Math 308, Sections 301, 302, Summer 2008 Review before Test I 06/09/2008

Transcription

1 Math 308, Sections 301, 302, Summer 2008 Review before Test I 06/09/2008

2 Chapter 1. Introduction Section 1.1 Background Definition Equation that contains some derivatives of an unknown function is called a differential equation. Definition A differential equation involving only ordinary derivatives with respect to a single variable is called an ordinary differential equations or ODE. A differential equation involving partial derivatives with respect to more then one variable is a partial differential equations or PDE. Definition The order of a differential equation is the order of the highest-order derivatives present in equation. Definition An ODE is linear if it has format a n (x) dn y dx n + a n 1(x) dn 1 y dx n 1 + a 1(x) dy dx + a 0(x)y = F(x), where a n (x), a n 1 (x),...,a 0 (x) and F(x) depend only on variable x. If an ODE is not linear, then we call it nonlinear.

3 Section 1.2 Solutions and Initial Value Problems A general form for an nth-order equation with variable x and unknown function y = y(x) can be expressed as ( F x,y, dy ) dx,, dn y dx n = 0, (1) where F is a function that depends on x, y, and the derivatives of y up to the order n. We assume that the equation holds for all x in an open interval I (a < x < b, where a or b could be infinite). d n ( y dx n = f x,y, dy ) dx,, dn 1 y dx n 1 = 0. (2)

4 Definition A function ϕ(x) that when substituted by y in equation (4) or (5) satisfies the equation for all x in the interval I is called an explicit solution to the equation on I. Sometimes the solution to the equation is defined implicitly. Definition A relation G(x,y) = 0 is said to be an implicit solution to equation (4) on the interval I if it defines one or more explicit solutions on I.

5 Definition A function ϕ(x) that when substituted by y in equation (4) or (5) satisfies the equation for all x in the interval I is called an explicit solution to the equation on I. Sometimes the solution to the equation is defined implicitly. Definition A relation G(x,y) = 0 is said to be an implicit solution to equation (4) on the interval I if it defines one or more explicit solutions on I. Example 8 (practice test) Show that the relation y = x + e y is an implicit solution to the equation (x + y 1)y = 1.

6 Definition By an initial value problem for an nth-order differential equation ( F x,y, dy ) dx,, dn y dx n = 0 we mean: Find a solution to the differential equation on an interval I that satisfies at x 0 the n initial conditions: dy y(x 0 ) = y 0, dx (x 0) = y 1,..., dn 1 y dx n 1 (x 0) = y n 1, where x 0 I and y 0, y 1,...,y n 1 are given constants. Theorem Given the initial value problem dy dx = f (x,y), y(0) = y 0, assume that f and f / y are continuous functions in a rectangle R = {(x,y) : a < x < b,c < y < d} that contains the point (x 0,y 0 ). Then the initial value problem has a unique solution ϕ(x) in some interval x 0 δ < x < x 0 + δ, where δ is a positive number.

7 Section 1.3 Direction Fields A first-order equation dy dx = f (x,y) specifies a slope at each point in the xy-plane where f is defined. A plot of short line segments drawn at various points in the xy-plane showing the slope of the solution curve there is called a direction field for the differential equation. Because the direction field gives the flow of solutions, it facilitates the drawing of any particular solution (such as the solution to an initial value problem). Method of Isoclines Definition An isocline for the differential equation y = f (x,y) is a set of points in the xy-plane where all the solutions have the same slope dy/dx; thus, it is a level curve for the function f (x,y).

8 Section 1.3 Direction Fields A first-order equation dy dx = f (x,y) specifies a slope at each point in the xy-plane where f is defined. A plot of short line segments drawn at various points in the xy-plane showing the slope of the solution curve there is called a direction field for the differential equation. Because the direction field gives the flow of solutions, it facilitates the drawing of any particular solution (such as the solution to an initial value problem). Method of Isoclines Definition An isocline for the differential equation y = f (x,y) is a set of points in the xy-plane where all the solutions have the same slope dy/dx; thus, it is a level curve for the function f (x,y). Example 4 (practice test) Draw the isoclines with their direction markers and sketch the solution to the initial value problem yy = x, y(0) = 4.

9 Chapter 2. First-Order Differential Equations Section 2.2 Separable Equations dy dx = g(x)h(y). (3) Lets multiply equation (5) by dx h(y) dy h(y) = g(x)dx dy h(y) = g(x)dx The solution to an equation (5) is H(y) = G(x) + C, here H(y) is an antiderivative of 1/h(y), G(x) is an antiderivative of g(x), C is a constant.

10 Chapter 2. First-Order Differential Equations Section 2.2 Separable Equations dy dx = g(x)h(y). (3) Lets multiply equation (5) by dx h(y) dy h(y) = g(x)dx dy h(y) = g(x)dx The solution to an equation (5) is H(y) = G(x) + C, here H(y) is an antiderivative of 1/h(y), G(x) is an antiderivative of g(x), C is a constant. Example 6 (practice test) Solve the IVP ydx + (1 + x)dy = 0 y(0) = 1.

