# 2.2 Separable Equations

Save this PDF as:

Size: px
Start display at page:

## Transcription

1 2.2 Separable Equations Separable Equations An equation y = f(x, y) is called separable provided algebraic operations, usually multiplication, division and factorization, allow it to be written in a separable form y = F (x)g(y) for some functions F and G. This class includes the quadrature equations y = F (x). Separable equations and associated solution methods were discovered by G. Leibniz in 1691 and formalized by J. Bernoulli in Finding a Separable Form Given differential equation y = f(x, y), invent values x 0, y 0 such that f(x 0, y 0 ) 0. Define F, G by the formulas (1) F (x) = f(x, y 0) f(x 0, y 0 ), G(y) = f(x 0, y). Because f(x 0, y 0 ) 0, then (1) makes sense. Test I below implies the following test. Theorem 2 (Separability Test) Let F and G be defined by equation (1). Compute F (x)g(y). Then (a) F (x)g(y) = f(x, y) implies y = f(x, y) is separable. (b) F (x)g(y) f(x, y) implies y = f(x, y) is not separable. Invention and Application. Initially, let (x 0, y 0 ) be (0, 0) or (1, 1) or some suitable pair, for which f(x 0, y 0 ) 0; then define F and G by (1). Multiply to test the equation F G = f. The algebra will discover a factorization f = F (x)g(y) without having to know algebraic tricks like factorizing multi-variable equations. But if F G f, then the algebra proves the equation is not separable. Non-Separability Tests Test I Equation y = f(x, y) is not separable if for some pair of points (x 0, y 0 ), (x, y) in the domain of f (2) f(x, y 0 )f(x 0, y) f(x 0, y 0 )f(x, y) 0. Test II The equation y = f(x, y) is not separable if either f x (x, y)/f(x, y) is non-constant in y or f y (x, y)/f(x, y) is non-constant in x. Illustration. Consider y = xy + y 2. Test I implies it is not separable, because f(x, 1)f(0, y) f(0, 1)f(x, y) = (x+1)y 2 (xy+y 2 ) = x(y 2 y) 0. Test II implies it is not separable, because f x /f = 1/(x + y) is not constant as a function of y.

2 74 Test I details. Assume f(x, y) = F (x)g(y), then equation (2) fails because each term on the left side of (2) evaluates to F (x)g(y 0 )F (x 0 )G(y) for all choices of (x 0, y 0 ) and (x, y) (hence contradiction 0 0). Test II details. Assume f(x, y) = F (x)g(y) and F, G are sufficiently differentiable. Then f x (x, y)/f(x, y) = F (x)/f (x) is independent of y and f y (x, y)/f(x, y) = G (y)/g(y) is independent of x. Separated Form Test A separated equation y /G(y) = F (x) is recognized by this test: Left Side Test. The left side of the equation has factor y and it is independent of symbol x. Right Side Test. The right side of the equation is independent of symbols y and y. Variables-Separable Method Determined by the method are the following kinds of solution formulas. Equilibrium Solutions. They are the constant solutions y = c of y = f(x, y). Find them by substituting y = c in y = f(x, y), followed by solving for c, then report the list of answers y = c for all values of c. Non-Equilibrium Solutions. For separable equation y = F (x)g(y), it is a solution y with G(y) 0. It is found by dividing by G(y) and applying the method of quadrature. The term equilibrium is borrowed from kinematics. Alternative terms are rest solution and stationary solution; all mean y = 0 in calculus terms. Spurious Solutions. If F (x)g(y) = 0 is solved instead of G(y) = 0, then both x and y solutions might be found. The x-solutions are ignored: they are not equilibrium solutions. Only solutions of the form y = constant are called equilibrium solutions. It is important to check the solution to a separable equation, because certain steps used to arrive at the solution may not be reversible. For most applications, the two kinds of solutions suffice to determine all possible solutions. In general, a separable equation may have non-unique solutions to some initial value problem. To prevent this from happening, it can be assumed that F, G and G are continuous; see the Picard- Lindelöf theorem, page 61. If non-uniqueness does occur, then often the equilibrium and non-equilibrium solutions can be pieced together to represent all solutions.

3 2.2 Separable Equations 75 Finding Equilibria The search for equilibria can be done without finding the separable form of y = f(x, y). It is enough to solve for y in the equation f(x, y) = 0, subject to the condition that x is arbitrary. An equilibrium solution y cannot depend upon x, because it is constant. If y turns out to depend on x, after solving f(x, y) = 0 for y, then this is sufficient evidence that y = f(x, y) is not separable. Some examples: y = y sin(x y) It is not separable. The solutions of y sin(x y) = 0 are y = 0 and x y = nπ for any integer n. The solution y = x nπ is non-constant, therefore the equation cannot be separable. y = xy(1 y 2 ) It is separable. The equation xy(1 y 2 ) = 0 has three constant solutions y = 0, y = 1, y = 1. The problem of finding equilibria is known to be technically unsolvable, that is, there is no proven algorithm for finding all the solutions of G(y) = 0. However, there are some very good numerical methods that apply, including Newton s method and the bisection method. Modern computer algebra systems make it practical to find both algebraic and numerical equilibrium solutions, in a single effort. Finding Non-Equilibrium Solutions A given solution y(x) satisfying G(y(x)) 0 throughout its domain of definition is called a non-equilibrium solution. Then division by G(y(x)) is allowed. The method of quadrature applies to the separated equation y /G(y(x)) = F (x). Some details: x x 0 y (t)dt G(y(t)) = x x 0 F (t)dt y(x) du y 0 G(u) = x x 0 F (t)dt ( y(x) = W 1 ) x x 0 F (t)dt Integrate both sides of the separated equation over x 0 t x. Apply on the left the change of variables u = y(t). Define y 0 = y(x 0 ). Define W (y) = y y 0 du/g(u). Take inverses to isolate y(x). The calculation produces a formula which is strictly speaking a candidate solution y. It does not prove that the formula works in the equation: checking the solution is required. Theoretical Inversion The function W 1 appearing in the last step above is generally not given by a formula. Therefore, W 1 rarely appears explicitly in applications or examples. It is the method that is memorized:

