 0 + n s 0 t n 1 e st dt. By repeated use of L Hôpital s rule we find lim t t n e st n! = lim t s > 0. Thus, L[t n ] = n s L[tn 1 ], s > 0. s n e st = 0 for Using induction on n = 0, 1, 2, one can easily establish that L[t n ] = n!, s > 0. sn+1 Using the above result together with the linearity property of L one can find the Laplace transform of any polynomial. The next two results are referred to as the first and second shift theorems. As with the linearity property, the shift theorems increase the number of functions for which we can easily find Laplace transforms. Theorem 19.1 (First Shifting Theorem) If f(t) is a piecewise continuous function for t 0 and has exponential order with f(t) Me at, t C, then for any real number α we have L[e αt f(t)] = F (s α), s > a + α where L[f(t)] = F (s). Theorem 19.2 (Second Shifting Theorem) If f(t) is a piecewise continuous function for t 0 and has exponential order with f(t) Me at, t C, then for any real number α 0 we have L[f(t α)h(t α)] = e αs F (s), s > a where L[f(t)] = F (s) and h(t) is the Heaviside step function. Example 19.1 Find (a) L[e 2t t 2 ] (b) L[e 3t cos 2t] (c) L 1 [ e 2s s 2 ].

144 19 FURTHER STUDIES OF LAPLACE TRANSFORM Solution. (a) By Theorem 19.1, we have L[e 2t t 2 ] = F (s 2) where L[t 2 ] = 2! = F (s), s > 0. s 3 Thus, L[e 2t t 2 ] = 2, s > 2. (s 2) 3 (b) As in part (a), we have L[e 3t cos 2t] = F (s 3) where L[cos 2t] = F (s). But L[cos 2t] =, s > 0. Thus, s s 2 +4 L[e 3t cos 2t] = (c) Since L[t] = 1 s 2, by Theorem 19.2, we have s 3 (s 3) 2 + 4, s > 3. e 2s s 2 = L[(t 2)h(t 2)]. Therefore, [ ] { e L 1 2s 0, 0 t < 2 s 2 = (t 2)h(t 2) = t 2, t 2 The following result relates the Laplace transform of derivatives and integrals to the Laplace transform of the function itself. Theorem 19.3 Suppose that f(t) is continuous for t 0 and f (t) is piecewise continuous and exponentially bounded. Then (a) f(t) is exponentially bounded. (b) L[f (t)] = sl[f(t)] f(0) = sf (s) f(0), s > max{a, 0} + 1. (c) L[f (t)] = s 2 L[f(t)] sf(0) f (0) = s 2 F (s) sf(0) f(0), s > max{a, 0}+1. (d) L [ ] t 0 f(u)du = L[f(t)] s = F (s) s, s > max{a, 0} + 1. The results in (b) and (c) of the previous theorem can be extended to higher order derivatives. Theorem 19.4 Let f(t), f (t),, f (n 1) (t) be continuous and f (n) (t) be piecewise continuous and exponentially bounded. Then L[f (n) (t)] = s n L[f(t)] s n 1 f(0) s n 2 f (0) f (n 1) (0), s > max{a, 0} + 1. We next illustrate the use of the previous theorem in solving initial value problems.

145 Example 19.2 Use Laplace transform technique to solve the initial value problem y 4y = e 3t, y(0) = 0, y (0) = 0. Solution. Taking the Laplace transform of the DE and using linearity we find L[y ] 4L[y] = L[e 3t ]. But L[y ] = s 2 L[y] sy(0) y (0) = s 2 L[y]. Letting L[y] = Y (s) we obtain Solving for Y (s) we find s 2 Y (s) 4Y (s) = 1 s 3. Y (s) = Using partial fraction decomposition 1 (s 3)(s 2)(s + 2). 1 (s 3)(s 2)(s + 2) = A s 3 + B s + 2 + C s 2 we find A = 1 5, B = 1 20, and C = 1 4. Thus, [ ] y(t) =L 1 1 = 1 [ ] 1 (s 3)(s 2)(s + 2) 5 L 1 + 1 [ ] 1 s 3 20 L 1 1 [ ] 1 s + 2 4 L 1 s 2 = 1 5 e3t + 1 20 e 2t 1 4 e2t, t 0 Example 19.3 Determine the constants α, β, y 0, and y 0 s so that Y (s) = transform of the solution to the initial value problem (s+1) 2 is the Laplace y + αy + βy = 0, y(0) = y 0, y (0) = y 0. Solution. Taking the Laplace transform of both sides we find Solving for Y (s) we find s 2 Y (s) sy 0 y 0 + αsy (s) αy 0 + βy (s) = 0. Y (s) = sy 0 + (y 0 + αy 0) s 2 + αs + β = s s 2 + 2s + 1. By identification we find α = 2, β = 1, y 0 = 1, and y 0 = 2

146 19 FURTHER STUDIES OF LAPLACE TRANSFORM Example 19.4 Solve the initial value problem y 4y + 9y = t, y(0) = 0, y (0) = 1. Solution. We apply Theorem 19.4 that gives the Laplace transform of a derivative. By the linearity property of the Laplace transform we can write Now since L[y ] 4L[y ] + 9L[y] = L[t]. L[y ] =s 2 L[y] sy(0) y (0) = s 2 Y (s) 1 L[y ] =sy (s) y(0) = sy (s) L[t] = 1 s 2 where L[y] = Y (s), we obtain Rearranging gives Thus, and s 2 Y (s) 1 4sY (s) + 9Y (s) = 1 s 2. (s 2 4s + 9)Y (s) = s2 + 1 s 2. Y (s) = s 2 + 1 s 2 (s 2 4s + 9) [ y(t) = L 1 s 2 ] + 1 s 2 (s 2 4s + 9) In the next section we will discuss a method for finding the inverse Laplace transform of the above expression.

