SECOND-ORDER LINEAR DIFFERENTIAL EQUATIONS
|
|
- Violet Stafford
- 7 years ago
- Views:
Transcription
1 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 in the study of the motion of a spring. In Additional Topics: Applications of Second-Order Differential Equations we will further pursue this application as well as the application to electric circuits. In this section we study the case where Gx 0, for all x, in Equation 1. Such equations are called homogeneous linear equations. Thus, the form of a second-order linear homogeneous differential equation is Px d y dx dy Qx dx Rxy 0 If Gx 0 for some x, Equation 1 is nonhomogeneous and is discussed in Additional Topics: Nonhomogeneous Linear Equations. Two basic facts enable us to solve homogeneous linear equations. The first of these says that if we know two solutions y 1 and y of such an equation, then the linear combination y c 1 y 1 c y is also a solution. 3 Theorem If y 1 x and y x are both solutions of the linear homogeneous equation () and and are any constants, then the function c 1 c yx c 1 y 1 x c y x is also a solution of Equation. y 1 y Proof Since and are solutions of Equation, we have Pxy 1 Qxy 1 Rxy 1 0 and Pxy Qxy Rxy 0 Therefore, using the basic rules for differentiation, we have Pxy Qxy Rxy Pxc 1 y 1 c y Qxc 1 y 1 c y Rxc 1 y 1 c y Pxc 1 y 1 c y Qxc 1 y 1 c y Rxc 1 y 1 c y c 1 Pxy 1 Qxy 1 Rxy 1 c Pxy Qxy Rxy c 1 0 c 0 0 Thus, y c 1 y 1 c y is a solution of Equation. The other fact we need is given by the following theorem, which is proved in more advanced courses. It says that the general solution is a linear combination of two linearly independent solutions y 1 and y. This means that neither y 1 nor y is a constant multiple of the other. For instance, the functions f x x and tx 5x are linearly dependent, but f x e x and tx xe x are linearly independent. 1
2 SECOND-ORDER LINEAR DIFFERENTIAL EQUATIONS 4 Theorem If y 1 and y are linearly independent solutions of Equation, and Px is never 0, then the general solution is given by yx c 1 y 1 x c y x c 1 c where and are arbitrary constants. Theorem 4 is very useful because it says that if we know two particular linearly independent solutions, then we know every solution. In general, it is not easy to discover particular solutions to a second-order linear equation. But it is always possible to do so if the coefficient functions P, Q, and R are constant functions, that is, if the differential equation has the form 5 ay by cy 0 where a, b, and c are constants and a 0. It s not hard to think of some likely candidates for particular solutions of Equation 5 if we state the equation verbally. We are looking for a function y such that a constant times its second derivative y plus another constant times y plus a third constant times y is equal to 0. We know that the exponential function y e rx (where r is a constant) has the property that its derivative is a constant multiple of itself: y re rx. Furthermore, y r e rx. If we substitute these expressions into Equation 5, we see that y e rx is a solution if ar e rx bre rx ce rx 0 or ar br ce rx 0 But e rx is never 0. Thus, y e rx is a solution of Equation 5 if r is a root of the equation 6 ar br c 0 Equation 6 is called the auxiliary equation (or characteristic equation) of the differential equation ay by cy 0. Notice that it is an algebraic equation that is obtained from the differential equation by replacing y by r, y by r, and y by 1. Sometimes the roots r 1 and r of the auxiliary equation can be found by factoring. In other cases they are found by using the quadratic formula: 7 r 1 b sb 4ac a r b sb 4ac a We distinguish three cases according to the sign of the discriminant b 4ac. CASE I b 4ac 0 In this case the roots r and of the auxiliary equation are real and distinct, so y 1 e r1x 1 r and are two linearly independent solutions of Equation 5. (Note that e r x y e r x is not a constant multiple of e r1x.) Therefore, by Theorem 4, we have the following fact. 8 If the roots r and of the auxiliary equation ar 1 r br c 0 are real and unequal, then the general solution of ay by cy 0 is y c 1 e r1x c e r x
