Worksheet # 2: Higher Order Linear ODEs (SOLUTIONS)
|
|
- Evelyn Hensley
- 5 years ago
- Views:
Transcription
1 Name: November 8, 011 Worksheet # : Higher Order Linear ODEs (SOLUTIONS) 1. A set of n-functions f 1, f,..., f n are linearly independent on an interval I if the only way that c 1 f 1 (t) + c f (t) c n f n (t) = 0 holds for all t in I is if c 1 = c =... = c n = 0. Otherwise, they are linearly dependent. (a) Prove that y 1 (t) = t, y (t) = 3t + t + 1, y 3 (t) = t + 1, y 4 = sin t are linearly dependent on (, ) by finding a combination with c 1,..., c 4 not all zero so that for all t. c 1 y 1 (t) + c y (t) + c 3 y 3 (t) + c 4 y(t) = 0 Solution. Set c 1 = 3, c = 1, c 3 = 1 and c 4 = 0, since 3(t ) + (3t + t + 1) (t + 1) + 0(sin t) = 0 for all t. Note there are other possible choices. (b) There is a Wronskian test for verifying that a set of functions is linearly independent. For 3 twice-differentiable functions, the Wronskian is y 1 (t) y (t) y 3 (t) W (y 1, y, y 3 )(t) = y 1(t) y (t) y 3(t) y 1 (t) y (t) y 3 (t) (The Wronskian for 4 or more functions is defined in a similar way (see page 1) but requires computation of determinant larger than 3 3). If W (y 1, y, y 3 )(t) 0 for some point t in I, then the functions must be linearly independent. If, on the other hand, the functions are linearly dependent then the Wronskian must be 0 for all t in I. i. Use the Wronskian to verify that y 1 = x, y = x and y 3 = x 3 are linearly independent on (, ). Solution. x x x 3 W (y 1, y, y 3 )(x) = 1 x 3x 0 6x = x(1x 6x ) x (6x) + x 3 () = x 3. Since this expression is not zero for example at x = 1 the functions are linearly independent. ii. The functions y 1 (t) = t + t, y (t) = t t, and y 3 (t) = t are linearly dependent on (, ). Find c 1, c, c 3 not all zero so that c 1 y 1 (t) + c y (t) + c 3 y 3 (t) = 0 holds for all t. Verify that the Wronskian is 0 for all t. Solution. Let c 1 =, c = 1, and c 3 = 3/. We have (t + t) + 1(t t) + 3 (t) = t t + t t + 3t = 0.
2 Therefore, the functions are linearly dependent. The Wronskian must be zero. We check that t + t t t t W (y 1, y, y 3 )(t) = t + 1 4t = (t + t)( 8) (t t)( 4) + t(4(t + 1) (4t 1)) = 8t 8t + 8t 4t + t(8t + 4 8t + ) = 1t + 1t = 0. Consider the homogeneous third order differential equation L[y] = y (3) + p(t)y + q(t)y + r(t)y = 0 Assume that p, q, r are all continuous functions on an open interval I. The general solution to this equation is y = c 1 y 1 + c y + c 3 y 3 where y 1, y, and y 3 are three times differentiable, linearly independent functions which are also solutions to the ODE. The solutions y 1, y, y 3 form a fundamental set. There is again a Wronskian test for linearly independence of solutions: y 1, y, y 3 are linearly independent solutions to the ODE if and only if y 1 (t) y (t) y 3 (t) W (y 1, y, y 3 )(t) = y 1(t) y (t) y 3(t) y 1 (t) y (t) y 3 (t) 0 for some point t in I. (a) Let p(t) = 6, q(t) = 11, and r(t) = 6. Verify that y 1 (t) = e t, y (t) = e t, y 3 (t) = e 3t are linearly independent solutions to the ODE. Construct the general solution and find the constants c 1, c, c 3 so that the initial conditions hold. Solution. The differential equation is Note that y(0) = 1, y (0) = 1, y (0) = 1 L[y] = y 6y + 11y 6y = 0 L[e rt ] = (r 3 6r + 11r 6)e rt so that if y = e rt is a solution r must be a root of the cubic equation This equation factors as r 3 6r + 11r 6 = 0 (r 1)(r )(r 3) = 0 There are three, distinct real roots r = 1,, 3. Each of these gives a solution to the differential equation: y 1 (t) = e t, y (t) = e t, and y 3 (t) = e 3t. We must verify that these are linearly independent. Compute the Wronskian: e t e t e 3t W (y 1, y, y 3 )(t) = e t e t 3e 3t e t 4e t 9e 3t = e t (18e 5t 1e 5t ) e t (9e 4t 3e 4t ) + e 3t (4e 3t e 3t ) = 6e 6t 6e 6t + e 6t = e 6t 0