4 76 Prepare a separable differential equation by transforming it to separated form. Then apply the method of quadrature. The theoretical basis for using W 1 is a calculus theorem which says that a strictly monotone continuous function has a continuous inverse. The fundamental theorem of calculus implies that W (y) is continuous with nonzero derivative W (y) = 1/G(y). Therefore, W (y) is strictly monotone. The cited calculus theorem implies that W (y) has a continuous inverse W 1. Explicit and Implicit Solutions The variables-separable method gives equilibrium solutions which are already explicit, that is: Definition 1 (Explicit Solution) A solution of y = f(x, y) is called explicit provided it is given by an equation y = an expression independent of y. To elaborate, on the left side must appear exactly the symbol y followed by an equal sign. Symbols y and = are followed by an expression which does not contain the symbol y. Definition 2 (Implicit Solution) A solution of y = f(x, y) is called implicit provided it is not explicit. Equations like 2y = x are not explicit (they are called implicit) because the coefficient of y on the left is not 1. Similarly, y = x + y 2 is not explicit because the right side contains symbol y. Equation y = e π is explicit because the right side fails to contain symbol y (symbol x may be absent). Applications can leave the non-equilibrium solutions in implicit form y(x) y 0 du/g(u) = x x 0 F (t)dt, with serious effort being expended to do the indicated integrations. In special cases, it is possible to find an explicit solution from the implicit one by algebraic methods. Students find the algebraic methods to be unmotivated tricks. Computer algebra systems can make this step look like science instead of art. Examples 3 Example (Non-separable Equation) Explain why yy = x y 2 is not separable.

5 2.2 Separable Equations 77 Solution: It is tempting to try manipulations like adding y 2 to both sides of the equation, in an attempt to obtain a separable form, but every such trick fails. The failure of such attempts is evidence that the equation is perhaps not separable. Failure of attempts does not prove non-separability. Test I applies to verify that the equation is not separable. Let f(x, y) = x/y y and choose x 0 = 0, y 0 = 1. Then f(x 0, y 0 ) 0. Compute as follows: LHS = f(x, y 0 )f(x 0, y) f(x 0, y 0 )f(x, y) Identity (2) left side. = f(x, 1)f(0, y) f(0, 1)f(x, y) Use x 0 = 0, y 0 = 1. = (x 1)( y) ( 1)(x/y y) Substitute f(x, y) = x/y y. = xy + x/y Simplify. This expression fails to be zero for all (x, y) (e.g., x = 1, y = 2), therefore the equation is not separable, by Test I. Test II also applies to verify the equation is not separable: f x /f = 1/yf = x y 2 is non-constant in x. 4 Example (Separated Form Test Failure) Given yy = 1 y 2, explain why the equivalent equation yy + y 2 = 1, obtained by adding y 2 across the equation, fails the separated form test, page 74. Solution: The test requires the left side of yy +y 2 = 1 to contain the factor y. It doesn t, so it fails the test. Yes, yy + y 2 = 1 does pass the other checkpoints of the test: the left side is independent of x and the right side is independent of y and y. 5 Example (Separated Equation) Find for (x+1)yy = x xy 2 a separated equation using the test, page 74. Solution: The equation usually reported is by factoring and division. yy (1 y)(1 + y) = x. It is found x + 1 The given equation is factored into (1 + x)yy = x(1 y)(1 + y). To pass the test, the objective is to move all factors containing only y to the left and all factors containing only x to the right. This is technically accomplished using division by (x + 1)(1 y)(1 + y). To the result of the division is applied the test on page 74: the left side contains factor y and otherwise involves the factor y/(1 y 2 ), which depends only on y; the right side is x/(x + 1), which depends only on x. In short, the candidate separated equation passes the test. There is another way to approach the problem, by writing the differential equation in standard form y = f(x, y) where f(x, y) = x(1 y 2 )/(1 + x). Then f(1, 0) = 1/2 0. Define F (x) = f(x, 0)/f(1, 0), G(y) = f(1, y). We verify F (x)g(y) = f(x, y). A separated form is then y /G(y) = F (x) or 2y /(1 y 2 ) = 2x/(1 + x). 6 Example (Equilibria) Given y = x(1 y)(1 + y), find all equilibria.

6 78 Solution: The constant solutions y = 1 and y = 1 are the equilibria, as will be justified. Equilibria are found by solving for constant value y in the equation x(1 y)(1 + y) = 0. The formal college algebra solutions are x = 0, y = 1 and y = 1. However, we do not seek these college algebra answers! Equilibria are the solutions y =constant such that x(1 y)(1 + y) = 0 for all x. The conditional for all x causes the algebra problem to reduce to just two equations: 0 = 0 (from x = 0) and (1 y)(1 + y) = 0 (from x 0). We report only the equilibrium solutions y = 1 and y = 1, which were found from the second equation (1 y)(1+y) = 0. 7 Example (Non-Equilibria) Given y = x 2 (1 + y), y(0) = y 0, find all nonequilibrium solutions. Solution: The unique solution is y = (1 + y 0 )e x3 /3 1. Details follow. The separable form y = F (x)g(y) is realized for F (x) = x 2 and G(y) = 1 + y. Sought are solutions with G(y) 0, or simply 1 + y 0. y = x 2 (1 + y) Original equation. y 1 + y = x2 Divide by 1 + y. Separated form found. y 1 + y dx = x 2 dx Method of quadrature. du 1 + u = x 2 dx Change variables u = y(x) on the left. ln 1 + y(x) = x 3 /3 + c Evaluate integrals. Implicit solution found. Applications might stop at this point and report the implicit solution. This illustration continues, to find the explicit solution y = (1 + y 0 )e x3 / y(x) = e x3 /3+c By definition, ln u = w means u = e w. 1 + y(x) = ke x3 /3+c Drop absolute value, k = ±1. y(x) = ke x3 /3+c 1 Candidate solution. Constants unresolved. The initial condition y(0) = y 0 is used to resolve the constants c and k. First, 1 + y 0 = e c from the first equation. Second, 1 + y 0 and 1 + y(x) must have the same sign (they are never zero), so k(1 + y 0 ) > 0. Hence, 1 + y 0 = ke c, which implies the solution is y = ke c e x3 /3 1 or y = (1 + y 0 )e x3 / Example (Equilibria) Given y = x sin(1 y) cos(1+y), find all equilibrium solutions. Solution: The infinite set of equilibria are justified below to be y = 1 + nπ, y = 1 + (2n + 1) π, n = 0, ±1, ±2,... 2