3 SECOND-ORDER LINEAR DIFFERENTIAL EQUATIONS 3 In Figure 1 the graphs of the basic solutions f x e x and tx e 3x of the differential equation in Example 1 are shown in black and red, respectively. Some of the other solutions, linear combinations of f and t, are shown in blue. 8 5f+g f+5g f+g f g _1 1 g-f f-g EXAMPLE 1 Solve the equation y y 6y 0. SOLUTION The auxiliary equation is r r 6 r r 3 0 whose roots are r, 3. Therefore, by (8) the general solution of the given differential equation is y c 1 e x c e 3x We could verify that this is indeed a solution by differentiating and substituting into the differential equation. FIGURE 1 _5 EXAMPLE Solve 3 d y. dx dy dx y 0 SOLUTION To solve the auxiliary equation 3r r 1 0 we use the quadratic formula: r 1 s13 6 Since the roots are real and distinct, the general solution is y c 1 e (1s13 )x6 (1s13 )x6 c e CASE II b 4ac 0 In this case r 1 r ; that is, the roots of the auxiliary equation are real and equal. Let s denote by r the common value of and r. Then, from Equations 7, we have r 1 9 r b a so ar b 0 We know that y is one solution of Equation 5. We now verify that y xe rx 1 e rx is also a solution: ay by cy are rx r xe rx be rx rxe rx cxe rx ar be rx ar br cxe rx 0e rx 0xe rx 0 The first term is 0 by Equations 9; the second term is 0 because r is a root of the auxiliary equation. Since y and y xe rx 1 e rx are linearly independent solutions, Theorem 4 provides us with the general solution. 10 If the auxiliary equation ar br c 0 has only one real root r, then the general solution of ay by cy 0 is y c 1 e rx c xe rx EXAMPLE 3 Solve the equation 4y 1y 9y 0. SOLUTION The auxiliary equation 4r 1r 9 0 can be factored as r 3 0
4 4 SECOND-ORDER LINEAR DIFFERENTIAL EQUATIONS Figure shows the basic solutions f x e 3x and tx xe 3x in Example 3 and some other members of the family of solutions. Notice that all of them approach 0 as x l. f FIGURE f-g 8 _5 5f+g f+5g _ f+g g-f g so the only root is r 3. By (10) the general solution is CASE III b 4ac 0 In this case the roots r 1 and r of the auxiliary equation are complex numbers. (See Additional Topics: Complex Numbers for information about complex numbers.) We can write y c 1 e 3x c xe 3x r 1 i r i ba where and are real numbers. [In fact,,.] Then, using Euler s equation e i cos i sin s4ac b a from Additional Topics: Complex Numbers, we write the solution of the differential equation as y C 1 e r1x C e r x C 1 e ix C e ix C 1 e x cos x i sin x C e x cos x i sin x e x C 1 C cos x ic 1 C sin x e x c 1 cos x c sin x where c 1 C 1 C, c ic 1 C. This gives all solutions (real or complex) of the differential equation. The solutions are real when the constants c 1 and c are real. We summarize the discussion as follows. 11 If the roots of the auxiliary equation ar br c 0 are the complex numbers r 1 i, r i, then the general solution of ay by cy 0 is y e x c 1 cos x c sin x Figure 3 shows the graphs of the solutions in Example 4, f x e 3x cos x and tx e 3x sin x, together with some linear combinations. All solutions approach 0 as x l. f+g f-g f _3 3 g EXAMPLE 4 Solve the equation y 6y 13y 0. SOLUTION The auxiliary equation is r 6r By the quadratic formula, the roots are r 6 s s16 By (11) the general solution of the differential equation is y e 3x c 1 cos x c sin x 3 i FIGURE 3 _3 INITIAL-VALUE AND BOUNDARY-VALUE PROBLEMS An initial-value problem for the second-order Equation 1 or consists of finding a solution y of the differential equation that also satisfies initial conditions of the form yx 0 y 0 yx 0 y 1 where y 0 and y 1 are given constants. If P, Q, R, and G are continuous on an interval and Px 0 there, then a theorem found in more advanced books guarantees the existence and uniqueness of a solution to this initial-value problem. Examples 5 and 6 illustrate the technique for solving such a problem.
5 SECOND-ORDER LINEAR DIFFERENTIAL EQUATIONS 5 EXAMPLE 5 Solve the initial-value problem Figure 4 shows the graph of the solution of the initial-value problem in Example 5. Compare with Figure 1. 0 y y 6y 0 y0 1 y0 0 SOLUTION From Example 1 we know that the general solution of the differential equation is yx c 1 e x c e 3x Differentiating this solution, we get yx c 1 e x 3c e 3x To satisfy the initial conditions we require that _ 0 FIGURE 4 1 y0 c 1 c 1 13 y0 c 1 3c 0 From (13) we have c 3c 1 and so (1) gives c 1 3c 1 1 c Thus, the required solution of the initial-value problem is c 5 y 3 5e x 5 e 3x The solution to Example 6 is graphed in Figure 5. It appears to be a shifted sine curve and, indeed, you can verify that another way of writing the solution is y s13 sinx where tan 3 _π 5 π EXAMPLE 6 Solve the initial-value problem y y 0 y0 y0 3 SOLUTION The auxiliary equation is r 1 0, or r 1, whose roots are i. Thus,, and since e 0x 1, the general solution is 0 1 yx c 1 cos x c sin x Since yx c 1 sin x c cos x the initial conditions become y0 c 1 y0 c 3 FIGURE 5 _5 Therefore, the solution of the initial-value problem is yx cos x 3sin x A boundary-value problem for Equation 1 consists of finding a solution y of the differential equation that also satisfies boundary conditions of the form yx 0 y 0 yx 1 y 1 In contrast with the situation for initial-value problems, a boundary-value problem does not always have a solution. EXAMPLE 7 Solve the boundary-value problem y y y 0 y0 1 y1 3 SOLUTION The auxiliary equation is r r 1 0 or r 1 0 whose only root is r 1. Therefore, the general solution is yx c 1 e x c xe x