3 So the solutions are linearly independent. The general solution is y(t) = c 1 e t + c e t + c 3 e 3t The first and second derivatives of this solution are y (t) = c 1 e t + c e t + 3c 3 e 3t y (t) = c 1 e t + 4c e t + 9c 3 e 3t To satisfy the initial conditions at t = 0 we need: 1 = c 1 + c + c 3 1 = c 1 + c + 3c 3 1 = c 1 + 4c + 9c 3 There are many ways to solve this system of algebraic equations. One easy way is to subtract the first equation from the third: 0 = 3c + 8c 3 So c = 8 3 c 3. We can then subtract the second equation from the first to get = c c 3 Substituting the c = 8 3 c 3: = 3 c 3 c 3 = 3 Then and plugging into the first equation c = 8 1 = c c 1 = 6. Therefore, the unique solution satisfying these initial conditions is y = 6e t 8e t + 3e 3t 4.1: #11 Verify that 1, cos t, sin t are solutions to y + y = 0 and compute their Wronskian. Solution. By substituting these functions into the equation, it is clear that each is a solution. We compute the Wronskian 1 cos t sin t 0 sin t cos t 0 cos t sin t = 1(sin t + cos t ) = : # 15 Verify that 1, x, x 3 are solutions to and compute their Wronskian. xy xy = 0
4 Solution. By substituting these functions into the equation, it is clear that each is a solution. For example x(x 3 ) x(x 3 ) = 3x(x ) 3x(x ) = 6x(x) 6x = 0. We compute the Wronskian 1 x x x 0 0 6x = 6x. 3. Consider the constant coefficient homogeneous, 3rd order ODE ay + by + cy + dy = 0. (a) Substitute y = e rt into the ODE to obtain a characteristic equation. You can always factor this equation into the form a(r r 1 )(r r )(r r 3 ) = 0. List all possible cases for the roots of r. For example, can this equation have three complex roots? One complex and two real? Repeated roots? Answer. The important fact is that complex roots of polynomial equations (with real coefficients) can only occur in complex conjugate pairs. So you cannot have three complex roots or one complex root. Since the roots are conjugates, complex roots in this case can t be repeated. The cases are I. Three real and distinct roots. II. Three real roots, but only two distinct. That is r 1, r, r 3 real with, say r = r 3. III. Three real roots, all repeated. IV. One real root and a complex conjugate pair of roots. (b) If the roots are all real and different, prove that y 1 = e r1t, y = e rt, y 3 = e r3t form a fundamental set. Solution. Use the operator notation If y = e rt, then it is clear that L[y] = ay + by + cy + dy. L[y] = (ar 3 + br + cr + d)e rt and so L[y] = 0 if and only if r is one of r 1, r, r 3. Thus y 1, y, y 3 are solutions. The Wronskian is e r 1t e r t e r 3t r 1 e r1t r e rt r 3 e r3t r 1e r 1t re r t r3e r 3t = (r r3 rr 3 )e (r1+r+r3)t (r 1 r3 r1r 3 )e (r1+r+r3)t + (r 1 r r 1r )e (r 1+r +r 3 )t = (r r 3 r r 3 r 1 r 3 r 1r 3 + r 1 r r 1r )e (r1+r+r3)t = (r r 1 )(r 3 r 1 )(r 3 r )e (r 1+r +r 3 )t This expression is clearly not zero since all of r 1, r and r 3 are distinct. Therefore, the solutions are linearly independent. (c) If the equation has one repeated real root r, prove that y 1 = e rt, y = te rt, y 3 = t e rt form a fundamental set.