7 2.2 Separable Equations 79 A separable form y = F (x)g(y) is verified by defining F (x) = x and G(y) = sin(1 y) cos(1 + y). Equilibria are found by solving for y in the equation G(y) = 0, which is sin(1 y) cos(1 + y) = 0. This equation is satisfied when the argument of the sine is an integer multiple of π or when the argument of the cosine is an odd integer multiple of π/2. The solutions are y 1 = 0, ±π, ±2π,... and 1 + y = ±π/2, ±3π/2,.... Multiple solutions and maple. Equations having multiple solutions may require CAS setup. Below, the first code fragment returns two solutions, y = 1 and y = 1 + π/2. The second returns all solutions. # The default returns two solutions G:=y->sin(1-y)*cos(1+y): solve(g(y)=0,y); # Special setup returns all solutions _EnvAllSolutions := true: G:=y->sin(1-y)*cos(1+y): solve(g(y)=0,y); 9 Example (Non-Equilibria) Given y = x 2 sin(y), y(0) = y 0, justify that all non-equilibrium solutions are given by 1 ( ) y = 2Arctan tan(y 0 /2)e x3 /3 + 2nπ. Solution: A separable form y = F (x)g(y) is defined by F (x) = x 2 and G(y) = sin(y). A non-equilibrium solution will satisfy G(y) 0, or simply sin(y) 0. Define n by y 0 /2 = Arctan(tan(y 0 /2)) + nπ, where Arctan(u) < π/2. Then y = x 2 sin(y) The original equation. csc(y)y = x 2 Separated form. Divided by sin(y) 0. csc(y)y dx = x 2 dx Quadrature using indefinite integrals. csc(u)du = x 2 dx Change variables u = y(x) on the left. ln csc y(x) cot y(x) = 1 3 x3 + c Integral tables. Implicit solution found. Trigonometric Identity. Integral tables make use of the identity tan(y/2) = csc y cot y, which is derived from the relations 2θ = y, 1 cos 2θ = 2 sin 2 θ, sin 2θ = 2 sin θ cos θ. Tables offer an alternate answer for the last integral above, ln tan(y/2). The solution obtained at this stage is called an implicit solution, because y has not been isolated. It is possible to solve for y in terms of x, an explicit solution. The details: csc y cot y = e x3 /3+c By definition, ln u = w means u = e w. csc y cot y = ke x3 /3+c Assign k = ±1 to drop absolute values. 1 While θ = arctan u gives any angle, θ = Arctan(u) gives θ < π/2.

8 80 1 cos y sin y = ke x3 /3+c Then k has the same sign as sin(y), because 1 cos y 0. tan(y/2) = ke x3 /3+c Use tan(y/2) = csc y cot y. ( ) y = 2Arctan ke x3 /3+c + 2nπ Candidate solution, n = 0, ±1, ±2,... Resolving the Constants. Constants c and k are uniquely resolved for a given initial condition y(0) = y 0. Values x = 0 and y = y 0 determine constant c by the equation tan(y 0 /2) = ke c (two equations back). The condition k sin(y 0 ) > 0 determines k, because sin y 0 and sin y have identical signs. If n is defined by y 0 /2 = Arctan(tan(y 0 /2)) + nπ and K = ke c = tan(y 0 /2), then the explicit solution is ( ) y = 2Arctan Ke x3 /3 + 2nπ, K = tan(y 0 /2). Trigonometric identities and maple. Using the identity csc y cot y = tan(y/2), maple finds the same relation. Complications occur without it. _EnvAllSolutions := true: solve(csc(y)-cot(y)=k*exp(x^3/3+c),y); solve(tan(y/2)=k*exp(x^3/3+c),y); 10 Example (Independent of x) Solve y = y(1 ln y), y(0) = y 0. Solution: There is just one equilibrium solution y = e Not included is y = 0, because y(1 ln y) is undefined for y 0. Details appear below for the explicit solution (which includes y = e) y = e 1 (1 ln y 0)e x. An equation y = f(x, y) independent of x has the form y = F (x)g(y) where F (x) = 1. Divide by G(y) to obtain a separated form y /G(y) = 1. In the present case, G(y) = y(1 ln y) is defined for y > 0. To require G(y) 0 means y > 0, y e. Non-equilibrium solutions will satisfy y > 0 and y e. y = 1 Separated form. Assume y > 0 and y e. y(1 ln y) y y(1 ln y) dx = dx Method of quadrature. du u = dx Substitute u = 1 ln y on the left. Chain rule (ln y) = y /y applied; du = y dx/y. ln 1 ln y(x) = x + c Evaluate the integral using u = 1 ln y. Implicit solution found. The remainder of the solution contains college algebra details, to find from the implicit solution all explicit solutions y. 1 ln y(x) = e x c Use ln u = w equivalent to u = e w. 1 ln y(x) = ke x c Drop absolute value, k = ±1. ln y(x) = 1 ke x c Solved for ln y.