147 f(t) h(t) = { 1, t 0 0, t < 0 F(s) 1 s, s > 0 t n, n = 1, 2, n! s n+1, s > 0 e αt 1 s α, s > α sin (ωt) cos (ωt) sinh (ωt) cosh (ωt) e αt f(t), with f(t) Me at ω s 2 +ω 2, s > 0 s s 2 +ω 2, s > 0 ω s 2 ω 2, s > ω s s 2 ω 2, s > ω F (s α), s > α + a e αt h(t) 1 s α, s > α e αt t n, n = 1, 2, n! (s α) n+1, s > α e αt sin (ωt) e αt cos (ωt) f(t α)h(t α), α 0 with f(t) Me at ω (s α) 2 +ω 2, s > α s α (s α) 2 +ω 2, s > α e αs F (s), s > a

148 19 FURTHER STUDIES OF LAPLACE TRANSFORM f(t) h(t α), α 0 tf(t) t s 2ω sin ωt F(s) (continued) e αs s, s > 0 -F (s) (s 2 +ω 2 ) 2, s > 0 1 1 [sin ωt ωt cos ωt], s > 0 2ω 3 (s 2 +ω 2 ) 2 f (t), with f(t) continuous sf (s) f(0) and f (t) Me at s > max{a, 0} + 1 f (t), with f (t) continuous s 2 F (s) sf(0) f (0) and f (t) Me at s > max{a, 0} + 1 f (n) (t), with f (n 1) (t) continuous s n F (s) s n 1 f(0) and f (n) (t) Me at -sf (n 2) (0) f (n 1) (0) s > max{a, 0} + 1 t 0 f(u)du, with f(t) Meat F (s) s Table L, s > max{a, 0} + 1

149 Practice Problems Problem 19.1 Use Table L to find L[2e t + 5]. Problem 19.2 Use Table L to find L[e 3t 3 h(t 1)]. Problem 19.3 Use Table L to find L[sin 2 ωt]. Problem 19.4 Use Table L to find L[sin 3t cos 3t]. Problem 19.5 Use Table L to find L[e 2t cos 3t]. Problem 19.6 Use Table L to find L[e 4t (t 2 + 3t + 5)]. Problem 19.7 Use Table L to find L 1 [ 10 s 2 +25 + 4 s 3 ]. Problem 19.8 Use Table L to find L 1 [ 5 (s 3) 4 ]. Problem 19.9 Use Table L to find L 1 [ e 2s s 9 ]. Problem 19.10 Use Table L to find L 1 [ e 3s (2s+7) s 2 +16 ]. Problem 19.11 Graph the function f(t) = h(t 1) + h(t 3) for t 0, where h(t) is the Heaviside step function, and use Table L to find L[f(t)]. Problem 19.12 Graph the function f(t) = t[h(t 1) h(t 3)] for t 0, where h(t) is the Heaviside step function, and use Table L to find L[f(t)]. Problem 19.13 Graph the function f(t) = 3[h(t 1) h(t 4)] for t 0, where h(t) is the Heaviside step function, and use Table L to find L[f(t)].

150 19 FURTHER STUDIES OF LAPLACE TRANSFORM Problem 19.14 Graph the function f(t) = 2 t [h(t 1) h(t 3)] for t 0, where h(t) is the Heaviside step function, and use Table L to find L[f(t)]. Problem 19.15 Graph the function f(t) = h(2 t) for t 0, where h(t) is the Heaviside step function, and use Table L to find L[f(t)]. Problem 19.16 Graph the function f(t) = h(t 1) + h(4 t) for t 0, where h(t) is the Heaviside step function, and use Table L to find L[f(t)]. Problem 19.17 The graph of f(t) is given below. Represent f(t) as a combination of Heaviside step functions, and use Table L to calculate the Laplace transform of f(t). Problem 19.18 The graph of f(t) is given below. Represent f(t) as a combination of Heaviside step functions, and use Table L to calculate the Laplace transform of f(t). Problem 19.19 Use Laplace transform technique to solve the initial value problem y + 4y = g(t), y(0) = 2 where 0, 0 t < 1 g(t) = 12, 1 t < 3 0, t 3.

20 The Laplace Transform and the Method of Partial Fractions In the last example of the previous section we encountered the equation [ y(t) = L 1 s 2 ] + 1 s 2 (s 2. 4s + 9) We would like to find an explicit expression for y(t). This can be done using the method of partial fractions ( ) which is the topic of this section. According to this method, finding L 1 N(s) D(s), where N(s) and D(s) are polynomials, require decomposing the rational function into a sum of simpler expressions whose inverse Laplace transform can be recognized from a table of Laplace transform pairs. The method of partial fractions decomposition consists of writing a rational function, i.e., function of the form R(s) = N(s) D(s) where N(s) and D(s) are polynomials, as a sum of simpler fractions called partial fractions. This can be done in the following way: Step 1. Use long division to find two polynomials r(s) and q(s) such that N(s) D(s) = q(s) + r(s) D(s). Note that if the degree of N(s) is smaller than that of D(s) then q(s) = 0 and r(s) = N(s). Step 2. Write D(s) as a product of factors of the form (as + b) n or (as 2 + bs + c) n where as 2 + bs + c is irreducible, i.e. as 2 + bs + c = 0 has no real zeros. 151

15220 THE LAPLACE TRANSFORM AND THE METHOD OF PARTIAL FRACTIONS Step 3. Decompose r(s) D(s) into a sum of partial fractions in the following way: (1) For each factor of the form (as + b) k write A 1 as + b + A 2 (as + b) 2 + + A k (as + b) k, where the numbers A 1, A 2,, A k are to be determined. (2) For each factor of the form (as 2 + bs + c) k write B 1 s + C 1 as 2 + bs + c + B 2s + C 2 (as 2 + bs + c) 2 + + B ks + C k (as 2 + bs + c) k, where the numbers B 1, B 2,, B k and C 1, C 2,, C k are to be determined. by D(s) and simplify. This leads to an expres- Step 4. Multiply both sides of r(s) D(s) sion of the form r(s) = a polynomial whose coefficients are combinations of A i, B i, and C i. Finally, we find the constants, A i, B i, and C i by equating the coefficients of like powers of x on both sides of the last equation. Example 20.1 Decompose into partial fractions R(s) = s3 +s 2 +2 s 2 1. Solution. Step 1. s3 +s 2 +2 = s + 1 + s+3 s 2 1 s 2 1. Step 2. s 2 1 = (s 1)(s + 1). s+3 Step 3. (s+1)(s 1) = A s+1 + B s 1. Step 4. Multiply both sides of the last equation by (s 1)(s + 1) to obtain s + 3 = A(s 1) + B(s + 1). Expand the right hand side, collect terms with the same power of s, and identify coefficients of the polynomials obtained on both sides: s + 3 = (A + B)s + (B A). Hence, A + B = 1 and B A = 3. Adding these two equations gives B = 2. Thus, A = 1 and so s 3 + s 2 + 2 s 2 = s + 1 1 1 s + 1 + 2 s 1 Now, after decomposing the rational function into a sum of partial fractions all we need to do is to find the Laplace transform of expressions of the form or Bs+C (as 2 +bs+c) n. A (s α) n