5 Solution. You can verify exactly as above that y 1 (t) = e rt is a solution. Compute L[y ] = a(te rt ) + b(te rt ) + c(te rt ) + dte rt = a(e rt + rte tr ) + b(e rt + rte tr ) + c(e rt + rte tr ) + dte rt = a(re rt + re tr + r te rt ) + b(re rt + re tr + r te rt ) + c(e rt + rte rt ) + dte rt = a(3r e rt + r 3 te rt ) + b(r + r t)e rt + c(1 + rt)e rt + dte rt = (ar 3 + br + cr + d)te rt + (3ar + br + c)e rt Now clearly ar 3 + br + cr + d = 0. I claim the second term is also zero. To see why, notice that ax 3 + bx + cx + d = a(x r) 3 Expanding the right hand side we see that ax 3 + bx + cx + d = a x 3 3 a r x + 3 a r x a r 3. Matching coefficients, we have that b = 3ar, c = 3ar, d = ar 3. Therefore and it follows that 3ar + br + c = 0 L[y ] = 0. The same trick will work to show that L[y 3 ] = 0 and y 1, y, y 3 are all solutions. We compute the Wronskian and carefully simplify to obtain e rt te rt t e rt re rt (1 + rt)e rt (t + rt )e rt r e rt r( + rt))e rt ( + 4rt + r t )e rt = e3rt 0 So the solutions y 1, y, y 3 are linearly independent on (, ) and hence form a fundamental set for the ODE. (d) Complex roots always occur in conjugate pairs λ ± µi. A complex conjugate pair gives rise to solutions y 1 (t) = e λt cos µt and y (t) = e λt sin µt. What is the fundamental solution set in this case? Be sure to verify that the solutions in your fundamental set are linearly independent. Solution. The characteristic equation has one real root, call it r 1, and a complex conjugate pair of roots r,3 = µ ± λt. The fundamental set in this case is y 1 (t) = e rt, y (t) = e λt cos µt, y 3 (t) = e λt sin µt. It is clear that y 1 (t) is a solution by the arguments of the previous problem(s). You do not have to check that y (t), y 3 (t) are solutions, since it is a given in the statement of the problem.
6 If you want to see why they are though, substitute y (t) into the equation and simplify to obtain: L[y ] = ( (a µ 3 + ( 3 a λ b λ c ) µ ) sin (µ t) + ( ( 3 a λ b) µ + a λ 3 + b λ + c λ + d ) cos (µ t) ) e λ t On the other hand, you can plug r back in to the characteristic equation and collect real and imaginary parts to see that 0 = ar 3 + br + cr + d = ( ( 3 a λ b) µ + a λ 3 + b λ + c λ + d ) + (( 3 a λ + b λ + c ) µ a µ 3) i. For this to hold, it must follow that ( 3 a λ + b λ + c ) µ a µ 3 = 0 ( 3 a λ b) µ + a λ 3 + b λ + c λ + d = 0 and y must be solution. A similar argument works for y 3. The Wronskian to expand is: e r t cos (µ t) e t λ sin (µ t) e t λ r e r t ( (cos (µ t) λ µ sin (µ t)) e t λ (sin (µ t) λ + µ cos (µ t)) e t λ r e r t cos (µ t) λ µ sin (µ t) λ µ cos (µ t) ) ( e t λ sin (µ t) λ + µ cos (µ t) λ µ sin (µ t) ) e t λ If you compute things correctly, you should find that W (y 1, y, y 3 )(t) = µ(λ rλ + r + µ )e tλ+rt Can this quantity ever be 0? Well, the exponential part is never 0 and µ 0, since r, r 3 must have an imaginary part. This leaves the possibility that λ r λ + r + µ = 0. But this is a quadratic equation in r and if you solve it, you will find that the roots are precisely r = µ ± λi. This cannot occur since by assumption r must be the real root. We therefore may conclude that W (y 1, y, y 3 )(t) 0 and so the solutions are linearly independent on (, ). Therefore, y 1, y, y 3 form a fundamental set. (e) Now provide the fundamental solution sets for all other cases you described in part (a). Solutions. The only remaining case is (II). Here we have the root r 1 and the twice repeated root r and these lead to the fundamental set y 1 (t) = e r 1t, y (t) = te r t, and y 3 (t) = te r 3t. You can verify that this is a fundamental set as we did above but I do not require you to do this. (f) 4.: # 9: Solve y + y = 0 y(0) = 0 y (0) = 1 y (0) =
