Introduction to First Order Equations Sections

Transcription

1 A B I L E N E C H R I S T I A N U N I V E R S I T Y Department of Mathematics Introduction to First Order Equations Sections Dr. John Ehrke Department of Mathematics Fall 2012

2 Course Goals The study of differential equations has three principal goals: Slide 2/24 Dr. John Ehrke Lecture 1 Fall 2012

3 Course Goals The study of differential equations has three principal goals: 1 To discover the underlying differential equation that describes a specific physical situation. Slide 2/24 Dr. John Ehrke Lecture 1 Fall 2012

4 Course Goals The study of differential equations has three principal goals: 1 To discover the underlying differential equation that describes a specific physical situation. 2 To find, either exactly or approximately the appropriate solution of that equation. Slide 2/24 Dr. John Ehrke Lecture 1 Fall 2012

5 Course Goals The study of differential equations has three principal goals: 1 To discover the underlying differential equation that describes a specific physical situation. 2 To find, either exactly or approximately the appropriate solution of that equation. 3 To interpret the solution that is found in the context of the model being studied. Slide 2/24 Dr. John Ehrke Lecture 1 Fall 2012

6 Course Goals The study of differential equations has three principal goals: 1 To discover the underlying differential equation that describes a specific physical situation. 2 To find, either exactly or approximately the appropriate solution of that equation. 3 To interpret the solution that is found in the context of the model being studied. There are three different techniques that we will employ this semester to address these goals: Slide 2/24 Dr. John Ehrke Lecture 1 Fall 2012

7 Course Goals The study of differential equations has three principal goals: 1 To discover the underlying differential equation that describes a specific physical situation. 2 To find, either exactly or approximately the appropriate solution of that equation. 3 To interpret the solution that is found in the context of the model being studied. There are three different techniques that we will employ this semester to address these goals: Analytic Solution techniques like separation of variables, integrating factor method, and variation of parameters. Slide 2/24 Dr. John Ehrke Lecture 1 Fall 2012

8 Course Goals The study of differential equations has three principal goals: 1 To discover the underlying differential equation that describes a specific physical situation. 2 To find, either exactly or approximately the appropriate solution of that equation. 3 To interpret the solution that is found in the context of the model being studied. There are three different techniques that we will employ this semester to address these goals: Analytic Solution techniques like separation of variables, integrating factor method, and variation of parameters. Geometric Slope fields, phase portraits, integral curves,... Slide 2/24 Dr. John Ehrke Lecture 1 Fall 2012

9 Course Goals The study of differential equations has three principal goals: 1 To discover the underlying differential equation that describes a specific physical situation. 2 To find, either exactly or approximately the appropriate solution of that equation. 3 To interpret the solution that is found in the context of the model being studied. There are three different techniques that we will employ this semester to address these goals: Analytic Solution techniques like separation of variables, integrating factor method, and variation of parameters. Geometric Slope fields, phase portraits, integral curves,... Numerical Euler s method, Runge-Kutta (approximation schemes) Slide 2/24 Dr. John Ehrke Lecture 1 Fall 2012

10 Motion of a Falling Body Example (Physical Description of the Problem) An object falls through the air toward the Earth. Assuming that the only forces acting on the object are gravity and air resistance, determine the velocity of the object as a function of time. Slide 3/24 Dr. John Ehrke Lecture 1 Fall 2012

11 Motion of a Falling Body Example (Physical Description of the Problem) An object falls through the air toward the Earth. Assuming that the only forces acting on the object are gravity and air resistance, determine the velocity of the object as a function of time. Solution: The physical law governing the motion of objects is Newton s second law which states the net forces exerted on an object are equal to the object s mass times its acceleration, given by which we can rewrite as using a = dv/dt as F = ma (1) F = m(dv/dt) (2) At this point, we have the beginnings of a real differential equation, but where do we go from here? Slide 3/24 Dr. John Ehrke Lecture 1 Fall 2012

12 So, we have a basic model m dv dt = F where m represents the mass of the object, dv/dt represents its acceleration, and F represents the total force on the object. To develop a better model, we need to explore the net forces exerted on the mass during free fall: Slide 4/24 Dr. John Ehrke Lecture 1 Fall 2012