153 Example[ 20.2 ] Find L 1 1 s(s 3). Solution. We write 1 s(s 3) = A s + B s 3. Multiply both sides by s(s 3) and simplify to obtain or 1 = A(s 3) + Bs 1 = (A + B)s 3A. Now equating the coefficients of like powers of s to obtain 3A = 1 and A+B = 0. Solving for A and B we find A = 1 3 and B = 1 3. Thus, [ ] L 1 1 = 1 [ ] 1 s(s 3) 3 L 1 + 1 [ ] 1 s 3 L 1 s 3 = 1 3 h(t) + 1 3 e3t, t 0 where h(t) is the Heaviside unit step function Example[ 20.3 ] Find L 1 3s+6. s 2 +3s Solution. We factor the denominator and split the integrand into partial fractions: 3s + 6 s(s + 3) = A s + B s + 3. Multiplying both sides by s(s + 3) to obtain 3s + 6 = A(s + 3) + Bs = (A + B)s + 3A Equating the coefficients of like powers of x to obtain 3A = 6 and A + B = 3. Thus, A = 2 and B = 1. Finally, [ ] [ ] [ ] 3s + 6 1 1 L 1 s 2 = 2L 1 + L 1 = 2h(t) + e 3t, t 0. + 3s s s + 3

15420 THE LAPLACE TRANSFORM AND THE METHOD OF PARTIAL FRACTIONS Example[ 20.4 ] Find L 1 s 2 +1. s(s+1) 2 Solution. We factor the denominator and split the rational function into partial fractions: s 2 + 1 s(s + 1) 2 = A s + B s + 1 + C (s + 1) 2. Multiplying both sides by s(s + 1) 2 and simplifying to obtain s 2 + 1 =A(s + 1) 2 + Bs(s + 1) + Cs =(A + B)s 2 + (2A + B + C)s + A. Equating coefficients of like powers of s we find A = 1, 2A + B + C = 0 and A + B = 1. Thus, B = 0 and C = 2. Now finding the inverse Laplace transform to obtain L 1 [ s 2 + 1 s(s + 1) 2 ] [ ] [ 1 = L 1 2L 1 s 1 (s + 1) 2 Example 20.5 Use Laplace transform to solve the initial value problem y + 3y + 2y = e t, y(0) = y (0) = 0. ] = h(t) 2te t, t 0. Solution. By the linearity property of the Laplace transform we can write Now since L[y ] + 3L[y ] + 2L[y] = L[e t ]. where L[y] = Y (s), we obtain L[y ] =s 2 L[y] sy(0) y (0) = s 2 Y (s) L[y ] =sy (s) y(0) = sy (s) L[e t ] = 1 s + 1 s 2 Y (s) + 3sY (s) + 2Y (s) = 1 s + 1.

155 Rearranging gives Thus, and (s 2 + 3s + 2)Y (s) = 1 s + 1. Y (s) = 1 (s + 1)(s 2 + 3s + 2). [ ] y(t) = L 1 1 (s + 1)(s 2. + 3s + 2) Using the method of partial fractions we can write Thus, y(t) = L 1 [ 1 s + 2 1 (s + 1)(s 2 + 3s + 2) = 1 s + 2 1 s + 1 + 1 (s + 1) 2. ] [ ] [ 1 L 1 + L 1 s + 1 1 (s + 1) 2 ] = e 2t e t + te t, t 0

15620 THE LAPLACE TRANSFORM AND THE METHOD OF PARTIAL FRACTIONS Practice Problems In Problems 20.1-20.4, give the form of the partial fraction expansion for F (s). You need not evaluate the constants in the expansion. However, if the denominator has an irreducible quadratic expression then use the completing the square process to write it as the sum/difference of two squares. Problem 20.1 Problem 20.2 Problem 20.3 Problem 20.4 Problem[ 20.5 Find L 1 1 (s+1) 3 ] Problem[ 20.6 ] Find L 1 2s 3. s 2 3s+2 Problem[ 20.7 ] Find L 1 4s 2 +s+1. s 3 +s Problem[ 20.8 ] Find L 1 s 2 +6s+8. s 4 +8s 2 +16. F (s) = s3 + 3s + 1 (s 1) 3 (s 2) 2. F (s) = s2 + 5s 3 (s 2 + 16)(s 2). F (s) = s 3 1 (s 2 + 1) 2 (s + 4) 2. F (s) = s4 + 5s 2 + 2s 9 (s 2 + 8s + 17)(s 2) 2. Problem 20.9 Use Laplace transform to solve the initial value problem y + 2y = 26 sin 3t, y(0) = 3.

157 Problem 20.10 Use Laplace transform to solve the initial value problem y + 2y = 4t, y(0) = 3. Problem 20.11 Use Laplace transform to solve the initial value problem y + 3y + 2y = 6e t, y(0) = 1, y (0) = 2. Problem 20.12 Use Laplace transform to solve the initial value problem y + 4y = cos 2t, y(0) = 1, y (0) = 1. Problem 20.13 Use Laplace transform to solve the initial value problem y 2y + y = e 2t, y(0) = 0, y (0) = 0. Problem 20.14 Use Laplace transform to solve the initial value problem y + 9y = g(t), y(0) = 1, y (0) = 3 where g(t) = { 6, 0 t < π 0, π t < Problem 20.15 Determine the constants α, β, y 0, and y 0 so that Y (s) = 2s 1 transform of the solution to the initial value problem s 2 +s+2 is the Laplace y + αy + βy = 0, y(0) = y 0, y (0) = y 0.