7 Solution. The characteristic equation is r 3 + r = r(r + 1) = 0. The roots of this equation are Thus the general solution is r 1 = 0 r,3 = ±i. y(t) = c 1 + c cos t + c 3 sin t. Compute y (t) and y (t): y (t) = c sin t + c 3 cos t y (t) = c cos t c 3 sin t Using the initial conditions, we obtain the equations c 1 + c = 0 c 3 = 1 c = So c 1 =, c =, and c 3 = 1 The solution is y(t) = cos t sin t (g) Problem 4. #3 : Solve y y + y y = 0 y(0) = y (0) = 1 y (0) = Solution. The characteristic equation is r 3 r + r 1 = 0. There are a number of ways you can find the roots of simple polynomial equations like this. One method is to guess a root, e.g. r = 1, verify that its a root, then do polynomial division to break the polynomial into a linear term and a quadratic term. Here r = 1 is a root and doing polynomial division, we factor this equation as (r 1)(r + 1) = 0. So the roots are r 1 = 1 and r,3 = ±i. The general solution is thus The derivatives are y(t) = c 1 e t + c cos t + c 3 sin t. y (t) = c 1 e t c sin t + c 3 cos t y (t) = c 1 e t c cos t c 3 sin t. Using the initial conditions, we obtain the coefficient equations c 1 + c = c 1 + c 3 = 1 c 1 c = If you add the first and last equation, you obtain c 1 = 0 so that c = by the first equation and c 3 = 1 by the second. The solution is y(t) = cos t sin t
8 4. Consider the the constant coefficient homogeneous n-th order ODE: a 0 y (n) + a 1 y (n 1) a n 1 y + a n y = 0. By substituting in y = e rt you will obtain a characteristic equation a 0 r n + a 1 r n a n 1 r + a n = a 0 (r r 1 )(r r ) (r r n ) = 0. There are many cases for the roots here and it can be extraordinarily difficult to factor higher order polynomial equations 1. You should use software to find roots for higher order cases. By finding all the roots you can generate a fundamental set of solutions exactly as we did in the second order and third order cases. Combinations of distinct real roots and complex roots work the same way as in the previous problem. For higher order equations you can have many repeated roots and also complex repeated roots. Consult the text page 8 to see how to handle these scenarios. Use these ideas to solve the following problems from the text. (a) 4.: # 19: Find the general solution to Solution. The characteristic equation is y (5) 3y (4) + 3y 3y + y = 0. r 5 3r 4 + 3r 3 3r + r = r(r 4 3r 3 + 3r 3r + ) = 0 It is clear that r = 0 is one root. You could use Maxima to further factor the polynomial or guess roots and do polynomial division. Obtain r(r 1)(r )(r + 1) = 0 so that the characteristic equation has roots r = 0, 1,, ±i. The general solution is y(t) = c 1 + c e t + c 3 e t + c 4 cos t + c 5 sin t. (b) 4.: # 1: Find the general solution of Solutions. The characteristic equation is y (8) + 8y (4) + 16y = 0. r 8 + 8r = 0 Again you might use software to factor this but I indicate a different approach. Let u = r 4. But you may also be able factor this by hand: r 8 + 8r = (r 4 + 4) = ( (r r + )(r + r + ) ) = 0 The roots of the first quadratic term are r 1, = 1 ± i and the roots of the second are r 3,4 = 1 ± i. Since the quadratic factors are squared, each of these roots is repeated. This gives the general solution y(t) = e t (c 1 cos t + c sin t) + e t (c 3 cos t + c 4 sin t) + e t t (c 5 cos t + c 6 sin t) + e t t (c 7 cos t + c 8 sin t) 1 There are formulas analogous to the quadratic formula which find all the roots of cubic and quartic polynomials. They are very complicated and it is worth a Google search to see just how involved the quartic formula is. An amazing fact is that it is impossible to write down general closed formulas for the roots of 5th and higher degree polynomials. It is not just that none have been found but that it cannot be done!