13 So, we have a basic model m dv dt = F where m represents the mass of the object, dv/dt represents its acceleration, and F represents the total force on the object. To develop a better model, we need to explore the net forces exerted on the mass during free fall: The force due to gravity is expressed by mg where g is the acceleration due to gravity. It is convenient to define v as positive when it is directed downward. Slide 4/24 Dr. John Ehrke Lecture 1 Fall 2012

14 So, we have a basic model m dv dt = F where m represents the mass of the object, dv/dt represents its acceleration, and F represents the total force on the object. To develop a better model, we need to explore the net forces exerted on the mass during free fall: The force due to gravity is expressed by mg where g is the acceleration due to gravity. It is convenient to define v as positive when it is directed downward. The force exerted by air (i.e. the air resistance or drag) on the body acts in opposition to the force exerted by gravity, so we can reasonably represent this force by γv where γ is the positive drag constant dependent on the shape of the object and the air density. We use the negative sign because air resistance opposes the motion of the falling body. Slide 4/24 Dr. John Ehrke Lecture 1 Fall 2012

15 So, we have a basic model m dv dt = F where m represents the mass of the object, dv/dt represents its acceleration, and F represents the total force on the object. To develop a better model, we need to explore the net forces exerted on the mass during free fall: The force due to gravity is expressed by mg where g is the acceleration due to gravity. It is convenient to define v as positive when it is directed downward. The force exerted by air (i.e. the air resistance or drag) on the body acts in opposition to the force exerted by gravity, so we can reasonably represent this force by γv where γ is the positive drag constant dependent on the shape of the object and the air density. We use the negative sign because air resistance opposes the motion of the falling body. Together, these two quantities define F more completely, and so we can update our model as follows: m dv dt = mg γv (3) Slide 4/24 Dr. John Ehrke Lecture 1 Fall 2012

16 Solving the Model In order to solve our model, (3), we will employ a technique called separation of variables, and requires us to treat dv and dt as differentials. Rewriting, (3), we obtain dv mg γv = dt m (4) Slide 5/24 Dr. John Ehrke Lecture 1 Fall 2012

17 Solving the Model In order to solve our model, (3), we will employ a technique called separation of variables, and requires us to treat dv and dt as differentials. Rewriting, (3), we obtain dv mg γv = dt m (4) Integrating the separated equation, we obtain dv dt mg γv = 1 = m γ ln mg γv = t m + c (5) Slide 5/24 Dr. John Ehrke Lecture 1 Fall 2012

18 Solving the Model In order to solve our model, (3), we will employ a technique called separation of variables, and requires us to treat dv and dt as differentials. Rewriting, (3), we obtain dv mg γv = dt m (4) Integrating the separated equation, we obtain dv dt mg γv = 1 = m γ ln mg γv = t m + c (5) Solving for v, we obtain where A = e γc has the same sign as (mg γv). v = mg γ A γ e γ(t/m) (6) Slide 5/24 Dr. John Ehrke Lecture 1 Fall 2012

19 Solving the Model In order to solve our model, (3), we will employ a technique called separation of variables, and requires us to treat dv and dt as differentials. Rewriting, (3), we obtain dv mg γv = dt m (4) Integrating the separated equation, we obtain dv dt mg γv = 1 = m γ ln mg γv = t m + c (5) Solving for v, we obtain where A = e γc has the same sign as (mg γv). v = mg γ A γ e γ(t/m) (6) In the context of the problem, what would you call the constant mg/γ? Slide 5/24 Dr. John Ehrke Lecture 1 Fall 2012

20 Terminology Looking at the falling body problem, if this were a specific case, we would be given the values of m, g, and γ. The determination of the integration constant c and therefore A requires we also know the initial velocity, v 0 of the object. We can now formally define the problem we ve been working on suitable for use in a differential equations class as m dv dt = mg γv, v(0) = v 0 (7) Initial Value Problem (IVP) The equation, (7) interprets the physical problem as a differential equation dependent on an initial condition. Such problems are called initial value problems or IVPs. Boundary Value Problem (BVP) If instead of specifying the initial velocity we specified both a starting velocity and ending velocity on some discrete time interval, we would obtain a boundary value problem or BVP. Slide 6/24 Dr. John Ehrke Lecture 1 Fall 2012