15820 THE LAPLACE TRANSFORM AND THE METHOD OF PARTIAL FRACTIONS

21 Laplace Transforms of Periodic Functions In many applications, the nonhomogeneous term in a linear differential equation is a periodic function. In this section, we derive a formula for the Laplace transform of such periodic functions. Recall that a function f(t) is said to be T periodic if T is the smallest positive interger such that f(t + T ) = f(t) whenever t and t + T are in the domain of f(t). For example, the sine and cosine functions are 2π periodic whereas the tangent and cotangent functions are π periodic. If f(t) is T periodic for t 0 then we define the function { f(t), 0 t T f T (t) = 0, t > T. The Laplace transform of this function is then L[f T (t)] = 0 f T (t)e st dt = T 0 f(t)e st dt. The Laplace transform of a T periodic function is given next. Theorem 21.1 If f(t) is a T periodic, piecewise continuous function for t 0 then L[f(t)] = L[f T (t)], s > 0. 1 e st Example 21.1 Determine the Laplace transform of the function 1, 0 t T 2 f(t) = f(t + T ) = f(t), t 0. T 0, 2 < t < T. 159

160 21 LAPLACE TRANSFORMS OF PERIODIC FUNCTIONS Solution. The graph of f(t) is shown in Figure 21.1. Figure 21.1 We have By Theorem 21.1, we have L[f T (t)] = T 2 0 e st dt st 1 e 2 =. s L[f(t)] = 1 e st 2 s 1 e st = 1 s(1 + e st 2 ), s > 0 Example 21.2 Find the Laplace transform of the sawtooth curve shown in Figure 21.2 Figure 21.2

161 Solution. The given function is periodic of period b. For the first period the function is defined by f b (t) = a t[h(t) h(t b)]. b So we have L[f b (t)] = = a b = a b b 0 e st a b tdt [ ts e st e st s 2 ] b ( 1 s 2 bse bs + e bs ) s 2. 0 Hence, L[f(t)] = L[f b(t)] 1 e bs = a b [ 1 e bs bse bs ] s 2 (1 e bs ) Example[ 21.3 ] Find L 1 1 e s. s 2 s(1 e s ) Solution. Note first that 1 s 2 e s s(1 e s ) = 1 e s se s s 2 (1 e s ) [ According to the previous example with a = 1 and b = 1 we find that L 1 1 s 2 is the sawtooth function shown in Figure 21.2 Linear Time Invariant Systems and the Transfer Function The Laplace transform is a powerful technique for analyzing linear time-invariant systems such as electrical circuits, harmonic oscillators, optical devices, and mechanical systems, to name just a few. A mathematical model described by a linear differential equation with constant coefficients of the form a n y (n) + a n 1 y (n 1) + + a 1 y + a 0 y = b m u (m) + b m 1 u (m 1) + + b 1 u + b 0 u is called a linear time invariant system. The function y(t) denotes the system output and the function u(t) denotes the system input. The system is called timeinvariant because the parameters of the system are not changing over time and an input now will give the same result as the same input later.. ] e s s(1 e s )

162 21 LAPLACE TRANSFORMS OF PERIODIC FUNCTIONS Applying the Laplace transform on the linear differential equation with null initial conditions we obtain a n s n Y (s)+a n 1 s n 1 Y (s)+ +a 0 Y (s) = b m s m U(s)+b m 1 s m 1 U(s)+ +b 0 U(s). The function Φ(s) = Y (s) U(s) = b ms m + b m 1 s m 1 + + b 1 s + b 0 a n s n + a n 1 s n 1 + + a 1 s + a 0 is called the system transfer function. That is, the transfer function of a linear time-invariant system is the ratio of the Laplace transform of its output to the Laplace transform of its input. Example 21.4 Consider the mathematical model described by the initial value problem my + γy + ky = f(t), y(0) = 0, y (0) = 0. The coefficients m, γ, and k describe the properties of some physical system, and f(t) is the input to the system. The solution y is the output at time t. Find the system transfer function. Solution. By taking the Laplace transform and using the initial conditions we obtain Thus, (ms 2 + γs + k)y (s) = F (s). Φ(s) = Y (s) F (s) = 1 ms 2 + γs + k (21.1) Parameter Identification One of the most useful applications of system transfer functions is for the parameter identification of a system. Example 21.5 Consider a spring-mass system governed by my + γy + ky = f(t), y(0) = 0, y (0) = 0. (21.2) Suppose we apply a unit step force f(t) = h(t) to the mass, initially at equilibrium, and you observe the system respond as y(t) = 1 2 e t cos t 1 2 e t sin t + 1 2. What are the physical parameters m, γ, and k?

163 Solution. Start with the model (21.2) with f(t) = h(t) and take the Laplace transform 1 of both sides, then solve to find Y (s) =. Since f(t) = h(t) we find s(ms 2 +γs+k) F (s) = 1 s. Hence Φ(s) = Y (s) F (s) = 1 ms 2 + γs + k. On the other hand, for the input f(t) = h(t) the corresponding observed output is Hence, Thus, y(t) = 1 2 e t cos t 1 2 e t sin t + 1 2. Y (s) =L[ 1 2 e t cos t 1 2 e t sin t + 1 2 ] = 1 s + 1 2 (s + 1) 2 + 1 1 1 2 (s + 1) 2 + 1 + 1 2s 1 = s(s 2 + 2s + 2) Φ(s) = Y (s) F (s) = 1 s 2 + 2s + 2. By comparison, we conclude that m = 1, γ = 2, and k = 2

164 21 LAPLACE TRANSFORMS OF PERIODIC FUNCTIONS Practice Problems Problem 21.1 Find the Laplace transform of the periodic function whose graph is shown. Problem 21.2 Find the Laplace transform of the periodic function whose graph is shown. Problem 21.3 Find the Laplace transform of the periodic function whose graph is shown. Problem 21.4 Find the Laplace transform of the periodic function whose graph is shown.