9 2.2 Separable Equations 81 y(x) = e 1 ke x c Candidate solution; c and k unresolved. To resolve the constants, start with y 0 > 0 and y 0 e. To determine k, use the requirement G(y) 0 to deduce that k(1 ln y(x)) > 0. At x = 0, it means k 1 ln y 0 = 1 ln y 0. Then k = 1 for 0 < y 0 < e and k = 1 otherwise. Let y = y 0, x = 0 to determine c through the equation 1 ln y 0 = e c. Combining with the value of k gives 1 ln y 0 = ke c. Assembling the answers for k and c produces the relations y = e 1 ke x c Candidate solution. = e 1 ke c e x Exponential rule e a+b = e a e b. = e 1 (1 ln y 0)e x Explicit solution. Used ke c = 1 ln y 0. Even though the solution has been found through legal methods, it remains to verify the solution. See the exercises. Exercises 2.2 Separated Form Test. Test the given equation by the separated form test on page 74. Report whether or not the equation passes or fails. In this test, it is not allowed to do algebraic operations on the equation test the equation as written. See also Examples 3 and 4, page y = 2 2. y = x 3. y + y = 2 4. y + 2y = x 5. yy = 2 x 6. 2yy = x + x 2 7. yy + sin(y ) = 2 x 8. 2yy + cos(y) = x 9. 2yy = y cos(y) + x 10. (2y + tan(y))y = x Separated Equation. Determine the separated form y /G(y) = F (x) for the given separable equation. See Example 5, page (1 + x)y = 2 + y 12. (1 + y)y = xy 13. y = x + xy (x + 1) y 1 + y = sin(x) (x + 2) xy = y sin(y) cos(x) 16. x 2 y = y cos(y) tan(x) 17. y 2 (x y)y = x2 y 2 x + y 18. xy 2 (x + y)y = y2 x xy 2 y = y x x y 20. xy 2 y = x2 xy x y x y Equilibrium solutions. Determine the equilibria for the given equation. See Examples 6 and y = xy(1 + y) 22. xy = y(1 y)

10 y = 1 + y 1 y 24. xy = y(1 y) 1 + y 25. y = (1 + x) tan(y) 26. y = y(1 + ln y) 27. y = xe y (1 + y) 28. xy = e y (1 y) 29. xy = e y (1 y 2 )(1 + y) xy = e y (1 y 3 )(1 + y 3 ) Non-Equilibrium Solutions. Find the non-equilibrium solutions for the given separable equation. See Examples 7 and 9 for details. 31. y = (xy) 1/3, y(0) = y y = (xy) 1/5, y(0) = y y = 1 + x y xy, y(0) = y y = 1 + x + 2y + 2xy, y(0) = y y = 36. y = (x + 1)y3 x 2 (y 3 y), y(0) = y 0. (x 1)y2 x 3 (y 3 + y), y(0) = y yy = x(1 y 2 ) 38. 2yy = x(1 + y 2 ) 39. (1 + x)y = 1 y, y(0) = y (1 x)y = 1 + y, y(0) = y tan(x)y = y, y(π/2) = y tan(x)y = 1 + y, y(π/2) = y xy = cos 2 (y), y(1) = y xy = sin 2 (y), y(0) = y x 2 16yy = x, y(5) = y x 2 1yy = x, y(2) = y y = x 2 (1 + y 2 ), y(0) = (1 x)y = x(1 + y 2 ), y(0) = 1. Independent of x. Solve the given equation, finding all solutions. See Example y = sin y, y(0) = y y = cos y, y(0) = y y = y(1 + ln y), y(0) = y y = y(2 + ln y), y(0) = y y = y(y 1)(y 2), y(0) = y y = y(y 1)(y + 1), y(0) = y y = y 2 + 2y + 5, y(0) = y y = y 2 + 2y + 7, y(0) = y 0. Details in the Examples. Collected here are verifications for details in the examples. Brief solutions for odd exercises appear in the Answers. 57. (Example 6) The equation x(1 y)(1 + y) = 0 was solved in the example, but x = 0 was ignored, and only y = 1 and y = 1 were reported. Why? 58. (Example 7) An absolute value equation u = w was replaced by u = kw where k = ±1. Justify the replacement using the definition u = u for u 0, u = u for u < (Example 7) Verify directly that y = (1+y 0 )e x3 /3 1 solves the initial value problem y = x 2 (1 + y), y(0) = y (Example 8) The relation y = 1 + nπ, n = 0, ±1, ±2,... describes the list..., 1 π, 1, 1 + π,.... Write the list for the relation y = 1 + (2n + 1) π (Example 8) Solve sin(u) = 0 and cos(v) = 0 for u and v. Supply graphs which show why there are infinity many solutions.

11 2.2 Separable Equations (Example 9) Explain why y 0 /2 does not equal Arctan(tan(y 0 /2)). Give a calculator example. 63. (Example 9) Establish the identity tan(y/2) = csc y cot y. 64. (Example 10) Let y 0 > 0. Verify that y = e 1 (1 ln y 0)e x solves y = y(1 ln y), y(0) = y 0.

### Learning Objectives for Math 165

Learning Objectives for Math 165 Chapter 2 Limits Section 2.1: Average Rate of Change. State the definition of average rate of change Describe what the rate of change does and does not tell us in a given

### 1 if 1 x 0 1 if 0 x 1

Chapter 3 Continuity In this chapter we begin by defining the fundamental notion of continuity for real valued functions of a single real variable. When trying to decide whether a given function is or

### CHAPTER 2. Inequalities

CHAPTER 2 Inequalities In this section we add the axioms describe the behavior of inequalities (the order axioms) to the list of axioms begun in Chapter 1. A thorough mastery of this section is essential

### VI. Transcendental Functions. x = ln y. In general, two functions f, g are said to be inverse to each other when the

VI Transcendental Functions 6 Inverse Functions The functions e x and ln x are inverses to each other in the sense that the two statements y = e x, x = ln y are equivalent statements In general, two functions

### TOPIC 3: CONTINUITY OF FUNCTIONS

TOPIC 3: CONTINUITY OF FUNCTIONS. Absolute value We work in the field of real numbers, R. For the study of the properties of functions we need the concept of absolute value of a number. Definition.. Let

### 6.6 The Inverse Trigonometric Functions. Outline

6.6 The Inverse Trigonometric Functions Tom Lewis Fall Semester 2015 Outline The inverse sine function The inverse cosine function The inverse tangent function The other inverse trig functions Miscellaneous

### Inverse Trigonometric Functions - Trigonometric Equations

Inverse Trigonometric Functions - Trigonometric Equations Dr. Philippe B. Laval Kennesaw STate University April 0, 005 Abstract This handout defines the inverse of the sine, cosine and tangent functions.