21 General Solution The solution in (6) is called the general solution of the differential equation since it contains all possible solutions representing an infinite family of curves. Particular Solution If we solve the IVP, then we obtain a specific representative of this infinite family called a particular solution. The particular solution depends on the nature of the initial condition, v 0. Equilibrium Solution The particular solution corresponding to v(0) = mg/γ is a solution that does not change over time. In the physical context this represents the terminal velocity of the falling body. We call such solutions equilibrium solutions because regardless of the values of v 0, the velocity v(t) approaches this solution as t. Parameters The values of m, γ are chosen based on the specifics of the physical problem and would be known prior to solving the equation. Such values are called parameters. Parameters are different from the independent variable t, and dependent variable v, and the constant g, since its value is determined a priori for any problem chosen. Slide 7/24 Dr. John Ehrke Lecture 1 Fall 2012

22 Falling Body Problem Solutions Figure: * Graph of v(t) for nine different initial velocities v 0. (g = 9.8m/sec 2, m/gamma = 5 sec) Slide 8/24 Dr. John Ehrke Lecture 1 Fall 2012

23 First Order Differential Equations The first few weeks of class will deal primarily with differential equations of first order dx = f (t, x) (8) dt where the unknown function, f (t, x) is related to the first derivative of the dependent variable in some way. Even though first order differential equations represent the simplest class of differential equations to solve, they vary wildly in not only the techniques required to solve them, but the difficulty they pose. Slide 9/24 Dr. John Ehrke Lecture 1 Fall 2012

24 First Order Differential Equations The first few weeks of class will deal primarily with differential equations of first order dx = f (t, x) (8) dt where the unknown function, f (t, x) is related to the first derivative of the dependent variable in some way. Even though first order differential equations represent the simplest class of differential equations to solve, they vary wildly in not only the techniques required to solve them, but the difficulty they pose. Rank the following equations in terms of difficulty. (a) y (x) = x y 2 Slide 9/24 Dr. John Ehrke Lecture 1 Fall 2012

25 First Order Differential Equations The first few weeks of class will deal primarily with differential equations of first order dx = f (t, x) (8) dt where the unknown function, f (t, x) is related to the first derivative of the dependent variable in some way. Even though first order differential equations represent the simplest class of differential equations to solve, they vary wildly in not only the techniques required to solve them, but the difficulty they pose. Rank the following equations in terms of difficulty. (a) y (x) = x y 2 (b) y (x) = y x 2 Slide 9/24 Dr. John Ehrke Lecture 1 Fall 2012

26 First Order Differential Equations The first few weeks of class will deal primarily with differential equations of first order dx = f (t, x) (8) dt where the unknown function, f (t, x) is related to the first derivative of the dependent variable in some way. Even though first order differential equations represent the simplest class of differential equations to solve, they vary wildly in not only the techniques required to solve them, but the difficulty they pose. Rank the following equations in terms of difficulty. (a) y (x) = x y 2 (b) y (x) = y x 2 (c) y (x) = x/y Slide 9/24 Dr. John Ehrke Lecture 1 Fall 2012

27 First Order Differential Equations The first few weeks of class will deal primarily with differential equations of first order dx = f (t, x) (8) dt where the unknown function, f (t, x) is related to the first derivative of the dependent variable in some way. Even though first order differential equations represent the simplest class of differential equations to solve, they vary wildly in not only the techniques required to solve them, but the difficulty they pose. Rank the following equations in terms of difficulty. (a) y (x) = x y 2 (b) y (x) = y x 2 Hardest, impossible to obtain an analytic solution, must be handled numerically. (c) y (x) = x/y Slide 9/24 Dr. John Ehrke Lecture 1 Fall 2012

28 First Order Differential Equations The first few weeks of class will deal primarily with differential equations of first order dx = f (t, x) (8) dt where the unknown function, f (t, x) is related to the first derivative of the dependent variable in some way. Even though first order differential equations represent the simplest class of differential equations to solve, they vary wildly in not only the techniques required to solve them, but the difficulty they pose. Rank the following equations in terms of difficulty. (a) y (x) = x y 2 (b) y (x) = y x 2 (c) y (x) = x/y Hardest, impossible to obtain an analytic solution, must be handled numerically. Linear equation, we will handle this soon. Not very difficult. Slide 9/24 Dr. John Ehrke Lecture 1 Fall 2012

29 First Order Differential Equations The first few weeks of class will deal primarily with differential equations of first order dx = f (t, x) (8) dt where the unknown function, f (t, x) is related to the first derivative of the dependent variable in some way. Even though first order differential equations represent the simplest class of differential equations to solve, they vary wildly in not only the techniques required to solve them, but the difficulty they pose. Rank the following equations in terms of difficulty. (a) y (x) = x y 2 (b) y (x) = y x 2 (c) y (x) = x/y Hardest, impossible to obtain an analytic solution, must be handled numerically. Linear equation, we will handle this soon. Not very difficult. Easiest, separable equation, really more of a Calculus II problem. Slide 9/24 Dr. John Ehrke Lecture 1 Fall 2012