9 (c) 4.: # 30: Solve y (4) + y = 0 y(0) = 0 y (0) = 0 y (0) = 1 y (0) = 0 Solution. The characteristic equation is The roots of this equation are r = 0 r 1, = 1 ± 1 i r 3,4 = 1 ± 1 i The general solution is ( ( y(t) = e t 1 c 1 cos t ) ( )) ( ( ) ( )) 1 + c sin t + e t 1 1 c 3 cos t + c 4 sin t. So y(0) = c 1 + c 3 = 0 Compute the first through third derivative of y and evaluate at t = 0. You can do this by hand or with a computer and use the initial conditions to obtain the equations c 1 + c 3 = 0 c 1 + c c 3 + c 4 = 0 c c + 4 = 1 c 1 + c + c 3 + c 4 = 0 One way to solve this system is to add the second and fourth equations to get c + c 4 = 0 c + c 4 = 0 Adding the third equation to this, you ll find that c = 1 and then that c 4 = 1. The fourth equation is now c 1 + c 3 = 0 and using the first equation it follows that c 1 = c 3 = 0. So the solution is ) (e 1 t y(t) = e 1 t ( ) t sin.
Exam II Review: Selected Solutions and Answers
November 9, 2011 Exam II Review: Selected Solutions and Answers NOTE: For additional worked problems see last year s review sheet and answers, the notes from class, and your text. Answers to problems from
More information6. Linear Differential Equations of the Second Order
September 26, 2012 6-1 6. Linear Differential Equations of the Second Order A differential equation of the form L(y) = g is called linear if L is a linear operator and g = g(t) is continuous. The most
More informationSecond Order Linear Equations
Second Order Linear Equations Linear Equations The most general linear ordinary differential equation of order two has the form, a t y t b t y t c t y t f t. 1 We call this a linear equation because the
More informationLinear Homogeneous ODEs of the Second Order with Constant Coefficients. Reduction of Order
Linear Homogeneous ODEs of the Second Order with Constant Coefficients. Reduction of Order October 2 6, 2017 Second Order ODEs (cont.) Consider where a, b, and c are real numbers ay +by +cy = 0, (1) Let
More informationChapter 4: Higher Order Linear Equations
Chapter 4: Higher Order Linear Equations MATH 351 California State University, Northridge April 7, 2014 MATH 351 (Differential Equations) Ch 4 April 7, 2014 1 / 11 Sec. 4.1: General Theory of nth Order
More informationSection 9.8 Higher Order Linear Equations
Section 9.8 Higher Order Linear Equations Key Terms: Higher order linear equations Equivalent linear systems for higher order equations Companion matrix Characteristic polynomial and equation A linear
More information144 Chapter 3. Second Order Linear Equations
144 Chapter 3. Second Order Linear Equations PROBLEMS In each of Problems 1 through 8 find the general solution of the given differential equation. 1. y + 2y 3y = 0 2. y + 3y + 2y = 0 3. 6y y y = 0 4.