### Some Notes on Taylor Polynomials and Taylor Series

Some Notes on Taylor Polynomials and Taylor Series Mark MacLean October 3, 27 UBC s courses MATH /8 and MATH introduce students to the ideas of Taylor polynomials and Taylor series in a fairly limited

### GRE Prep: Precalculus

GRE Prep: Precalculus Franklin H.J. Kenter 1 Introduction These are the notes for the Precalculus section for the GRE Prep session held at UCSD in August 2011. These notes are in no way intended to teach

### The Derivative. Philippe B. Laval Kennesaw State University

The Derivative Philippe B. Laval Kennesaw State University Abstract This handout is a summary of the material students should know regarding the definition and computation of the derivative 1 Definition

### Homework # 3 Solutions

Homework # 3 Solutions February, 200 Solution (2.3.5). Noting that and ( + 3 x) x 8 = + 3 x) by Equation (2.3.) x 8 x 8 = + 3 8 by Equations (2.3.7) and (2.3.0) =3 x 8 6x2 + x 3 ) = 2 + 6x 2 + x 3 x 8

### First Order Non-Linear Equations

First Order Non-Linear Equations We will briefly consider non-linear equations. In general, these may be much more difficult to solve than linear equations, but in some cases we will still be able to solve

### Differentiation and Integration

This material is a supplement to Appendix G of Stewart. You should read the appendix, except the last section on complex exponentials, before this material. Differentiation and Integration Suppose we have

### Solving DEs by Separation of Variables.

Solving DEs by Separation of Variables. Introduction and procedure Separation of variables allows us to solve differential equations of the form The steps to solving such DEs are as follows: dx = gx).

### Precalculus Workshop - Functions

Introduction to Functions A function f : D C is a rule that assigns to each element x in a set D exactly one element, called f(x), in a set C. D is called the domain of f. C is called the codomain of f.

### Math 113 HW #9 Solutions

Math 3 HW #9 Solutions 4. 50. Find the absolute maximum and absolute minimum values of on the interval [, 4]. f(x) = x 3 6x 2 + 9x + 2 Answer: First, we find the critical points of f. To do so, take the

### Calculus 1: Sample Questions, Final Exam, Solutions

Calculus : Sample Questions, Final Exam, Solutions. Short answer. Put your answer in the blank. NO PARTIAL CREDIT! (a) (b) (c) (d) (e) e 3 e Evaluate dx. Your answer should be in the x form of an integer.

### Lecture 5 : Continuous Functions Definition 1 We say the function f is continuous at a number a if

Lecture 5 : Continuous Functions Definition We say the function f is continuous at a number a if f(x) = f(a). (i.e. we can make the value of f(x) as close as we like to f(a) by taking x sufficiently close

### 5 Indefinite integral

5 Indefinite integral The most of the mathematical operations have inverse operations: the inverse operation of addition is subtraction, the inverse operation of multiplication is division, the inverse

### Implicit Differentiation

Implicit Differentiation Sometimes functions are given not in the form y = f(x) but in a more complicated form in which it is difficult or impossible to express y explicitly in terms of x. Such functions

### 2 Integrating Both Sides

2 Integrating Both Sides So far, the only general method we have for solving differential equations involves equations of the form y = f(x), where f(x) is any function of x. The solution to such an equation

### Solutions to Homework 10

Solutions to Homework 1 Section 7., exercise # 1 (b,d): (b) Compute the value of R f dv, where f(x, y) = y/x and R = [1, 3] [, 4]. Solution: Since f is continuous over R, f is integrable over R. Let x

### 1. First-order Ordinary Differential Equations

Advanced Engineering Mathematics 1. First-order ODEs 1 1. First-order Ordinary Differential Equations 1.1 Basic concept and ideas 1.2 Geometrical meaning of direction fields 1.3 Separable differential

### Section 4.4. Using the Fundamental Theorem. Difference Equations to Differential Equations

Difference Equations to Differential Equations Section 4.4 Using the Fundamental Theorem As we saw in Section 4.3, using the Fundamental Theorem of Integral Calculus reduces the problem of evaluating a

### Know the following (or better yet, learn a couple and be able to derive the rest, quickly, from your knowledge of the trigonometric functions):

OCR Core Module Revision Sheet The C exam is hour 0 minutes long. You are allowed a graphics calculator. Before you go into the exam make sureyou are fully aware of the contents of theformula booklet you

### 14.1. Basic Concepts of Integration. Introduction. Prerequisites. Learning Outcomes. Learning Style

Basic Concepts of Integration 14.1 Introduction When a function f(x) is known we can differentiate it to obtain its derivative df. The reverse dx process is to obtain the function f(x) from knowledge of

### Differential Equations

Differential Equations A differential equation is an equation that contains an unknown function and one or more of its derivatives. Here are some examples: y = 1, y = x, y = xy y + 2y + y = 0 d 3 y dx

### Algebra. Exponents. Absolute Value. Simplify each of the following as much as possible. 2x y x + y y. xxx 3. x x x xx x. 1. Evaluate 5 and 123

Algebra Eponents Simplify each of the following as much as possible. 1 4 9 4 y + y y. 1 5. 1 5 4. y + y 4 5 6 5. + 1 4 9 10 1 7 9 0 Absolute Value Evaluate 5 and 1. Eliminate the absolute value bars from

### Section 6-4 Product Sum and Sum Product Identities

480 6 TRIGONOMETRIC IDENTITIES AND CONDITIONAL EQUATIONS Section 6-4 Product Sum and Sum Product Identities Product Sum Identities Sum Product Identities Our work with identities is concluded by developing

### 9.1 Trigonometric Identities

9.1 Trigonometric Identities r y x θ x -y -θ r sin (θ) = y and sin (-θ) = -y r r so, sin (-θ) = - sin (θ) and cos (θ) = x and cos (-θ) = x r r so, cos (-θ) = cos (θ) And, Tan (-θ) = sin (-θ) = - sin (θ)

### 106 Chapter 5 Curve Sketching. If f(x) has a local extremum at x = a and. THEOREM 5.1.1 Fermat s Theorem f is differentiable at a, then f (a) = 0.