30 Creation and Annihilation Operators Consider the two first order equations ( ) d dx + x y = 0 (9) ( ) d dx x y = 0 (10) The equations (9) and (10) are called the annihilation and creation operators, respectively. These operators are useful in the study of quantum harmonic oscillators. In particular, the annihilation operator decreases the number of particles in a given state by one, hence its name. You will explore the creation operator in your first homework set. Example Obtain an analytic solution of the annihilator equation, (3), and completely describe the family of solutions. Slide 10/24 Dr. John Ehrke Lecture 1 Fall 2012

31 Solution ( ) d dx + x y = 0 = = = dy dx = xy dy = x dx y dy y = x dx = ln y = x2 2 + c = y = e x2 /2 + c = y = e c e x2 /2 = y(x) = Ae x2 /2 Remarks What values is A allowed to take? Well, clearly A > 0, since A = e c. Technically also A < 0 works, since really we should have written, y = Ae x2 /2 y = ±Ae x2 /2. What about A = 0? A = 0 should be included also since this would give the solution y = 0, which is indeed a solution to the ODE. (It satisfies, equation (9).) This is sometime referred to as the lost solution because it is lost in the process of separating the variables when we divided by y. Note that if A = 1/ 2π, then the solution is the standard normal pdf. Slide 11/24 Dr. John Ehrke Lecture 1 Fall 2012

32 Polling Question #1 Solve the separable IVP A. y(x) = xe x y(x) = x dy dx y = 2x2 y, y(1) = 1. B. y(x) = x e ex2 C. y(x) = x 2 D. y(x) = 2x 1 x Slide 12/24 Dr. John Ehrke Lecture 1 Fall 2012

33 Geometry of Solutions Last time, we looked at an analytic method: separation of variables, and some of the issues that arise in solving differential equations analytically. In this lecture, we will consider the geometric view when solving differential equations. Analytic: Geometric: Slide 13/24 Dr. John Ehrke Lecture 1 Fall 2012

34 Geometry of Solutions Last time, we looked at an analytic method: separation of variables, and some of the issues that arise in solving differential equations analytically. In this lecture, we will consider the geometric view when solving differential equations. Analytic: Geometric: y (x) = f (x, y) Slide 13/24 Dr. John Ehrke Lecture 1 Fall 2012

35 Geometry of Solutions Last time, we looked at an analytic method: separation of variables, and some of the issues that arise in solving differential equations analytically. In this lecture, we will consider the geometric view when solving differential equations. Analytic: y (x) = f (x, y) Geometric: direction field (also called a slope field) Slide 13/24 Dr. John Ehrke Lecture 1 Fall 2012

36 Geometry of Solutions Last time, we looked at an analytic method: separation of variables, and some of the issues that arise in solving differential equations analytically. In this lecture, we will consider the geometric view when solving differential equations. Analytic: y (x) = f (x, y) Geometric: direction field (also called a slope field) y(x), solution Slide 13/24 Dr. John Ehrke Lecture 1 Fall 2012

37 Geometry of Solutions Last time, we looked at an analytic method: separation of variables, and some of the issues that arise in solving differential equations analytically. In this lecture, we will consider the geometric view when solving differential equations. Analytic: y (x) = f (x, y) y(x), solution Geometric: direction field (also called a slope field) integral curve Slide 13/24 Dr. John Ehrke Lecture 1 Fall 2012

38 Geometry of Solutions Last time, we looked at an analytic method: separation of variables, and some of the issues that arise in solving differential equations analytically. In this lecture, we will consider the geometric view when solving differential equations. Analytic: y (x) = f (x, y) y(x), solution Geometric: direction field (also called a slope field) integral curve y 1 (x 0 ) = f (x 0, y 1 ) Slide 13/24 Dr. John Ehrke Lecture 1 Fall 2012

39 Geometry of Solutions Last time, we looked at an analytic method: separation of variables, and some of the issues that arise in solving differential equations analytically. In this lecture, we will consider the geometric view when solving differential equations. Analytic: y (x) = f (x, y) y(x), solution y 1 (x 0 ) = f (x 0, y 1 ) Geometric: direction field (also called a slope field) integral curve slope of the direction field at (x 0, y 1 ) Slide 13/24 Dr. John Ehrke Lecture 1 Fall 2012