6 6 SECOND-ORDER LINEAR DIFFERENTIAL EQUATIONS Figure 6 shows the graph of the solution of the boundary-value problem in Example 7. 5 _1 5 _5 FIGURE 6 The boundary conditions are satisfied if y0 c 1 1 y1 c 1 e 1 c e 1 3 The first condition gives c 1 1, so the second condition becomes e1 c e1 3 Solving this equation for c by first multiplying through by e, we get 1 c 3e so c 3e 1 Thus, the solution of the boundary-value problem is Summary: Solutions of ay by c 0 y e x 3e 1xe x Roots of ar br c 0 r 1, r real and distinct r 1 r r r 1, r complex: i General solution y c 1 e r1x c e r x y c 1 e rx c xe rx y e x c 1 cos x c sin x EXERCISES 1 13 Solve the differential equation. 1. y 6y 8y 0. y 4y 8y 0 3. y 8y 41y 0 4. y y y 0 5. y y y y 5y 7. 4y y y 4y 9y y y y 4y d y d y 1. dt 6 dy 4y 0 dt dy y 0 dt dt 13. ; Graph the two basic solutions of the differential equation and several other solutions. What features do the solutions have in common? d y 15. dx dy dx y A d y dt Click here for answers. d y dx dy dt y 0 dy dx 5y Solve the initial-value problem. 17. y 5y 3y 0, y0 3, 18. y 3y 0, y0 1, y0 3 d y dx y y 4y y 0, y0 1, y0 1.5 S Click here for solutions. 8 dy dx 16y 0 0. y 5y 3y 0, y0 1, 1. y 16y 0, y4 3,. y y 5y 0, y 0, 3. y y y 0, y0, 4. y 1y 36y 0, y1 0, 5 3 Solve the boundary-value problem, if possible. 5. 4y y 0, y0 3, 6. y y 0, y0 1, 7. y 3y y 0, y0 1, 8. y 100y 0, y0, 9. y 6y 5y 0, y0 1, 30. y 6y 9y 0, y0 1, 31. y 4y 13y 0, y0, 3. 9y 18y 10y 0, y0 0, 33. Let L be a nonzero real number. (a) Show that the boundary-value problem y y 0, y0 0, yl 0 has only the trivial solution y 0 for the cases and. (b) For the case, find the values of for which this problem has a nontrivial solution and give the corresponding solution y 4 y1 y4 4 y y0 1 y3 0 y 5 y0 4 y1 1 y y1 0 y 1 y If a, b, and c are all positive constants and yx is a solution of the differential equation ay by cy 0, show that lim x l yx 0.
7 SECOND-ORDER LINEAR DIFFERENTIAL EQUATIONS 7 ANSWERS S Click here for solutions. 1. y c 1e 4x c e x 3. y e4x c 1 cos 5x c sin 5x 5. y c 1e x c xe x 7. y c 1 cosx c sinx 9. y c 1 c e x4 11. y c 1e (1s )t (1s )t c e 13. y e t [c 1 cos(s3t) c sin(s3t)] g f _0. 1 _40 All solutions approach 0 as x l and approach as x l. 17. y e 3x e x 19. y e x/ xe x 1. y 3 cos 4x sin 4x 3. y e x cos x 3 sin x y 3 cos( 1 x) 4 sin( 1 x) y e x3 e No solution 31. y e x cos 3x e 33. (b), n a positive integer; y C sinnxl n L e x 1 e 3 sin 3x
8 8 SECOND-ORDER LINEAR DIFFERENTIAL EQUATIONS SOLUTIONS 1. The auxiliary equation is r 6r +8=0 (r 4)(r ) = 0 r =4, r =. Then by (8) the general solution is y = c 1e 4x + c e x. 3. The auxiliary equation is r +8r +41=0 r = 4 ± 5i. Then by (11) the general solution is y = e 4x (c 1 cos 5x + c sin 5x). 5. The auxiliary equation is r r +1=(r 1) =0 r =1. Then by (10), the general solution is y = c 1e x + c xe x. 7. The auxiliary equation is 4r +1=0 r = ± 1 i,soy = c 1 cos 1 x + c sin 1 x. 9. The auxiliary equation is 4r + r = r(4r +1)=0 r =0, r = 1 4,soy = c1 + ce x/ The auxiliary equation is r r 1=0 r =1±,soy = c 1e (1+ )t + c e (1 )t. 13. The auxiliary equation is r + r +1=0 r = 1 ± c i,soy = e t/ 1 cos t + c sin. t 15. r 8r +16=(r 4) =0so y = c 1e 4x + c xe 4x. The graphs are all asymptotic to the x-axis as x, and as x the solutions tend to ±. 17. r +5r +3=(r +3)(r +1)=0,sor = 3, r = 1 and the general solution is y = c 1e 3x/ + c e x.then y(0) = 3 c 1 + c =3and y 0 (0) = 4 3 c 1 c = 4,soc 1 =and c =1. Thus the solution to the initial-value problem is y =e 3x/ + e x r 4r +1=(r 1) =0 r = 1 and the general solution is y = c1ex/ + c xe x/.theny(0) = 1 c 1 =1and y 0 (0) = c1 + c = 1.5,soc = and the solution to the initial-value problem is y = e x/ xe x/. 1. r +16=0 r = ±4i and the general solution is y = e 0x (c 1 cos 4x + c sin 4x) =c 1 cos 4x + c sin 4x. Then y π 4 = 3 c1 = 3 c 1 =3and y 0 π 4 =4 4c =4 c = 1,sothe solution to the initial-value problem is y =3cos4x sin 4x. 3. r +r +=0 r = 1 ± i and the general solution is y = e x (c 1 cos x + c sin x). Then=y(0) = c 1 and 1=y 0 (0) = c c 1 c =3and the solution to the initial-value problem is y = e x ( cos x +3sinx). 5. 4r +1=0 r = ± 1 i and the general solution is y = c1 cos 1 x + c sin 1 x.then3=y(0) = c 1 and 4 =y(π) =c, so the solution of the boundary-value problem is y =3cos 1 x 4sin 1 x. 7. r 3r +=(r )(r 1) = 0 r =1, r =and the general solution is y = c 1 e x + c e x.then 1=y(0) = c 1 + c and 0=y(3) = c 1 e 3 + c e 6 so c =1/(1 e 3 ) and c 1 = e 3 /(e 3 1). The solution of the boundary-value problem is y = ex+3 e ex 1 e. 3