5 Curve Sketching Whether we are interested in a function as a purely mathematical object or in connection with some application to the real world, it is often useful to know what the graph of the function

### Notes and questions to aid A-level Mathematics revision

Notes and questions to aid A-level Mathematics revision Robert Bowles University College London October 4, 5 Introduction Introduction There are some students who find the first year s study at UCL and

### Derivatives: rules and applications (Stewart Ch. 3/4) The derivative f (x) of the function f(x):

Derivatives: rules and applications (Stewart Ch. 3/4) The derivative f (x) of the function f(x): f f(x + h) f(x) (x) = lim h 0 h (for all x for which f is differentiable/ the limit exists) Property:if

### Taylor Polynomials and Taylor Series Math 126

Taylor Polynomials and Taylor Series Math 26 In many problems in science and engineering we have a function f(x) which is too complicated to answer the questions we d like to ask. In this chapter, we will

### Techniques of Integration

CHPTER 7 Techniques of Integration 7.. Substitution Integration, unlike differentiation, is more of an art-form than a collection of algorithms. Many problems in applied mathematics involve the integration

### Section 2.7 One-to-One Functions and Their Inverses

Section. One-to-One Functions and Their Inverses One-to-One Functions HORIZONTAL LINE TEST: A function is one-to-one if and only if no horizontal line intersects its graph more than once. EXAMPLES: 1.

### Sept 20, 2011 MATH 140: Calculus I Tutorial 2. ln(x 2 1) = 3 x 2 1 = e 3 x = e 3 + 1

Sept, MATH 4: Calculus I Tutorial Solving Quadratics, Dividing Polynomials Problem Solve for x: ln(x ) =. ln(x ) = x = e x = e + Problem Solve for x: e x e x + =. Let y = e x. Then we have a quadratic

### TOPIC 4: DERIVATIVES

TOPIC 4: DERIVATIVES 1. The derivative of a function. Differentiation rules 1.1. The slope of a curve. The slope of a curve at a point P is a measure of the steepness of the curve. If Q is a point on the

### Course outline, MA 113, Spring 2014 Part A, Functions and limits. 1.1 1.2 Functions, domain and ranges, A1.1-1.2-Review (9 problems)

Course outline, MA 113, Spring 2014 Part A, Functions and limits 1.1 1.2 Functions, domain and ranges, A1.1-1.2-Review (9 problems) Functions, domain and range Domain and range of rational and algebraic

### 1 Introduction to Differential Equations

1 Introduction to Differential Equations A differential equation is an equation that involves the derivative of some unknown function. For example, consider the equation f (x) = 4x 3. (1) This equation

### 6.5 Trigonometric formulas

100 CHAPTER 6. TRIGONOMETRIC FUNCTIONS 6.5 Trigonometric formulas There are a few very important formulas in trigonometry, which you will need to know as a preparation for Calculus. These formulas are

### Elementary Number Theory We begin with a bit of elementary number theory, which is concerned

CONSTRUCTION OF THE FINITE FIELDS Z p S. R. DOTY Elementary Number Theory We begin with a bit of elementary number theory, which is concerned solely with questions about the set of integers Z = {0, ±1,

### Second Order Linear Partial Differential Equations. Part I

Second Order Linear Partial Differential Equations Part I Second linear partial differential equations; Separation of Variables; - point boundary value problems; Eigenvalues and Eigenfunctions Introduction

### Module MA1S11 (Calculus) Michaelmas Term 2016 Section 3: Functions

Module MA1S11 (Calculus) Michaelmas Term 2016 Section 3: Functions D. R. Wilkins Copyright c David R. Wilkins 2016 Contents 3 Functions 43 3.1 Functions between Sets...................... 43 3.2 Injective

### WEEK #5: Trig Functions, Optimization

WEEK #5: Trig Functions, Optimization Goals: Trigonometric functions and their derivatives Optimization Textbook reading for Week #5: Read Sections 1.8, 2.10, 3.3 Trigonometric Functions From Section 1.8

### y cos 3 x dx y cos 2 x cos x dx y 1 sin 2 x cos x dx y 1 u 2 du u 1 3u 3 C

Trigonometric Integrals In this section we use trigonometric identities to integrate certain combinations of trigonometric functions. We start with powers of sine and cosine. EXAMPLE Evaluate cos 3 x dx.

### x 2 + y 2 = 25 and try to solve for y in terms of x, we get 2 new equations y = 25 x 2 and y = 25 x 2.

Lecture : Implicit differentiation For more on the graphs of functions vs. the graphs of general equations see Graphs of Functions under Algebra/Precalculus Review on the class webpage. For more on graphing

PYTHAGOREAN TRIPLES KEITH CONRAD 1. Introduction A Pythagorean triple is a triple of positive integers (a, b, c) where a + b = c. Examples include (3, 4, 5), (5, 1, 13), and (8, 15, 17). Below is an ancient

### x 2 + y 2 = 1 y 1 = x 2 + 2x y = x 2 + 2x + 1

Implicit Functions Defining Implicit Functions Up until now in this course, we have only talked about functions, which assign to every real number x in their domain exactly one real number f(x). The graphs

### Homework #1 Solutions

MAT 303 Spring 203 Homework # Solutions Problems Section.:, 4, 6, 34, 40 Section.2:, 4, 8, 30, 42 Section.4:, 2, 3, 4, 8, 22, 24, 46... Verify that y = x 3 + 7 is a solution to y = 3x 2. Solution: From

### M344 - ADVANCED ENGINEERING MATHEMATICS Lecture 9: Orthogonal Functions and Trigonometric Fourier Series

M344 - ADVANCED ENGINEERING MATHEMATICS ecture 9: Orthogonal Functions and Trigonometric Fourier Series Before learning to solve partial differential equations, it is necessary to know how to approximate

### Chapter 7 Outline Math 236 Spring 2001

Chapter 7 Outline Math 236 Spring 2001 Note 1: Be sure to read the Disclaimer on Chapter Outlines! I cannot be responsible for misfortunes that may happen to you if you do not. Note 2: Section 7.9 will

### MAT2400 Analysis I. A brief introduction to proofs, sets, and functions

MAT2400 Analysis I A brief introduction to proofs, sets, and functions In Analysis I there is a lot of manipulations with sets and functions. It is probably also the first course where you have to take