165 Problem 21.5 State the period of the function f(t) and find its Laplace transform where sin t, 0 t < π f(t) = f(t + 2π) = f(t), t 0. 0, π t < 2π. Problem 21.6 State the period of the function f(t) = 1 e t, 0 t < 2, find its Laplace transform. Problem 21.7 Using Example 21.3 find Problem 21.8 Consider the initial value problem [ s L 1 2 s e s ] s 3 + s(1 e s. ) ay + by + cy = f(t), y(0) = y (0) = 0, t > 0. f(t + 2) = f(t), and Suppose that the transfer function of this system is given by Φ(s) = 1 2s 2 +5s+2. (a) What are the constants a, b, and c? (b) If f(t) = e t, determine F (s), Y (s), and y(t). Problem 21.9 Consider the initial value problem ay + by + cy = f(t), y(0) = y (0) = 0, t > 0. Suppose that an input f(t) = t, when applied to the above system produces the output y(t) = 2(e t 1) + t(e t + 1), t 0. (a) What is the system transfer function? (b) What will be the output if the Heaviside unit step function f(t) = h(t) is applied to the system?

166 21 LAPLACE TRANSFORMS OF PERIODIC FUNCTIONS Problem 21.10 Consider the initial value problem where y + y + y = f(t), y(0) = y (0) = 0, 1, 0 t 1 f(t) = 1, 1 < t < 2. (a) Determine the system transfer function Φ(s). (b) Determine Y (s). Problem 21.11 Consider the initial value problem f(t + 2) = f(t) y 4y = e t + t, y(0) = y (0) = y (0) = 0. (a) Determine the system transfer function Φ(s). (b) Determine Y (s). Problem 21.12 Consider the initial value problem y + by + cy = h(t), y(0) = y 0, y (0) = y 0, t > 0. Suppose that L[y(t)] = Y (s) = y 0. s2 +2s+1 s 3 +3s 2 +2s. Determine the constants b, c, y 0, and

22 Convolution Integrals We start this section with the following problem. Example 22.1 A spring-mass system with a forcing function f(t) is modeled by the following initial-value problem mx + kx = f(t), x(0) = x 0, x (0) = x 0. Find solution to this initial value problem using the Laplace transform method. Solution. Apply Laplace transform to both sides of the equation to obtain ms 2 X(s) msx 0 mx 0 + kx(s) = F (s). Solving the above algebraic equation for X(s) we find X(s) = F (s) ms 2 + k + msx 0 ms 2 + k + mx 0 ms 2 + k = 1 F (s) m s 2 + k m + sx 0 s 2 + k m + x 0 s 2 + k m (22.1) Apply the inverse Laplace transform to obtain x(t) =L 1 [X(s)] { } { = 1 F (s) m L 1 s 2 + k + x 0 L 1 s m s 2 + k m } + x 0L 1 { 1 s 2 + k m } ( ) = {F 1 ( ) 1 k m k m L 1 (s) s 2 + k + x 0 cos t + x 0 m k sin t m m } 167

168 22 CONVOLUTION INTEGRALS } Finding L {F 1 1 (s),i.e., the inverse Laplace transform of a product, requires the use of the concept of convolution, a topic we discuss in this section s 2 + k m Convolution integrals are useful when finding the inverse Laplace transform of products H(s) = F (s)g(s). They are defined as follows: The convolution of two scalar piecewise continuous functions f(t) and g(t) defined for t 0 is the integral (f g)(t) = t Example 22.2 Find f g where f(t) = e t and g(t) = sin t. Solution. Using integration by parts twice we arrive at (f g)(t) = = 1 2 t 0 0 f(t s)g(s)ds. e (t s) sin sds [ e (t s) (sin s cos s) = e t 2 + 1 (sin t cos t) 2 Next, we state several properties of convolution product, which resemble those of ordinary product. Theorem 22.1 Let f(t), g(t), and k(t) be three piecewise continuous scalar functions defined for t 0 and c 1 and c 2 are arbitrary constants. Then (i) f g = g f (Commutative Law) (ii) (f g) k = f (g k) (Associative Law) (iii) f (c 1 g + c 2 k) = c 1 f g + c 2 f k (Distributive Law) Example 22.3 Express the solution to the initial value problem y +αy = g(t), y(0) = y 0 in terms of a convolution integral. Solution. Solving this initial value problem by the method of integrating factor we find y(t) = e αt y 0 + t 0 e α(t s) g(s)ds = e αt y 0 + e αt g(t) The following theorem, known as the Convolution Theorem, provides a way for finding the Laplace transform of a convolution integral and also finding the inverse Laplace transform of a product. ] t 0

169 Theorem 22.2 If f(t) and g(t) are piecewise continuous for t 0, and of exponential order a then Thus, (f g)(t) = L 1 [F (s)g(s)]. L[(f g)(t)] = L[f(t)]L[g(t)] = F (s)g(s). Example 22.4 Use the convolution theorem to find the inverse Laplace transform of Solution. Note that H(s) = H(s) = 1 (s 2 + a 2 ) 2. ( ) ( ) 1 1 s 2 + a 2 s 2 + a 2. So, in this case we have, F (s) = G(s) = 1 Thus, (f g)(t) = 1 a 2 t 0 s 2 +a 2 so that f(t) = g(t) = 1 a sin (at). sin (at as) sin (as)ds = 1 (sin (at) at cos (at)) 2a3 Convolution integrals are useful in solving initial value problems with forcing functions. Example 22.5 Solve the initial value problem 4y + y = g(t), y(0) = 3, y (0) = 7 Solution. Take the Laplace transform of all the terms and plug in the initial conditions to obtain 4(s 2 Y (s) 3s + 7) + Y (s) = G(s) or Solving for Y (s) we find (4s 2 + 1)Y (s) 12s + 28 = G(s). 12s 28 Y (s) = 4 ( s 2 + 1 ) + G(s) 4 4 ( s 2 + 1 ) 4 = 3s s 2 + ( 1 2 ) 2 14 1 2 s 2 + ( 1 2 ) 2 + 1 2 G(s) 1 2 s 2 + ( 1 2 ) 2