9 SECOND-ORDER LINEAR DIFFERENTIAL EQUATIONS 9 9. r 6r +5=0 r =3± 4i and the general solution is y = e 3x (c 1 cos 4x + c sin 4x). But1=y(0) = c 1 and =y(π) =c 1e 3π c 1 =/e 3π, so there is no solution. 31. r +4r +13=0 r = ± 3i and the general solution is y = e x (c 1 cos 3x + c sin 3x). But =y(0) = c 1 and 1=y π = e π ( c ), so the solution to the boundary-value problem is y = e x ( cos 3x e π sin 3x). 33. (a) Case 1 (λ =0): y 00 + λy =0 y 00 =0which has an auxiliary equation r =0 r =0 y = c 1 + c x where y(0) = 0 and y(l) =0. Thus, 0=y(0) = c 1 and 0=y(L) =c L c 1 = c =0. Thus, y =0. Case (λ <0): y 00 + λy =0has auxiliary equation r = λ r = ± λ (distinct and real since λ<0) y = c 1 e λx + c e λx where y(0) = 0 and y(l) =0.Thus,0=y(0) = c 1 + c ( )and 0=y(L) =c 1e λl + c e λl ( ). Multiplying ( )bye λl andsubtracting( )givesc e λl e λl c 1 =0from ( ). Thus, y =0for the cases λ =0and λ<0. =0 c =0and thus (b) y 00 + λy =0has an auxiliary equation r + λ =0 r = ±i λ y = c 1 cos λx+ c sin λx where y(0) = 0 and y(l) =0.Thus,0=y(0) = c 1 and 0=y(L) =c sin λl since c 1 =0.Sincewe cannot have a trivial solution, c 6= 0and thus sin λl =0 λl = nπ where n is an integer λ = n π /L and y = c sin(nπx/l) where n is an integer.
Second-Order Linear Differential Equations
Second-Order Linear Differential Equations A second-order linear differential equation has the form 1 Px d 2 y dx 2 dy Qx dx Rxy Gx where P, Q, R, and G are continuous functions. We saw in Section 7.1
More informationNonhomogeneous Linear Equations
Nonhomogeneous Linear Equations In this section we learn how to solve second-order nonhomogeneous linear differential equations with constant coefficients, that is, equations of the form ay by cy G x where
More informationA Brief Review of Elementary Ordinary Differential Equations
1 A Brief Review of Elementary Ordinary Differential Equations At various points in the material we will be covering, we will need to recall and use material normally covered in an elementary course on
More informationMath 2280 - Assignment 6
Math 2280 - Assignment 6 Dylan Zwick Spring 2014 Section 3.8-1, 3, 5, 8, 13 Section 4.1-1, 2, 13, 15, 22 Section 4.2-1, 10, 19, 28 1 Section 3.8 - Endpoint Problems and Eigenvalues 3.8.1 For the eigenvalue
More informationCHAPTER 2. Eigenvalue Problems (EVP s) for ODE s
A SERIES OF CLASS NOTES FOR 005-006 TO INTRODUCE LINEAR AND NONLINEAR PROBLEMS TO ENGINEERS, SCIENTISTS, AND APPLIED MATHEMATICIANS DE CLASS NOTES 4 A COLLECTION OF HANDOUTS ON PARTIAL DIFFERENTIAL EQUATIONS
More informationtegrals as General & Particular Solutions
tegrals as General & Particular Solutions dy dx = f(x) General Solution: y(x) = f(x) dx + C Particular Solution: dy dx = f(x), y(x 0) = y 0 Examples: 1) dy dx = (x 2)2 ;y(2) = 1; 2) dy ;y(0) = 0; 3) dx
More informationSecond Order Linear Differential Equations
CHAPTER 2 Second Order Linear Differential Equations 2.. Homogeneous Equations A differential equation is a relation involving variables x y y y. A solution is a function f x such that the substitution
More informationDifferentiation 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
More informationLS.6 Solution Matrices
LS.6 Solution Matrices In the literature, solutions to linear systems often are expressed using square matrices rather than vectors. You need to get used to the terminology. As before, we state the definitions
More information1. 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
More informationChapter 4. Linear Second Order Equations. ay + by + cy = 0, (1) where a, b, c are constants. The associated auxiliary equation is., r 2 = b b 2 4ac 2a
Chapter 4. Linear Second Order Equations ay + by + cy = 0, (1) where a, b, c are constants. ar 2 + br + c = 0. (2) Consequently, y = e rx is a solution to (1) if an only if r satisfies (2). So, the equation
More informationOn closed-form solutions to a class of ordinary differential equations