40 Drawing a Slope Field The natural question is: how might one draw a direction field for an associated ODE? Computer Way: Human Way: 1 Pick (x, y) with some sort of spacing condition. Slide 14/24 Dr. John Ehrke Lecture 1 Fall 2012

41 Drawing a Slope Field The natural question is: how might one draw a direction field for an associated ODE? Computer Way: Human Way: 1 Pick (x, y) with some sort of spacing condition. 2 Calculate f (x, y) at that point. Slide 14/24 Dr. John Ehrke Lecture 1 Fall 2012

42 Drawing a Slope Field The natural question is: how might one draw a direction field for an associated ODE? Computer Way: Human Way: 1 Pick (x, y) with some sort of spacing condition. 2 Calculate f (x, y) at that point. 3 Draw in a line element. Slide 14/24 Dr. John Ehrke Lecture 1 Fall 2012

43 Drawing a Slope Field The natural question is: how might one draw a direction field for an associated ODE? Computer Way: 1 Pick (x, y) with some sort of spacing condition. Human Way: 1 Pick a slope c. 2 Calculate f (x, y) at that point. 3 Draw in a line element. Slide 14/24 Dr. John Ehrke Lecture 1 Fall 2012

44 Drawing a Slope Field The natural question is: how might one draw a direction field for an associated ODE? Computer Way: 1 Pick (x, y) with some sort of spacing condition. 2 Calculate f (x, y) at that point. Human Way: 1 Pick a slope c. 2 Set f (x, y) = c, this is called an isocline or level curve from Calculus III. 3 Draw in a line element. Slide 14/24 Dr. John Ehrke Lecture 1 Fall 2012

45 Drawing a Slope Field The natural question is: how might one draw a direction field for an associated ODE? Computer Way: 1 Pick (x, y) with some sort of spacing condition. 2 Calculate f (x, y) at that point. 3 Draw in a line element. Human Way: 1 Pick a slope c. 2 Set f (x, y) = c, this is called an isocline or level curve from Calculus III. 3 All points (x, y) that lie on f (x, y) = c, should have line elements with slope c. Slide 14/24 Dr. John Ehrke Lecture 1 Fall 2012

46 Method of Isoclines We will work out a few examples to get used to this method, called the method of isoclines for drawing a direction field. Example Draw the isoclines and a typical solution curve for f (x, y) = x/y. This was the easy problem identified earlier. Slide 15/24 Dr. John Ehrke Lecture 1 Fall 2012

47 Method of Isoclines We will work out a few examples to get used to this method, called the method of isoclines for drawing a direction field. Example Draw the isoclines and a typical solution curve for f (x, y) = x/y. This was the easy problem identified earlier. Analytic Solution: dy/dx = x/y = y dy = x dx = y dy = x dx = y 2 /2 = x 2 /2 + c = x 2 + y 2 = 2c = x 2 + y 2 = r 2 ( let 2c = r 2 ) So the solutions are circles of arbitrary radius. Slide 15/24 Dr. John Ehrke Lecture 1 Fall 2012

48 Isocline Diagram Slide 16/24 Dr. John Ehrke Lecture 1 Fall 2012

49 Slide 17/24 Dr. John Ehrke Lecture 1 Fall 2012

50 First Order Linear Equations In this lecture, we will consider first order linear ODEs of the form a(x)y + b(x)y = c(x) (11) The standard form for a first order linear ODE is obtained by dividing through by a(x) to create new coefficients p(x) and q(x). Equation (11) becomes y + p(x)y = q(x) (12) Remarks Slide 18/24 Dr. John Ehrke Lecture 1 Fall 2012

51 First Order Linear Equations In this lecture, we will consider first order linear ODEs of the form a(x)y + b(x)y = c(x) (11) The standard form for a first order linear ODE is obtained by dividing through by a(x) to create new coefficients p(x) and q(x). Equation (11) becomes y + p(x)y = q(x) (12) Remarks The equation (11) can always be solved! Slide 18/24 Dr. John Ehrke Lecture 1 Fall 2012