More informationDifferential Equations Practice: 2nd Order Linear: Nonhomogeneous Equations: Undetermined Coefficients Page 1
Differential Equations Practice: 2nd Order Linear: Nonhomogeneous Equations: Undetermined Coefficients Page 1 Questions Example (3.5.3) Find a general solution of the differential equation y 2y 3y = 3te
More informationSecond Order Linear Equations
October 13, 2016 1 Second And Higher Order Linear Equations In first part of this chapter, we consider second order linear ordinary linear equations, i.e., a differential equation of the form L[y] = d
More informationLinear algebra and differential equations (Math 54): Lecture 20
Linear algebra and differential equations (Math 54): Lecture 20 Vivek Shende April 7, 2016 Hello and welcome to class! Last time We started discussing differential equations. We found a complete set of
More informationSecond-Order Linear ODEs
Second-Order Linear ODEs A second order ODE is called linear if it can be written as y + p(t)y + q(t)y = r(t). (0.1) It is called homogeneous if r(t) = 0, and nonhomogeneous otherwise. We shall assume
More informationLinear Second Order ODEs
Chapter 3 Linear Second Order ODEs In this chapter we study ODEs of the form (3.1) y + p(t)y + q(t)y = f(t), where p, q, and f are given functions. Since there are two derivatives, we might expect that
More informationExam Basics. midterm. 1 There will be 9 questions. 2 The first 3 are on pre-midterm material. 3 The next 1 is a mix of old and new material.
Exam Basics 1 There will be 9 questions. 2 The first 3 are on pre-midterm material. 3 The next 1 is a mix of old and new material. 4 The last 5 questions will be on new material since the midterm. 5 60
More informationHomogeneous Equations with Constant Coefficients
Homogeneous Equations with Constant Coefficients MATH 365 Ordinary Differential Equations J. Robert Buchanan Department of Mathematics Spring 2018 General Second Order ODE Second order ODEs have the form
More informationChapter 13: General Solutions to Homogeneous Linear Differential Equations
Worked Solutions 1 Chapter 13: General Solutions to Homogeneous Linear Differential Equations 13.2 a. Verifying that {y 1, y 2 } is a fundamental solution set: We have y 1 (x) = cos(2x) y 1 (x) = 2 sin(2x)
More informationLecture Notes for Math 251: ODE and PDE. Lecture 13: 3.4 Repeated Roots and Reduction Of Order
Lecture Notes for Math 251: ODE and PDE. Lecture 13: 3.4 Repeated Roots and Reduction Of Order Shawn D. Ryan Spring 2012 1 Repeated Roots of the Characteristic Equation and Reduction of Order Last Time:
More informationSecond Order Differential Equations Lecture 6
Second Order Differential Equations Lecture 6 Dibyajyoti Deb 6.1. Outline of Lecture Repeated Roots; Reduction of Order Nonhomogeneous Equations; Method of Undetermined Coefficients Variation of Parameters
More informationSection 3.1 Second Order Linear Homogeneous DEs with Constant Coefficients
Section 3. Second Order Linear Homogeneous DEs with Constant Coefficients Key Terms/ Ideas: Initial Value Problems Homogeneous DEs with Constant Coefficients Characteristic equation Linear DEs of second
More informationHigher Order Linear Equations Lecture 7
Higher Order Linear Equations Lecture 7 Dibyajyoti Deb 7.1. Outline of Lecture General Theory of nth Order Linear Equations. Homogeneous Equations with Constant Coefficients. 7.2. General Theory of nth
More informationMath 3313: Differential Equations Second-order ordinary differential equations
Math 3313: Differential Equations Second-order ordinary differential equations Thomas W. Carr Department of Mathematics Southern Methodist University Dallas, TX Outline Mass-spring & Newton s 2nd law Properties
More informationHomework #6 Solutions
Problems Section.1: 6, 4, 40, 46 Section.:, 8, 10, 14, 18, 4, 0 Homework #6 Solutions.1.6. Determine whether the functions f (x) = cos x + sin x and g(x) = cos x sin x are linearly dependent or linearly
More informationMATH 308 Differential Equations
MATH 308 Differential Equations Summer, 2014, SET 6 JoungDong Kim Set 6: Section 3.3, 3.4, 3.5, 3.6 Section 3.3 Complex Roots of the Characteristic Equation Recall that a second order ODE with constant
More informationMath 215/255: Elementary Differential Equations I Harish N Dixit, Department of Mathematics, UBC
Math 215/255: Elementary Differential Equations I Harish N Dixit, Department of Mathematics, UBC First Order Equations Linear Equations y + p(x)y = q(x) Write the equation in the standard form, Calculate
More informationA: Brief Review of Ordinary Differential Equations
A: Brief Review of Ordinary Differential Equations Because of Principle # 1 mentioned in the Opening Remarks section, you should review your notes from your ordinary differential equations (odes) course
More informationWhat if the characteristic equation has a double root?