International Journal of Advanced Mathematical Sciences, 2 (1 (2014 57-70 c Science Publishing Corporation www.sciencepubco.com/index.php/ijams doi: 10.14419/ijams.v2i1.1556 Research Paper On closed-form
More informationSecond 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
More informationHomework #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
More informationGeneral Theory of Differential Equations Sections 2.8, 3.1-3.2, 4.1
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 General Theory of Differential Equations Sections 2.8, 3.1-3.2, 4.1 Dr. John Ehrke Department of Mathematics Fall 2012 Questions
More informationLimits 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
More informationThe Method of Partial Fractions Math 121 Calculus II Spring 2015
Rational functions. as The Method of Partial Fractions Math 11 Calculus II Spring 015 Recall that a rational function is a quotient of two polynomials such f(x) g(x) = 3x5 + x 3 + 16x x 60. The method
More informationAn Introduction to Partial Differential Equations
An Introduction to Partial Differential Equations Andrew J. Bernoff LECTURE 2 Cooling of a Hot Bar: The Diffusion Equation 2.1. Outline of Lecture An Introduction to Heat Flow Derivation of the Diffusion
More information19.7. Applications of Differential Equations. Introduction. Prerequisites. Learning Outcomes. Learning Style
Applications of Differential Equations 19.7 Introduction Blocks 19.2 to 19.6 have introduced several techniques for solving commonly-occurring firstorder and second-order ordinary differential equations.
More informationHomework #2 Solutions
MAT Spring Problems Section.:, 8,, 4, 8 Section.5:,,, 4,, 6 Extra Problem # Homework # Solutions... Sketch likely solution curves through the given slope field for dy dx = x + y...8. Sketch likely solution
More informationSeparable First Order Differential Equations
Separable First Order Differential Equations Form of Separable Equations which take the form = gx hy or These are differential equations = gxĥy, where gx is a continuous function of x and hy is a continuously
More informationZeros of a Polynomial Function
Zeros of a Polynomial Function An important consequence of the Factor Theorem is that finding the zeros of a polynomial is really the same thing as factoring it into linear factors. In this section we
More informationSecond Order Linear Nonhomogeneous Differential Equations; Method of Undetermined Coefficients. y + p(t) y + q(t) y = g(t), g(t) 0.
Second Order Linear Nonhomogeneous Differential Equations; Method of Undetermined Coefficients We will now turn our attention to nonhomogeneous second order linear equations, equations with the standard
More information4.5 Linear Dependence and Linear Independence
4.5 Linear Dependence and Linear Independence 267 32. {v 1, v 2 }, where v 1, v 2 are collinear vectors in R 3. 33. Prove that if S and S are subsets of a vector space V such that S is a subset of S, then
More informationMATH 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
More information2 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
More informationReducibility of Second Order Differential Operators with Rational Coefficients
Reducibility of Second Order Differential Operators with Rational Coefficients Joseph Geisbauer University of Arkansas-Fort Smith Advisor: Dr. Jill Guerra May 10, 2007 1. INTRODUCTION In this paper we
More information2.2 Separable Equations
2.2 Separable Equations 73 2.2 Separable Equations An equation y = f(x, y) is called separable provided algebraic operations, usually multiplication, division and factorization, allow it to be written
More informationTOPIC 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
More informationDifferential Equations I MATB44H3F. Version September 15, 2011-1949
Differential Equations I MATB44H3F Version September 15, 211-1949 ii Contents 1 Introduction 1 11 Preliminaries 1 12 Sample Application of Differential Equations 2 2 First Order Ordinary Differential Equations
More informationr (t) = 2r(t) + sin t θ (t) = r(t) θ(t) + 1 = 1 1 θ(t) 1 9.4.4 Write the given system in matrix form x = Ax + f ( ) sin(t) x y 1 0 5 z = dy cos(t)
Solutions HW 9.4.2 Write the given system in matrix form x = Ax + f r (t) = 2r(t) + sin t θ (t) = r(t) θ(t) + We write this as ( ) r (t) θ (t) = ( ) ( ) 2 r(t) θ(t) + ( ) sin(t) 9.4.4 Write the given system
More informationHigher Order Equations
Higher Order Equations We briefly consider how what we have done with order two equations generalizes to higher order linear equations. Fortunately, the generalization is very straightforward: 1. Theory.