11 Section 2.3 Linear Equations A linear first-order equation is an equation that can be expressed in the form a 1 (x) dy dx + a 0(x)y = b(x), (4) where a 0 (x), a 1 (x), b(x) depend only on x. We will assume that a 0 (x), a 1 (x), b(x) are continuous functions of x on an interval I. Assume that a 1 (x) is never zero on I. We can rewrite (4) in the standard form dy + P(x)y = Q(x), (5) dx where P(x) = a 0 (x)/a 1 (x) and Q(x) = b(x)/a 1 (x) are continuous on I.

12 To solve the first order linear equation (a) Write the equation in the standard form (5). (b) Find the integrating factor µ(x) solving differential equation (c) Integrate the equation dµ dx P(x)µ = 0. d [µy] = µq(x) dx and solve for y by dividing by µ(x).

13 To solve the first order linear equation (a) Write the equation in the standard form (5). (b) Find the integrating factor µ(x) solving differential equation (c) Integrate the equation dµ dx P(x)µ = 0. d [µy] = µq(x) dx and solve for y by dividing by µ(x). Example 7 (practice test) Find the general solution to the equation (x 2 1)y + 2xy + 3 = 0.

14 Existence and uniqueness of solution

15 Existence and uniqueness of solution Theorem Suppose P(x) and Q(x) are continuous on some interval I that contains the point x 0. Then for any choice of initial value y 0, there exists a unique solution y(x) on I to the initial value problem y + P(x)y = Q(x), y(x 0 ) = y 0. (6)

16 Existence and uniqueness of solution Theorem Suppose P(x) and Q(x) are continuous on some interval I that contains the point x 0. Then for any choice of initial value y 0, there exists a unique solution y(x) on I to the initial value problem y + P(x)y = Q(x), y(x 0 ) = y 0. (6) Example 5 (practice test) Determine (without solving the problem) an interval in which the solution to the initial value problem y + y tanx = sec x, y(0) = 3 is certain to exist.

17 Chapter 3. Mathematical methods and numerical methods involving first order equations. Section 3.2 Compartmental analysis The basic one-compartment system consists of a function x(t) that represents the amount of a substance in the compartment at time t, an input rate at which the substance enters the compartment, and an output rate at which the substance leaves the compartment Because the derivative of x with respect to t can be interpreted as the rate of change in the amount of the substance in the compartment with respect to time, the one-compartment system suggests dx dt = input rate output rate as a mathematical model for the process.

18 Chapter 3. Mathematical methods and numerical methods involving first order equations. Section 3.2 Compartmental analysis The basic one-compartment system consists of a function x(t) that represents the amount of a substance in the compartment at time t, an input rate at which the substance enters the compartment, and an output rate at which the substance leaves the compartment Because the derivative of x with respect to t can be interpreted as the rate of change in the amount of the substance in the compartment with respect to time, the one-compartment system suggests dx dt = input rate output rate as a mathematical model for the process. Mixing problem.

19 Chapter 3. Mathematical methods and numerical methods involving first order equations. Section 3.2 Compartmental analysis The basic one-compartment system consists of a function x(t) that represents the amount of a substance in the compartment at time t, an input rate at which the substance enters the compartment, and an output rate at which the substance leaves the compartment Because the derivative of x with respect to t can be interpreted as the rate of change in the amount of the substance in the compartment with respect to time, the one-compartment system suggests dx dt = input rate output rate as a mathematical model for the process. Mixing problem. Example 1 (practice test) A large tank initially contains 10 L of fresh water. A brine containing 20 g/l of salt flows into the tank at a rate of 3 L/min. The solution inside the tank is kept well stirred and flows out of the tank at the rate 2 L/min. Determine the concentration of salt in the tank as a function of time.

20 Section 3.3 Heating and cooling of buildings. Let T(t) represent the temperature inside the building at time t and view the building as a single compartment. We will consider three main factors that affect the temperature inside the building. heat produced by people, lights, and machines inside the building. This causes a rate of increase in temperature that we will denote by H(t). the heating (or cooling) supplied by the furnace (or air conditioner). This rate of increase (or decrease) in temperature will be represented by U(t). the effect of the outside temperature M(t) on the temperature inside the building.

21 We have dt = K(M(t) T(t)) + U(t) + H(t) dt where H(t) 0 and U(t) > 0 for furnace heating and U(t) < 0 for air conditioning cooling. T(t) = e Kt { e Kt [KM(t) + U(t) + H(t)]dt + C } If M(t) = M 0, H(t) = U(t) = 0, and T(0) = T 0, then T(t) = M 0 + (T 0 M 0 )e K(t t 0), here K is a constant.

22 We have dt = K(M(t) T(t)) + U(t) + H(t) dt where H(t) 0 and U(t) > 0 for furnace heating and U(t) < 0 for air conditioning cooling. T(t) = e Kt { e Kt [KM(t) + U(t) + H(t)]dt + C } If M(t) = M 0, H(t) = U(t) = 0, and T(0) = T 0, then T(t) = M 0 + (T 0 M 0 )e K(t t 0), here K is a constant. Example 3. (practice test) An object with temperature is placed in a freezer whose temperature is Assume that the temperature of the freezer remains essentially constant.