52 First Order Linear Equations In this lecture, we will consider first order linear ODEs of the form a(x)y + b(x)y = c(x) (11) The standard form for a first order linear ODE is obtained by dividing through by a(x) to create new coefficients p(x) and q(x). Equation (11) becomes y + p(x)y = q(x) (12) Remarks The equation (11) can always be solved! If q(x) = 0, (11) is called homogeneous. Slide 18/24 Dr. John Ehrke Lecture 1 Fall 2012

53 First Order Linear Equations In this lecture, we will consider first order linear ODEs of the form a(x)y + b(x)y = c(x) (11) The standard form for a first order linear ODE is obtained by dividing through by a(x) to create new coefficients p(x) and q(x). Equation (11) becomes y + p(x)y = q(x) (12) Remarks The equation (11) can always be solved! If q(x) = 0, (11) is called homogeneous. If q(x) 0, (11) is called inhomogeneous. Slide 18/24 Dr. John Ehrke Lecture 1 Fall 2012

54 First Order Linear Equations In this lecture, we will consider first order linear ODEs of the form a(x)y + b(x)y = c(x) (11) The standard form for a first order linear ODE is obtained by dividing through by a(x) to create new coefficients p(x) and q(x). Equation (11) becomes y + p(x)y = q(x) (12) Remarks The equation (11) can always be solved! If q(x) = 0, (11) is called homogeneous. If q(x) 0, (11) is called inhomogeneous. The equation (11) arises in a variety of physical situations, i.e. the process of conduction and diffusion. Slide 18/24 Dr. John Ehrke Lecture 1 Fall 2012

55 Conduction Diffusion Model The conduction diffusion model is based on a physical law known as Newtons law of cooling which states that the rate of change of the temperature of an object placed into a medium (air, liquid, gel, etc...) is directly proportional to the difference in temperature between that object and its environment. Slide 19/24 Dr. John Ehrke Lecture 1 Fall 2012

56 Conduction Diffusion Model The conduction diffusion model is based on a physical law known as Newtons law of cooling which states that the rate of change of the temperature of an object placed into a medium (air, liquid, gel, etc...) is directly proportional to the difference in temperature between that object and its environment. In this model, we let T(t) be the temperature of the object at time t placed into a medium having temperature given by T e (t). Slide 19/24 Dr. John Ehrke Lecture 1 Fall 2012

57 Conduction Diffusion Model The conduction diffusion model is based on a physical law known as Newtons law of cooling which states that the rate of change of the temperature of an object placed into a medium (air, liquid, gel, etc...) is directly proportional to the difference in temperature between that object and its environment. In this model, we let T(t) be the temperature of the object at time t placed into a medium having temperature given by T e (t). If T > T e, then dt dt < 0 and so T(t) is a decreasing function and describes how the body cools. Slide 19/24 Dr. John Ehrke Lecture 1 Fall 2012

58 Conduction Diffusion Model The conduction diffusion model is based on a physical law known as Newtons law of cooling which states that the rate of change of the temperature of an object placed into a medium (air, liquid, gel, etc...) is directly proportional to the difference in temperature between that object and its environment. In this model, we let T(t) be the temperature of the object at time t placed into a medium having temperature given by T e (t). If T > T e, then dt dt < 0 and so T(t) is a decreasing function and describes how the body cools. Conversely, if T < T e, then dt dt > 0 and so T(t) is an increasing function and describes how the body heats. Slide 19/24 Dr. John Ehrke Lecture 1 Fall 2012

59 Newton s Law of Cooling Newton s law of cooling states that the rate of change in the temperature T(t) of a body is proportional to the difference between the temperature of the object s environment, T e and the temperature of the body. That is, dt dt = k (T e(t) T(t)) (13) where k is the conductivity constant specific to the medium in which the object is placed. 1 The units for k are inverse time. Why? Slide 20/24 Dr. John Ehrke Lecture 1 Fall 2012

60 Newton s Law of Cooling Newton s law of cooling states that the rate of change in the temperature T(t) of a body is proportional to the difference between the temperature of the object s environment, T e and the temperature of the body. That is, dt dt = k (T e(t) T(t)) (13) where k is the conductivity constant specific to the medium in which the object is placed. 1 The units for k are inverse time. Why? 2 The value of k is positive. Why? Slide 20/24 Dr. John Ehrke Lecture 1 Fall 2012