More information4.3 Lagrange Approximation
206 CHAP. 4 INTERPOLATION AND POLYNOMIAL APPROXIMATION Lagrange Polynomial Approximation 4.3 Lagrange Approximation Interpolation means to estimate a missing function value by taking a weighted average
More informationHomework # 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
More information3.2 The Factor Theorem and The Remainder Theorem
3. The Factor Theorem and The Remainder Theorem 57 3. The Factor Theorem and The Remainder Theorem Suppose we wish to find the zeros of f(x) = x 3 + 4x 5x 4. Setting f(x) = 0 results in the polynomial
More information19.6. Finding a Particular Integral. Introduction. Prerequisites. Learning Outcomes. Learning Style
Finding a Particular Integral 19.6 Introduction We stated in Block 19.5 that the general solution of an inhomogeneous equation is the sum of the complementary function and a particular integral. We have
More informationJUST THE MATHS UNIT NUMBER 1.8. ALGEBRA 8 (Polynomials) A.J.Hobson
JUST THE MATHS UNIT NUMBER 1.8 ALGEBRA 8 (Polynomials) by A.J.Hobson 1.8.1 The factor theorem 1.8.2 Application to quadratic and cubic expressions 1.8.3 Cubic equations 1.8.4 Long division of polynomials
More informationTaylor and Maclaurin Series
Taylor and Maclaurin Series In the preceding section we were able to find power series representations for a certain restricted class of functions. Here we investigate more general problems: Which functions
More informationCOMPLEX NUMBERS. a bi c di a c b d i. a bi c di a c b d i For instance, 1 i 4 7i 1 4 1 7 i 5 6i
COMPLEX NUMBERS _4+i _-i FIGURE Complex numbers as points in the Arg plane i _i +i -i A complex number can be represented by an expression of the form a bi, where a b are real numbers i is a symbol with
More informationCoffeyville Community College #MATH 202 COURSE SYLLABUS FOR DIFFERENTIAL EQUATIONS. Ryan Willis Instructor
Coffeyville Community College #MATH 202 COURSE SYLLABUS FOR DIFFERENTIAL EQUATIONS Ryan Willis Instructor COURSE NUMBER: MATH 202 COURSE TITLE: Differential Equations CREDIT HOURS: 3 INSTRUCTOR: OFFICE
More informationChapter 20. Vector Spaces and Bases
Chapter 20. Vector Spaces and Bases In this course, we have proceeded step-by-step through low-dimensional Linear Algebra. We have looked at lines, planes, hyperplanes, and have seen that there is no limit
More informationClass Meeting # 1: Introduction to PDEs
MATH 18.152 COURSE NOTES - CLASS MEETING # 1 18.152 Introduction to PDEs, Fall 2011 Professor: Jared Speck Class Meeting # 1: Introduction to PDEs 1. What is a PDE? We will be studying functions u = u(x
More informationx 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
More informationLinearly Independent Sets and Linearly Dependent Sets
These notes closely follow the presentation of the material given in David C. Lay s textbook Linear Algebra and its Applications (3rd edition). These notes are intended primarily for in-class presentation
More informationCalculus 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.
More informationNumerical Methods for Solving Systems of Nonlinear Equations
Numerical Methods for Solving Systems of Nonlinear Equations by Courtney Remani A project submitted to the Department of Mathematical Sciences in conformity with the requirements for Math 4301 Honour s
More informationcorrect-choice plot f(x) and draw an approximate tangent line at x = a and use geometry to estimate its slope comment The choices were:
Topic 1 2.1 mode MultipleSelection text How can we approximate the slope of the tangent line to f(x) at a point x = a? This is a Multiple selection question, so you need to check all of the answers that
More information1.7. Partial Fractions. 1.7.1. Rational Functions and Partial Fractions. A rational function is a quotient of two polynomials: R(x) = P (x) Q(x).
.7. PRTIL FRCTIONS 3.7. Partial Fractions.7.. Rational Functions and Partial Fractions. rational function is a quotient of two polynomials: R(x) = P (x) Q(x). Here we discuss how to integrate rational
More informationCHAPTER 17 CALCULUS. Based on the original work by. George B. Thomas, Jr. Massachusetts Institute of Technology. as revised by
CHAPTER 17 THOMAS CALCULUS Twelfth Edition Based on the iginal wk by Gege B. Thomas, Jr. Massachusetts Institute of Technology as revised by Maurice D. Weir Naval Postgraduate School Joel Hass University
More informationby the matrix A results in a vector which is a reflection of the given
Eigenvalues & Eigenvectors Example Suppose Then So, geometrically, multiplying a vector in by the matrix A results in a vector which is a reflection of the given vector about the y-axis We observe that
More information1 Homework 1. [p 0 q i+j +... + p i 1 q j+1 ] + [p i q j ] + [p i+1 q j 1 +... + p i+j q 0 ]
1 Homework 1 (1) Prove the ideal (3,x) is a maximal ideal in Z[x]. SOLUTION: Suppose we expand this ideal by including another generator polynomial, P / (3, x). Write P = n + x Q with n an integer not
More informationZeros of Polynomial Functions
Zeros of Polynomial Functions The Rational Zero Theorem If f (x) = a n x n + a n-1 x n-1 + + a 1 x + a 0 has integer coefficients and p/q (where p/q is reduced) is a rational zero, then p is a factor of
More informationSECOND DERIVATIVE TEST FOR CONSTRAINED EXTREMA
SECOND DERIVATIVE TEST FOR CONSTRAINED EXTREMA This handout presents the second derivative test for a local extrema of a Lagrange multiplier problem. The Section 1 presents a geometric motivation for the