### 2 2 [ 1, 1]. Thus there exists a function sin 1 or arcsin. sin x arcsin x(= sin 1 x) arcsin(x) = y sin(y) = x. arcsin(sin x) = x if π 2 x π 2

7.4 Inverse Trigonometric Functions From looking at the graphs of the trig functions, we see that they fail the horizontal line test spectacularly. However, if you restrict their domain, you can find an

### Blue Pelican Calculus First Semester

Blue Pelican Calculus First Semester Teacher Version 1.01 Copyright 2011-2013 by Charles E. Cook; Refugio, Tx Edited by Jacob Cobb (All rights reserved) Calculus AP Syllabus (First Semester) Unit 1: Function

### Review for Calculus Rational Functions, Logarithms & Exponentials

Definition and Domain of Rational Functions A rational function is defined as the quotient of two polynomial functions. F(x) = P(x) / Q(x) The domain of F is the set of all real numbers except those for

### Inverse Trigonometric Functions

Section 3 Inverse Trigonometric Functions 00 Kiryl Tsishchanka Inverse Trigonometric Functions DEFINITION: The inverse sine function, denoted by sin x or arcsin x), is defined to be the inverse of the

### y cos 3 x dx y cos 2 x cos x dx y 1 sin 2 x cos x dx

Trigonometric Integrals In this section we use trigonometric identities to integrate certain combinations of trigonometric functions. We start with powers of sine and cosine. EXAMPLE Evaluate cos 3 x dx.

### Limits and Continuity

Math 20C Multivariable Calculus Lecture Limits and Continuity Slide Review of Limit. Side limits and squeeze theorem. Continuous functions of 2,3 variables. Review: Limits Slide 2 Definition Given a function

### Practice Problems for Midterm 2

Practice Problems for Midterm () For each of the following, find and sketch the domain, find the range (unless otherwise indicated), and evaluate the function at the given point P : (a) f(x, y) = + 4 y,

### SECOND-ORDER LINEAR DIFFERENTIAL EQUATIONS

SECOND-ORDER LINEAR DIFFERENTIAL EQUATIONS A second-order linear differential equation has the form 1 Px d y dx dy Qx dx Rxy Gx where P, Q, R, and G are continuous functions. Equations of this type arise

### CHAPTER 2 FOURIER SERIES

CHAPTER 2 FOURIER SERIES PERIODIC FUNCTIONS A function is said to have a period T if for all x,, where T is a positive constant. The least value of T>0 is called the period of. EXAMPLES We know that =

### Chapter Usual types of questions Tips What can go ugly 1 Algebraic Fractions

C3 Cheat Sheet Last Updated: 9 th Nov 2015 Chapter Usual types of questions Tips What can go ugly 1 Algebraic Fractions Almost always adding or subtracting fractions. Factorise everything in each fraction

### Definition of Vertical Asymptote The line x = a is called a vertical asymptote of f (x) if at least one of the following is true: f (x) =

Vertical Asymptotes Definition of Vertical Asymptote The line x = a is called a vertical asymptote of f (x) if at least one of the following is true: lim f (x) = x a lim f (x) = lim x a lim f (x) = x a

### MATH 425, PRACTICE FINAL EXAM SOLUTIONS.

MATH 45, PRACTICE FINAL EXAM SOLUTIONS. Exercise. a Is the operator L defined on smooth functions of x, y by L u := u xx + cosu linear? b Does the answer change if we replace the operator L by the operator

### Chapter 5: Trigonometric Functions of Real Numbers

Chapter 5: Trigonometric Functions of Real Numbers 5.1 The Unit Circle The unit circle is the circle of radius 1 centered at the origin. Its equation is x + y = 1 Example: The point P (x, 1 ) is on the

### Recognizing Types of First Order Differential Equations E. L. Lady

Recognizing Types of First Order Differential Equations E. L. Lady Every first order differential equation to be considered here can be written can be written in the form P (x, y)+q(x, y)y =0. This means

### MATH 132: CALCULUS II SYLLABUS

MATH 32: CALCULUS II SYLLABUS Prerequisites: Successful completion of Math 3 (or its equivalent elsewhere). Math 27 is normally not a sufficient prerequisite for Math 32. Required Text: Calculus: Early

### SOLUTIONS TO HOMEWORK ASSIGNMENT #5, Math 253

SOLUTIONS TO HOMEWORK ASSIGNMENT #5, Math 53. For what values of the constant k does the function f(x, y) =kx 3 + x +y 4x 4y have (a) no critical points; (b) exactly one critical point; (c) exactly two

### M3 PRECALCULUS PACKET 1 FOR UNIT 5 SECTIONS 5.1 TO = to see another form of this identity.

M3 PRECALCULUS PACKET FOR UNIT 5 SECTIONS 5. TO 5.3 5. USING FUNDAMENTAL IDENTITIES 5. Part : Pythagorean Identities. Recall the Pythagorean Identity sin θ cos θ + =. a. Subtract cos θ from both sides

### Finding Antiderivatives and Evaluating Integrals

Chapter 5 Finding Antiderivatives and Evaluating Integrals 5. Constructing Accurate Graphs of Antiderivatives Motivating Questions In this section, we strive to understand the ideas generated by the following

### IB Mathematics SL :: Checklist. Topic 1 Algebra The aim of this topic is to introduce students to some basic algebraic concepts and applications.

Topic 1 Algebra The aim of this topic is to introduce students to some basic algebraic concepts and applications. Check Topic 1 Book Arithmetic sequences and series; Sum of finite arithmetic series; Geometric

### EXAM. Practice Questions for Exam #2. Math 3350, Spring April 3, 2004 ANSWERS

EXAM Practice Questions for Exam #2 Math 3350, Spring 2004 April 3, 2004 ANSWERS i Problem 1. Find the general solution. A. D 3 (D 2)(D 3) 2 y = 0. The characteristic polynomial is λ 3 (λ 2)(λ 3) 2. Thus,

### South Carolina College- and Career-Ready (SCCCR) Pre-Calculus

South Carolina College- and Career-Ready (SCCCR) Pre-Calculus Key Concepts Arithmetic with Polynomials and Rational Expressions PC.AAPR.2 PC.AAPR.3 PC.AAPR.4 PC.AAPR.5 PC.AAPR.6 PC.AAPR.7 Standards Know

### (x) = lim. x 0 x. (2.1)

Differentiation. Derivative of function Let us fi an arbitrarily chosen point in the domain of the function y = f(). Increasing this fied value by we obtain the value of independent variable +. The value

### Quotient Rings and Field Extensions

Chapter 5 Quotient Rings and Field Extensions In this chapter we describe a method for producing field extension of a given field. If F is a field, then a field extension is a field K that contains F.