170 22 CONVOLUTION INTEGRALS Hence, y(t) = 3 cos ( ) ( ) t t 14 sin + 1 2 2 2 t 0 ( s sin g(t s)ds. 2) So, once we decide on a g(t) all we need to do is to evaluate the integral and we ll have the solution Example 22.6 Use Laplace transform to solve for y(t) : y(t) t 0 e (t λ) y(λ)dλ = t. Solution. Note that the given equation reduces to e t y(t) = y(t) t. Taking Laplace transform of both sides we find Y (s) s 1 = Y (s) 1. Solving for Y (s) we find Y (s) = s 2 Using partial fractions decomposition we can write s 1 s 2 (s 2) = 1 1 1 4 s + 2 s 2 + 4 (s 2). s 1 s 2 (s 2). Hence, y(t) = 1 4 + t 2 + 1 4 e2t, t 0 Example 22.7 Use Laplace transform to solve for y(t) : t y(t) = t 2 (1 e t ). Solution. Taking Laplace transform of both sides we find Y (s) = 2 2. This implies s 2 s 3 (s+1) 3 Y (s) = 2 s 2s2. Using partial fractions decomposition we can write (s+1) 3 Hence, s 2 (s + 1) 3 = 1 s + 1 2 (s + 1) 2 + 1 (s + 1) 3. y(t) = 2 2(e t 2te t + t2 2 e t ) = 2 (1 (1 2t + t22 ) )e t, t 0

171 Example 22.8 Solve the following initial value problem. y y = t 0 (t λ)e λ dλ, y(0) = 1. Solution. Note that y y = t e t. Taking Lalplace transform of both sides we find sy ( 1) Y = 1 1 1 s 2 s 1. This implies that Y (s) = s 1 + 1. Using partial s 2 (s 1) 2 fractions decomposition we can write Thus, 1 s 2 (s 1) 2 = 2 s + 1 s 2 2 s 1 + 1 (s 1) 2. Y (s) = 1 s 1 + 2 s + 1 s 2 2 s 1 + 1 (s 1) 2 = 2 s + 1 s 2 3 s 1 + 1 (s 1) 2. Finally, y(t) = 2 + t 3e t + te t, t 0

172 22 CONVOLUTION INTEGRALS Practice Problems Problem 22.1 Consider the functions f(t) = g(t) = h(t), t 0 where h(t) is the Heaviside unit step function. Compute f g in two different ways. (a) By directly evaluating the integral. (b) By computing L 1 [F (s)g(s)] where F (s) = L[f(t)] and G(s) = L[g(t)]. Problem 22.2 Consider the functions f(t) = e t and g(t) = e 2t, t 0. Compute f g in two different ways. (a) By directly evaluating the integral. (b) By computing L 1 [F (s)g(s)] where F (s) = L[f(t)] and G(s) = L[g(t)]. Problem 22.3 Consider the functions f(t) = sin t and g(t) = cos t, t 0. Compute f g in two different ways. (a) By directly evaluating the integral. (b) By computing L 1 [F (s)g(s)] where F (s) = L[f(t)] and G(s) = L[g(t)]. Problem 22.4 Compute and graph f g where f(t) = h(t) and g(t) = t[h(t) h(t 2)]. Problem 22.5 Compute and graph f g where f(t) = h(t) h(t 1) and g(t) = h(t 1) 2h(t 2)]. Problem 22.6 Compute t t t. Problem 22.7 Compute h(t) e t e 2t. Problem 22.8 Compute t e t e t. Problem 22.9 n functions {}}{ Suppose it is known that h(t) h(t) h(t) = Ct 8. Determine the constants C and the positive integer n. Problem 22.10 Use Laplace transform to solve for y(t) : t 0 sin (t λ)y(λ)dλ = t 2.

173

174 CHEAT SHEETS Cheat Sheets

Answer Key Section 1 1.1 (a) 16 m/sec (b) 80 m. 1.2(a) 1.5 meters (b) 7 m/sec (c) 9.8 m/sec 2. 1.3 The acceleration due to gravity is y (t) = 50 ft/sec 2. The initial velocity is v(0) = y (0) = 72 ft/sec. The initial height is y(0) = 40 ft. 1.4 The height of the object at any time t is given by y(t) = 16t 2 + 400. The object hits the ground when y(t) = 0. Solving the equation 16t 2 + 400 = 0 we find t = 5 sec. The velocity of the object at the time of impact is v(5) = y (5) = 32(5) = 160 ft/sec. 1.5 400 ft. 1.6 v(0) = 160 ft/sec and maximum height is 400 ft. 1.7 The impact time is t = 2y0 g. v( 2y0 g ) = 2gy 0. 1.8 ɛ = π 4. 1.9 (a) Order 2 (b) Order 1 (c) Order 3. 1.10 (a) First order (b), (c), and (d) second order. 1.11 u is the dependent variable whereas x and y are the independent variables. 1.12 (a) ODE (b) PDE (c) ODE. 181

182 ANSWER KEY 1.13 y(t) = t3 3 + C 1t 2 + C 2 t + C 3. 1.14 The solution is y(t) = 50e 3t. The domain is the set of all real numbers, i.e., (, ). 1.15 λ = ±3. 1.16 m = 2 or m = 1. 1.17 Substituting y(t) = y (t) = y (t) = e t into the equation we find ( y 2 + 2 ) ( y + 1 + 2 ) ( y =e t 2 + 2 ) ( e t + 1 + 2 ) e t t t t t =e t 2e t 2 t et + e t + 2 t et = 0. 1.18 Finding the first and the second derivatives of y we obtain and y (t) = C 1 ω sin ωt + C 2 ω cos ωt y (t) = C 1 ω 2 cos ωt C 2 ω 2 sin ωt Substituting this into the equation to obtain d 2 y dt 2 + ω2 y = C 1 ω 2 cos ωt C 2 ω 2 sin ωt + ω 2 (C 1 cos ωt + C 2 sin ωt) =0. 1.19 y(0) = 2 and k = 4. 1.20 n = 1 or n = 2. 1.21 (b) y(t) = e 2t + e 2t (c) y(t) = 2e 2t. 1.22 y(0) = 1, m = 2, y(t) = t + 2. 1.23 t 0 = 0, m = 1, y(t) = t2 2 1. 1.24 Finding the second derivative and substituting into the equation we find y + 4y = 4e 2t + 4e 2t = 8e 2t 0.