More informationTechniques 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
More informationThe Heat Equation. Lectures INF2320 p. 1/88
The Heat Equation Lectures INF232 p. 1/88 Lectures INF232 p. 2/88 The Heat Equation We study the heat equation: u t = u xx for x (,1), t >, (1) u(,t) = u(1,t) = for t >, (2) u(x,) = f(x) for x (,1), (3)
More informationContinued Fractions and the Euclidean Algorithm
Continued Fractions and the Euclidean Algorithm Lecture notes prepared for MATH 326, Spring 997 Department of Mathematics and Statistics University at Albany William F Hammond Table of Contents Introduction
More informationa 1 x + a 0 =0. (3) ax 2 + bx + c =0. (4)
ROOTS OF POLYNOMIAL EQUATIONS In this unit we discuss polynomial equations. A polynomial in x of degree n, where n 0 is an integer, is an expression of the form P n (x) =a n x n + a n 1 x n 1 + + a 1 x
More informationMATH 381 HOMEWORK 2 SOLUTIONS
MATH 38 HOMEWORK SOLUTIONS Question (p.86 #8). If g(x)[e y e y ] is harmonic, g() =,g () =, find g(x). Let f(x, y) = g(x)[e y e y ].Then Since f(x, y) is harmonic, f + f = and we require x y f x = g (x)[e
More informationSolutions 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
More informationA First Course in Elementary Differential Equations. Marcel B. Finan Arkansas Tech University c All Rights Reserved
A First Course in Elementary Differential Equations Marcel B. Finan Arkansas Tech University c All Rights Reserved 1 Contents 1 Basic Terminology 4 2 Qualitative Analysis: Direction Field of y = f(t, y)
More informationSection 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
More information100. In general, we can define this as if b x = a then x = log b
Exponents and Logarithms Review 1. Solving exponential equations: Solve : a)8 x = 4! x! 3 b)3 x+1 + 9 x = 18 c)3x 3 = 1 3. Recall: Terminology of Logarithms If 10 x = 100 then of course, x =. However,
More informationQuotient 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.
More informationLinear Algebra Notes for Marsden and Tromba Vector Calculus
Linear Algebra Notes for Marsden and Tromba Vector Calculus n-dimensional Euclidean Space and Matrices Definition of n space As was learned in Math b, a point in Euclidean three space can be thought of
More informationC. Complex Numbers. 1. Complex arithmetic.
C. Complex Numbers. Complex arithmetic. Most people think that complex numbers arose from attempts to solve quadratic equations, but actually it was in connection with cubic equations they first appeared.
More informationMore Quadratic Equations
More Quadratic Equations Math 99 N1 Chapter 8 1 Quadratic Equations We won t discuss quadratic inequalities. Quadratic equations are equations where the unknown appears raised to second power, and, possibly
More informationFind all of the real numbers x that satisfy the algebraic equation:
Appendix C: Factoring Algebraic Expressions Factoring algebraic equations is the reverse of expanding algebraic expressions discussed in Appendix B. Factoring algebraic equations can be a great help when
More information3.2 Sources, Sinks, Saddles, and Spirals
3.2. Sources, Sinks, Saddles, and Spirals 6 3.2 Sources, Sinks, Saddles, and Spirals The pictures in this section show solutions to Ay 00 C By 0 C Cy D 0. These are linear equations with constant coefficients
More informationCore Maths C1. Revision Notes
Core Maths C Revision Notes November 0 Core Maths C Algebra... Indices... Rules of indices... Surds... 4 Simplifying surds... 4 Rationalising the denominator... 4 Quadratic functions... 4 Completing the
More informationPRACTICE FINAL. Problem 1. Find the dimensions of the isosceles triangle with largest area that can be inscribed in a circle of radius 10cm.
PRACTICE FINAL Problem 1. Find the dimensions of the isosceles triangle with largest area that can be inscribed in a circle of radius 1cm. Solution. Let x be the distance between the center of the circle
More informationLecture 3 : The Natural Exponential Function: f(x) = exp(x) = e x. y = exp(x) if and only if x = ln(y)
Lecture 3 : The Natural Exponential Function: f(x) = exp(x) = Last day, we saw that the function f(x) = ln x is one-to-one, with domain (, ) and range (, ). We can conclude that f(x) has an inverse function
More informationy 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.
More informationMATH 304 Linear Algebra Lecture 20: Inner product spaces. Orthogonal sets.
MATH 304 Linear Algebra Lecture 20: Inner product spaces. Orthogonal sets. Norm The notion of norm generalizes the notion of length of a vector in R n. Definition. Let V be a vector space. A function α
More information3.3. Solving Polynomial Equations. Introduction. Prerequisites. Learning Outcomes
Solving Polynomial Equations 3.3 Introduction Linear and quadratic equations, dealt within Sections 3.1 and 3.2, are members of a class of equations, called polynomial equations. These have the general
More informationMethods of Solution of Selected Differential Equations Carol A. Edwards Chandler-Gilbert Community College
Methods of Solution of Selected Differential Equations Carol A. Edwards Chandler-Gilbert Community College Equations of Order One: Mdx + Ndy = 0 1. Separate variables. 2. M, N homogeneous of same degree:
More information1 Lecture: Integration of rational functions by decomposition
Lecture: Integration of rational functions by decomposition into partial fractions Recognize and integrate basic rational functions, except when the denominator is a power of an irreducible quadratic.