### 1. Introduction identity algbriac factoring identities

1. Introduction An identity is an equality relationship between two mathematical expressions. For example, in basic algebra students are expected to master various algbriac factoring identities such as

### CALCULUS 2. 0 Repetition. tutorials 2015/ Find limits of the following sequences or prove that they are divergent.

CALCULUS tutorials 5/6 Repetition. Find limits of the following sequences or prove that they are divergent. a n = n( ) n, a n = n 3 7 n 5 n +, a n = ( n n 4n + 7 ), a n = n3 5n + 3 4n 7 3n, 3 ( ) 3n 6n

### Tangent and normal lines to conics

4.B. Tangent and normal lines to conics Apollonius work on conics includes a study of tangent and normal lines to these curves. The purpose of this document is to relate his approaches to the modern viewpoints

### Complex Numbers and the Complex Exponential

Complex Numbers and the Complex Exponential Frank R. Kschischang The Edward S. Rogers Sr. Department of Electrical and Computer Engineering University of Toronto September 5, 2005 Numbers and Equations

### MATHEMATICS (CLASSES XI XII)

MATHEMATICS (CLASSES XI XII) General Guidelines (i) All concepts/identities must be illustrated by situational examples. (ii) The language of word problems must be clear, simple and unambiguous. (iii)

### 1.2. Mathematical Models: A Catalog of Essential Functions

1.2. Mathematical Models: A Catalog of Essential Functions Mathematical model A mathematical model is a mathematical description (often by means of a function or an equation) of a real-world phenomenon

### The graph of. horizontal line between-1 and 1 that the sine function is not 1-1 and therefore does not have an inverse.

Inverse Trigonometric Functions The graph of If we look at the graph of we can see that if you draw a horizontal line between-1 and 1 that the sine function is not 1-1 and therefore does not have an inverse.

### Core Maths C3. Revision Notes

Core Maths C Revision Notes October 0 Core Maths C Algebraic fractions... Cancelling common factors... Multipling and dividing fractions... Adding and subtracting fractions... Equations... 4 Functions...

### Lesson A - Natural Exponential Function and Natural Logarithm Functions

A- Lesson A - Natural Exponential Function and Natural Logarithm Functions Natural Exponential Function In Lesson 2, we explored the world of logarithms in base 0. The natural logarithm has a base of e.

### CHAPTER 6: RATIONAL NUMBERS AND ORDERED FIELDS

CHAPTER 6: RATIONAL NUMBERS AND ORDERED FIELDS LECTURE NOTES FOR MATH 378 (CSUSM, SPRING 2009). WAYNE AITKEN 1. Introduction In this chapter we construct the set of rational numbers Q using equivalence

### APPLICATIONS OF DIFFERENTIATION

4 APPLICATIONS OF DIFFERENTIATION APPLICATIONS OF DIFFERENTIATION So far, we have been concerned with some particular aspects of curve sketching: Domain, range, and symmetry (Chapter 1) Limits, continuity,

### Trigonometry provides methods to relate angles and lengths but the functions we define have many other applications in mathematics.

MATH19730 Part 1 Trigonometry Section2 Trigonometry provides methods to relate angles and lengths but the functions we define have many other applications in mathematics. An angle can be measured in degrees

### Trigonometric Identities

Trigonometric Identities Dr. Philippe B. Laval Kennesaw STate University April 0, 005 Abstract This handout dpresents some of the most useful trigonometric identities. It also explains how to derive new

### Section 0.4 Inverse Trigonometric Functions

Section 0.4 Inverse Trigonometric Functions Short Recall from Trigonometry Definition: A function f is periodic of period T if f(x + T ) = f(x) for all x such that x and x+t are in the domain of f. The

### Fourier Series Chapter 3 of Coleman

Fourier Series Chapter 3 of Coleman Dr. Doreen De eon Math 18, Spring 14 1 Introduction Section 3.1 of Coleman The Fourier series takes its name from Joseph Fourier (1768-183), who made important contributions

### MTH 233 Calculus I. Part 1: Introduction, Limits and Continuity

MTH 233 Calculus I Tan, Single Variable Calculus: Early Transcendentals, 1st ed. Part 1: Introduction, Limits and Continuity 0 Preliminaries It is assumed that students are comfortable with the material

### Lies My Calculator and Computer Told Me

Lies My Calculator and Computer Told Me 2 LIES MY CALCULATOR AND COMPUTER TOLD ME Lies My Calculator and Computer Told Me See Section.4 for a discussion of graphing calculators and computers with graphing

### SAT Subject Math Level 2 Facts & Formulas

Numbers, Sequences, Factors Integers:..., -3, -2, -1, 0, 1, 2, 3,... Reals: integers plus fractions, decimals, and irrationals ( 2, 3, π, etc.) Order Of Operations: Arithmetic Sequences: PEMDAS (Parentheses

### 5.1 Radical Notation and Rational Exponents

Section 5.1 Radical Notation and Rational Exponents 1 5.1 Radical Notation and Rational Exponents We now review how exponents can be used to describe not only powers (such as 5 2 and 2 3 ), but also roots

### 1. Periodic Fourier series. The Fourier expansion of a 2π-periodic function f is:

CONVERGENCE OF FOURIER SERIES 1. Periodic Fourier series. The Fourier expansion of a 2π-periodic function f is: with coefficients given by: a n = 1 π f(x) a 0 2 + a n cos(nx) + b n sin(nx), n 1 f(x) cos(nx)dx