183 Thus, y(t) = e 2t is not a solution to the given equation. 1.25 (a) Substituting y and y into the equation we find 3Ce 3t + 2 + 3[Ce 3t + 2t + 1] = 3Ce 3t + 3Ce 3t + 6t + 5 = 6t + 5. (b) C = 2.

184 ANSWER KEY Section 2 2.1 y 0 = 3 and p(t) = 2t. 2.2 (a) (, ) (b) (, ). 2.3 (a) 3 < t < (b) 2 < t < 2 (c) < t < 2. 2.4 (a) r = 3 and C = 1 (b) (, 0) (c) (, ). 2.5 The interval of existence is 0 < t < if y(1) = 1 and < t < 0 if y( 1) = 1. 2.6 (a) 5 < t < 0 (b) 0 < t < 4 (c) < t < 5 (d) 4 < t <. 2.7 (a) 1 < t < 3 (b) No such solution since t > 0 (c) 3 < t <. 2.8 (a) If p(t) and g(t) are continuous functions in the open interval I = (a, b) and t 0 a point inside I then the IVP y + p(t)y = g(t), y(t 0 ) = y 0 has a unique solution y(t) defined on I. (b) Since p(t) = t 2 and g(t) = e t3, the IVP has a unique solution. 2.9 (a) Non-linear (b) Linear and homogeneous (c) Linear and non-homogeneous (d) Linear and homogeneous. 2.10 0 < t < 5. 2.11 α = ln 2 and y 0 = 8. 2.12(a) (iv) (b) (i) (c) (iii) (d) (ii). 2.13 A non-linear first order ordinary differential equation. 2.14 We have y = t+y t = 1 + y t. Thus, y 1 t y = 1 This is a first-order linear non-homogeneous equation.

185 Section 3 3.1 y(t) = e 1 t2 3.2 y(t) = 2e 1 et. 3.3 (a)z + p(t)z = 0, z(t 0 ) = y(t 0 ) α (b) y(t) = (y 0 α)e t t 0 p(s)ds + α. 3.4 y(t) = e t2 + 3. 3.5 α = 2 and y 0 = 0.25. 3.6 (a) n = 3 and y 0 = 1 (b) y( 1) = 4 e 1. 3.7 p(t) = 1 t. 3.8 p(t) = 3 t. 3.9 y(t) = t 3. 3.10 y(t) = Ce t2. 3.11 f(t) = 4e 2t. 3.12 y(t) = Ce 2t + C. 3.13 y = 1 2 (1 e t2 ). 3.14 y(t) = 3t 1 9 + e 2t + Ce 3t. 3.15 y(t) = 3 sin t + 3 cos t t + C t. 3.16 y(t) = 1 13 (3 sin (3t) + 2 cos (3t)) + Ce 2t. 3.17 α = 2. 3.18 p(t) = 2 and g(t) = 2t + 3. 3.19 y 0 = 1 and g(t) = 2e t + cos t + sin t.

186 ANSWER KEY 3.20 lim t y(t) = 1. 3.21 3.22 3.23 { y(t) = y(t) = y(t) = { 2t + 1 if 0 t 1 t + 2 t if 1 < t 2. 1 + 2e cos t 1 if 0 t π 1 + 2 ( 1 e + e) e cos t if π < t 2π. { 1 + 2e t if 0 t 1 (2 + e)e t if t > 1. 3.24 y(t) = The graph of y(t) is shown below 3e (t2 t) if 0 t 1 3 if 1 < t 3 t if 3 < t 4. It follows that lim t y(t) =. The function y(t) is not differentiable at t = 1 and t = 3 on the domain t > 0. 3.25 y(t) = ( te t + e t ) 1.

187 Section 4 4.1 (a) y = 1 3 (1 2t cos y) = f(t, y), (b) D = {(t, y) : < t <, < y < } (c) R = {(t, y) : < t <, < y < }. 4.2 (a) y = 1 3t (1 2 cos y) = f(t, y) (b) D = {(t, y) : < t < 0, 0 < t <, < y < } (c) R = {(t, y) : 0 < t <, < y < }. 4.3 (a) y = 2t 1+y 3 = f(t, y) (b) D = {(t, y) : < t <, < y < 1, 1 < y < } (c) R = {(t, y) : < t <, 1 < y < }. 4.4 (a) y = t2 e y y 2 9 = f(t, y) (b) D = {(t, y) : < t <, < y < 3, 3 < y < 3, 3 < y < } (c) D = {(t, y) : < t <, 3 < y < 3}. 4.5 (a) y = 2+tan t cos y = f(t, y) (b) D = {(t, y) : t (2n + 1) π 2, y (2m + 1) π 2, where n and m are integers} (c) R = {(t, y) : π 2 < t < π 2, π 2 < y < π 2 }. 4.6 y 1 = t(t 4)(y+1)(y 2), y(2) = 0. 4.7 Since f(t, y) = y t and f y (t, y) = 1 t, both functions are continuous in the region D = {(t, y) : 0 < t <, < y < } Since (0, 2) is not in D, Theorem 4.1 can not be applied in this case. 4.8 The function f(t, y) = t y t+y is continuous everywhere except along the line t + y = 0. (a) and (b): Since both (0, 0) and (1, 1) lie on this line, we cannot conclude existence from Theorem 4.1. (c) The point ( 1, 1) is not on that line so we can find a small rectangle around this point where Theorem 4.1 guarantees the existence of a solution. Furthermore, since f y (t, y) = 2t is continuous at ( 1, 1), the solution is unique. (t+y) 2 4.9 The equation is of the form y = f(t, y) = y t + 2. The function f is continuous outside the line t = 0. The initial value point is (0,1), so there is no rectangle containing it in which f is continuous, and the conditions of Theorem 4.1 are not satisfied.