More informationSolving Cubic Polynomials
Solving Cubic Polynomials 1.1 The general solution to the quadratic equation There are four steps to finding the zeroes of a quadratic polynomial. 1. First divide by the leading term, making the polynomial
More informationwww.mathsbox.org.uk ab = c a If the coefficients a,b and c are real then either α and β are real or α and β are complex conjugates
Further Pure Summary Notes. Roots of Quadratic Equations For a quadratic equation ax + bx + c = 0 with roots α and β Sum of the roots Product of roots a + b = b a ab = c a If the coefficients a,b and c
More informationAn important theme in this book is to give constructive definitions of mathematical objects. Thus, for instance, if you needed to evaluate.
Chapter 10 Series and Approximations An important theme in this book is to give constructive definitions of mathematical objects. Thus, for instance, if you needed to evaluate 1 0 e x2 dx, you could set
More informationThe 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
More informationu dx + y = 0 z x z x = x + y + 2 + 2 = 0 6) 2
DIFFERENTIAL EQUATIONS 6 Many physical problems, when formulated in mathematical forms, lead to differential equations. Differential equations enter naturally as models for many phenomena in economics,
More informationDECOMPOSING SL 2 (R)
DECOMPOSING SL 2 R KEITH CONRAD Introduction The group SL 2 R is not easy to visualize: it naturally lies in M 2 R, which is 4- dimensional the entries of a variable 2 2 real matrix are 4 free parameters
More information1 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
More informationLinear algebra and the geometry of quadratic equations. Similarity transformations and orthogonal matrices
MATH 30 Differential Equations Spring 006 Linear algebra and the geometry of quadratic equations Similarity transformations and orthogonal matrices First, some things to recall from linear algebra Two
More informationInner product. Definition of inner product
Math 20F Linear Algebra Lecture 25 1 Inner product Review: Definition of inner product. Slide 1 Norm and distance. Orthogonal vectors. Orthogonal complement. Orthogonal basis. Definition of inner product
More informationRecognizing 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
More informationThis unit will lay the groundwork for later units where the students will extend this knowledge to quadratic and exponential functions.
Algebra I Overview View unit yearlong overview here Many of the concepts presented in Algebra I are progressions of concepts that were introduced in grades 6 through 8. The content presented in this course
More informationTo differentiate logarithmic functions with bases other than e, use
To ifferentiate logarithmic functions with bases other than e, use 1 1 To ifferentiate logarithmic functions with bases other than e, use log b m = ln m ln b 1 To ifferentiate logarithmic functions with
More informationLecture 5 Principal Minors and the Hessian
Lecture 5 Principal Minors and the Hessian Eivind Eriksen BI Norwegian School of Management Department of Economics October 01, 2010 Eivind Eriksen (BI Dept of Economics) Lecture 5 Principal Minors and
More informationSMT 2014 Algebra Test Solutions February 15, 2014
1. Alice and Bob are painting a house. If Alice and Bob do not take any breaks, they will finish painting the house in 20 hours. If, however, Bob stops painting once the house is half-finished, then the
More informationPYTHAGOREAN TRIPLES KEITH CONRAD
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
More informationZero: If P is a polynomial and if c is a number such that P (c) = 0 then c is a zero of P.
MATH 11011 FINDING REAL ZEROS KSU OF A POLYNOMIAL Definitions: Polynomial: is a function of the form P (x) = a n x n + a n 1 x n 1 + + a x + a 1 x + a 0. The numbers a n, a n 1,..., a 1, a 0 are called
More informationRepresentation of functions as power series
Representation of functions as power series Dr. Philippe B. Laval Kennesaw State University November 9, 008 Abstract This document is a summary of the theory and techniques used to represent functions
More informationAutonomous Equations / Stability of Equilibrium Solutions. y = f (y).
Autonomous Equations / Stabilit of Equilibrium Solutions First order autonomous equations, Equilibrium solutions, Stabilit, Longterm behavior of solutions, direction fields, Population dnamics and logistic
More informationNumerical Solution of Differential Equations
Numerical Solution of Differential Equations Dr. Alvaro Islas Applications of Calculus I Spring 2008 We live in a world in constant change We live in a world in constant change We live in a world in constant
More informationApplications of Second-Order Differential Equations
Applications of Second-Order Differential Equations Second-order linear differential equations have a variety of applications in science and engineering. In this section we explore two of them: the vibration
More informationDiscrete Mathematics: Homework 7 solution. Due: 2011.6.03
EE 2060 Discrete Mathematics spring 2011 Discrete Mathematics: Homework 7 solution Due: 2011.6.03 1. Let a n = 2 n + 5 3 n for n = 0, 1, 2,... (a) (2%) Find a 0, a 1, a 2, a 3 and a 4. (b) (2%) Show that
More informationSolving 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).
More informationy 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.
More informationTHE COMPLEX EXPONENTIAL FUNCTION
Math 307 THE COMPLEX EXPONENTIAL FUNCTION (These notes assume you are already familiar with the basic properties of complex numbers.) We make the following definition e iθ = cos θ + i sin θ. (1) This formula
More information