61 Newton s Law of Cooling Newton s law of cooling states that the rate of change in the temperature T(t) of a body is proportional to the difference between the temperature of the object s environment, T e and the temperature of the body. That is, dt dt = k (T e(t) T(t)) (13) where k is the conductivity constant specific to the medium in which the object is placed. 1 The units for k are inverse time. Why? 2 The value of k is positive. Why? 3 The conduction-diffusion model, (13), is a classic example of a first order linear model. Written in standard form, it is given by T (t) + k T(t) = k T e (t) (14) Slide 20/24 Dr. John Ehrke Lecture 1 Fall 2012

62 Newton s Law of Cooling Newton s law of cooling states that the rate of change in the temperature T(t) of a body is proportional to the difference between the temperature of the object s environment, T e and the temperature of the body. That is, dt dt = k (T e(t) T(t)) (13) where k is the conductivity constant specific to the medium in which the object is placed. 1 The units for k are inverse time. Why? 2 The value of k is positive. Why? 3 The conduction-diffusion model, (13), is a classic example of a first order linear model. Written in standard form, it is given by T (t) + k T(t) = k T e (t) (14) The big question at this point, is how do we solve such equations? The answer involves a technique called the integrating factor method. Slide 20/24 Dr. John Ehrke Lecture 1 Fall 2012

63 The Integrating Factor Method Our goal is to solve the general linear first order equation y + p(x)y = q(x). (15) We seek a function u(x) to multiply through (15) simplifying the left hand side into a single expression. What conditions, should such a u(x) satisfy? Slide 21/24 Dr. John Ehrke Lecture 1 Fall 2012

64 The Integrating Factor Method Our goal is to solve the general linear first order equation y + p(x)y = q(x). (15) We seek a function u(x) to multiply through (15) simplifying the left hand side into a single expression. What conditions, should such a u(x) satisfy? Solution: Multiply both sides of (15) by u(x) and we obtain, u(x)y + u(x)p(x)y = u(x)q(x). Using the product rule in reverse, we would like to combine the left hand side as (u(x) y) = u(x)q(x) but in order to do so, u(x) must satisfy the separable differential equation u (x) = u(x) p(x). Solving this equation yields the general form of the integrating factor [ ] u(x) = exp p(x) dx (16) Slide 21/24 Dr. John Ehrke Lecture 1 Fall 2012

65 Solution of the Conduction-Diffusion Model Example Obtain a particular solution for the conduction-diffusion model, dt dt + k T(t) = k Te(t) (17) where T(t) and T e(t) are both functions of time t, and k > 0 is the conductivity constant. Assume T(0) = T 0. Consider what happens as t. Slide 22/24 Dr. John Ehrke Lecture 1 Fall 2012

66 Solution of the Conduction-Diffusion Model Example Obtain a particular solution for the conduction-diffusion model, dt dt + k T(t) = k Te(t) (17) where T(t) and T e(t) are both functions of time t, and k > 0 is the conductivity constant. Assume T(0) = T 0. Consider what happens as t. Solution: Our integrating factor is given by [ ] u(x) = exp k dt = exp(kt) = e kt. Multiplying through by u(x) we obtain, e kt T(t) + ke kt T(t) = kt e(t)e kt. Simplifying the left hand side of the above equation via the product rule yields, ( e kt T(t)) = kte(t)e kt. Slide 22/24 Dr. John Ehrke Lecture 1 Fall 2012

67 Transient and Steady-State Solutions Continuing our solution from the previous slide, we integrate both sides of the equation to obtain, e kt T(t) = t 0 ke ks T e(s) ds + c. Note that we include the constant of integration so we might satisfy the initial value T(0) = T 0 explicitly. Exponentiating both sides gives t T(t) = e kt ke ks T e(s) ds + ce }{{} kt 0 }{{} transient solution steady-state solution 1 The value of c is clearly T 0 when t = 0, so our full solution is given by t T(t) = e kt ke ks T e(s) ds + T 0e kt. 0 2 The steady-state solution is the part of the solution which is a response to the system being forced externally. The transient solution (which eventually goes away) is the baseline behavior of the system as it is the solution of the equation in the case when T e(t) = 0. Slide 23/24 Dr. John Ehrke Lecture 1 Fall 2012

68 Polling Question #1 Solve the first order linear equation A. c(90 + t) 3 + 2t B. 2(90 + t) 4 + c C t + c(90 + t) 3 D. 720t + 2t 4 + c x (t) + 3x 90 + t x = 8. Slide 24/24 Dr. John Ehrke Lecture 1 Fall 2012