23 We have dt = K(M(t) T(t)) + U(t) + H(t) dt where H(t) 0 and U(t) > 0 for furnace heating and U(t) < 0 for air conditioning cooling. T(t) = e Kt { e Kt [KM(t) + U(t) + H(t)]dt + C } If M(t) = M 0, H(t) = U(t) = 0, and T(0) = T 0, then T(t) = M 0 + (T 0 M 0 )e K(t t 0), here K is a constant. Example 3. (practice test) An object with temperature is placed in a freezer whose temperature is Assume that the temperature of the freezer remains essentially constant. 1. If the object is cooled to after 8 min, what will its temperature be after 18 min?

24 We have dt = K(M(t) T(t)) + U(t) + H(t) dt where H(t) 0 and U(t) > 0 for furnace heating and U(t) < 0 for air conditioning cooling. T(t) = e Kt { e Kt [KM(t) + U(t) + H(t)]dt + C } If M(t) = M 0, H(t) = U(t) = 0, and T(0) = T 0, then T(t) = M 0 + (T 0 M 0 )e K(t t 0), here K is a constant. Example 3. (practice test) An object with temperature is placed in a freezer whose temperature is Assume that the temperature of the freezer remains essentially constant. 1. If the object is cooled to after 8 min, what will its temperature be after 18 min? 2. When will its temperature be 60 0?

25 Section 3.4 Newtonian mechanics Mechanics is the study of the motion of objects and the effect of forces acting on those objects. Newtonian or classical, mechanics deals with the motion of ordinary objects that is, objects that are large compared to an atom and slow moving compared with the speed of light. Newton s laws of motion: 1. When a body is subject to no resultant external force, it moves with a constant velocity. 2. When a body is subject to one or more external forces, the time rate of change of the body s momentum is equal to the vector sum of the external forces acting on it. 3. When one body interacts with a second body, the force of the first body on the second is equal in magnitude, but opposite in direction, to the force of the second body on the first.

26 Procedure for Newtonian models 1. Determine all relevant forces acting on the object being studied. It is helpful to draw a simple diagram of the object that depicts these forces. 2. Choose an appropriate axis or coordinate system in which to represent the motion of the object and the forces acting on it. 3. Apply Newton s second law to determine the equations of motion for the object. g = 32 ft/sec 2 = 9.81 m/sec 2

27 An object of mass m is given an initial downward velocity v 0 and allowed to fall under the influence of gravity. Assuming the gravitational force is constant and the force due to air resistance is proportional to the velocity of the object with some coefficient b (b 0), determine the equation of motion for this body. A model for the velocity of the falling body is expressed by the IVP m dv dt = mg bv, v(0) = v 0 v(t) = mg ( b + v 0 mg ) e bt/m b ) (1 e b m t ) x(t) = mg b t + m b ( v0 mg b

28 An object of mass m is given an initial downward velocity v 0 and allowed to fall under the influence of gravity. Assuming the gravitational force is constant and the force due to air resistance is proportional to the velocity of the object with some coefficient b (b 0), determine the equation of motion for this body. A model for the velocity of the falling body is expressed by the IVP m dv dt = mg bv, v(0) = v 0 v(t) = mg ( b + v 0 mg ) e bt/m b ) (1 e b m t ) x(t) = mg b t + m b ( v0 mg b Example An object of mass 5 kg is given the initial downward velocity of 50 m/sec and then allowed to fall under the influence of gravity. Assume that the force in newtons due to air resistance is 10v, where v is the velocity of the object in m/sec.

29 An object of mass m is given an initial downward velocity v 0 and allowed to fall under the influence of gravity. Assuming the gravitational force is constant and the force due to air resistance is proportional to the velocity of the object with some coefficient b (b 0), determine the equation of motion for this body. A model for the velocity of the falling body is expressed by the IVP m dv dt = mg bv, v(0) = v 0 v(t) = mg ( b + v 0 mg ) e bt/m b ) (1 e b m t ) x(t) = mg b t + m b ( v0 mg b Example An object of mass 5 kg is given the initial downward velocity of 50 m/sec and then allowed to fall under the influence of gravity. Assume that the force in newtons due to air resistance is 10v, where v is the velocity of the object in m/sec. 1. Determine the equation of motion of the object.

30 An object of mass m is given an initial downward velocity v 0 and allowed to fall under the influence of gravity. Assuming the gravitational force is constant and the force due to air resistance is proportional to the velocity of the object with some coefficient b (b 0), determine the equation of motion for this body. A model for the velocity of the falling body is expressed by the IVP m dv dt = mg bv, v(0) = v 0 v(t) = mg ( b + v 0 mg ) e bt/m b ) (1 e b m t ) x(t) = mg b t + m b ( v0 mg b Example An object of mass 5 kg is given the initial downward velocity of 50 m/sec and then allowed to fall under the influence of gravity. Assume that the force in newtons due to air resistance is 10v, where v is the velocity of the object in m/sec. 1. Determine the equation of motion of the object. 2. After how many seconds will the velocity of the object be 70 m/sec?