188 ANSWER KEY 4.10 The equation is of the form y = f(t, y) = y sin y + t. the function f is continuous in the whole plane, and so is its partial derivative f y (t, y) = sin y + y cos y. In particular, any rectangle around the initial value point will satisfy the conditions of Theorem 4.1.

189 Section 5 5.1 y 3 = 3 2 et2 + C. 5.2 y(t) = Ce t2 2 2t. 5.3 y(t) = Ct 2 + 4. 5.4 y(t) = 2Ce4t 1+Ce 4t and y(t) 2. 5.5 y(t) = 5 4 cos (2t). 5.6 y(t) = ( 2 cos t + 4). 5.7 y(t) = 2t + 3 1. 5.8 y(t) = 2 4t 2 +1. 5.9 y(t) = tan ( t π 2 ), π 2 < t < π 2. 5.10 y(t) = 3 e t2 3+e t2. 5.11 y 2 + cos y + cos t + t2 2 = 2. 5.12 y 0 = 1 2, α = 1 2, n = 3. 5.13 3y 2 y + (cos y)y + 2t = 0, y(2) = 0. 5.14 y 0 = 1 8. 5.15 y sin y t 3 + 1 = 0. 5.16 y(t) = Ce2t (t+1) 2. 5.17 tan y = t 2 + 1 4 sin 2t + 1. 5.18 y(t) = ln(2 e t ), t < ln 2. 5.19 (a) e y = t3 3 1 t + C (b) e y + t3 3 + 1 t = e 1 + 4 3.

190 ANSWER KEY Section 6 6.1 f t = ye ty, f y = ln y + 1 + te ty. 6.2 f t = 1 t + 2t y 5, f y = 1 y t2 +1 (y 5) 2. 6.3 df dt = 24 sin2 t 4 sin t 12. 6.4 y(t) = 4t t 3 + 6. 6.5 y(t) = tan ( t 3 + t + π 4 ). 6.6 The given differential equation is not exact. 6.7 e t+y + t 3 + y 2 = 1. 6.8 y sin (t + y) + t2 2 + ty + y2 2 = 0. 6.9 m = 4, n = 3, α = 4 3. 6.10 N(t, y) = 2y sin tdt = 2y cos t + h(y). 6.11 y 0 = ±2, M(t, y) = 3t 2 + e t, N(t, y) = t 3 + 2y. 6.12 y(0) = 2, a = 1, b = 2. 6.13 The differential equation is exact. 6.14 t 2 + 3t + y 2 2y = C. 6.15 B = 1, 1 2 e2ty + t2 2 = C. 6.16 ty 2 y 2 e y + 2ye y 2e y = C. 6.17 (a) M N y (t, y) = p(t) 0 and t (t, y) = 0 (b) y(t) = e p(t)dt e p(t)dt g(t)dt+ Ce p(t)dt. 6.18 y(t) = t ln t + 7t. 6.19 ty 3 t 2 y = C.

6.20 (a) 2t 3 y 2 + 4ty 8t + 4y = 6 (b) 6t 4ty + 8t 2 + 5y 2 + 2y = 30. 191

192 ANSWER KEY Section 7 7.1 y 2 = t 2 + Ct 3. 7.2 y 4 = 2e t2 + Ce 2t2. 7.3 y(t) = 1 t(4.5 t). 7.4 y(t) = 2t t 2 +1. 7.5 y(t) = (1 + e t ) 1. 7.6 y(t) = (t + 1 3 + Ce3t ) 1 3 and y(t) 0. 7.7 y(t) = (e t2 t 0 e s2 ds + Ce t2 ) 1 + t. 7.8 z z = 1. 7.9 (a) z + 2ɛz = 2σ (b) y(t) = ( σ ɛ + Ce 2ɛt ) 1 2 and y(t) 0. 7.10 (a) z = f(z) z t (b) 2 arctan ( y ) t = ln t 2 (1 + ( y ) ) 2 t. 7.11 y(t) = ( 3te t + Ce t ) 1 3 and y(t) 0. 7.12 y(t) = (2t + Ct 2 ) 1 2 and y(t) 0. 7.13 y(t) = ( t 2 ln t + t 2 + Ct) 1 and y(t) 0. 7.14 y(t) = ( 1 3 + Ce 3t ) 1 + 2. 7.15 y(t) = ( t 4 + Ct 3 ) 1 + 2 t.

193 Section 8 8.1 (a) We have (b) We have 8.2 (A) y = sin t (B) y = y(2 y) (c) y = t (D) y = 1 y

194 ANSWER KEY 8.3 (a) No (b) No (c) Yes (d) Yes 8.4 (a) y(t) 1, y(t) 2 (b) y(t) 1, y 2 (c) y 1, y 2, y 3 8.5 One answer is the differential equation: y = (y 1) 2 8.6 y = C, where C > 0. 8.7 One answer is the DE y = sin (2πy) 8.8 An answer is y = y(2 y) 8.9 (a) y = 1 stable, y = 1 unstable (b) y = 1 is neither. This is a semistable equilibrium. 8.10 The equilibrium solution at y = 1 is asymptotically stable where as the equilibrium solution at y = 0 is unstable. 8.11 8.12 The direction field is given below.

195 The equilibrium solution y = 1 is unstable; y = 0 is semi-stable; y = 1 is asymptotically stable. 8.13 The direction field is given below. The equilibrium solution y = 4 is unstable while y = 0 is asymptotically stable. If y(0) = 5 then lim t y(t) =. If y(0) = 4 then lim t y(t) = 4. If y(0) = 3 then lim t y(t) = 0 8.14 y(t) = 0 and y(t) = π are two equilibrium solutions. According to the direction field shown below we conclude that lim y(t) = π t

196 ANSWER KEY