MA 360 Lecture 17 - Summary of Recurrence Relations Friday, November 30, 018. Objectives: Prove basic facts about basic recurrence relations. Last time, we looked at the relational formula for a sequence
More informationMATH 308 Differential Equations
MATH 308 Differential Equations Summer, 2014, SET 5 JoungDong Kim Set 5: Section 3.1, 3.2 Chapter 3. Second Order Linear Equations. Section 3.1 Homogeneous Equations with Constant Coefficients. In this
More informationDIFFERENTIAL EQUATIONS REVIEW. Here are notes to special make-up discussion 35 on November 21, in case you couldn t make it.
DIFFERENTIAL EQUATIONS REVIEW PEYAM RYAN TABRIZIAN Here are notes to special make-up discussion 35 on November 21, in case you couldn t make it. Welcome to the special Friday after-school special of That
More informationCalculus IV - HW 3. Due 7/ Give the general solution to the following differential equations: y = c 1 e 5t + c 2 e 5t. y = c 1 e 2t + c 2 e 4t.
Calculus IV - HW 3 Due 7/13 Section 3.1 1. Give the general solution to the following differential equations: a y 25y = 0 Solution: The characteristic equation is r 2 25 = r 5r + 5. It follows that the
More informationLecture 10 - Second order linear differential equations
Lecture 10 - Second order linear differential equations In the first part of the course, we studied differential equations of the general form: = f(t, y) In other words, is equal to some expression involving
More informationStudy guide - Math 220
Study guide - Math 220 November 28, 2012 1 Exam I 1.1 Linear Equations An equation is linear, if in the form y + p(t)y = q(t). Introducing the integrating factor µ(t) = e p(t)dt the solutions is then in
More informationMA 266 Review Topics - Exam # 2 (updated)
MA 66 Reiew Topics - Exam # updated Spring First Order Differential Equations Separable, st Order Linear, Homogeneous, Exact Second Order Linear Homogeneous with Equations Constant Coefficients The differential
More informationNonhomogeneous Equations and Variation of Parameters
Nonhomogeneous Equations Variation of Parameters June 17, 2016 1 Nonhomogeneous Equations 1.1 Review of First Order Equations If we look at a first order homogeneous constant coefficient ordinary differential
More informationMath53: Ordinary Differential Equations Autumn 2004
Math53: Ordinary Differential Equations Autumn 2004 Unit 2 Summary Second- and Higher-Order Ordinary Differential Equations Extremely Important: Euler s formula Very Important: finding solutions to linear
More informationChapter 3. Reading assignment: In this chapter we will cover Sections dx 1 + a 0(x)y(x) = g(x). (1)
Chapter 3 3 Introduction Reading assignment: In this chapter we will cover Sections 3.1 3.6. 3.1 Theory of Linear Equations Recall that an nth order Linear ODE is an equation that can be written in the
More informationLINEAR EQUATIONS OF HIGHER ORDER. EXAMPLES. General framework
Differential Equations Grinshpan LINEAR EQUATIONS OF HIGHER ORDER. EXAMPLES. We consider linear ODE of order n: General framework (1) x (n) (t) + P n 1 (t)x (n 1) (t) + + P 1 (t)x (t) + P 0 (t)x(t) = 0
More informationMATH 307 Introduction to Differential Equations Autumn 2017 Midterm Exam Monday November
MATH 307 Introduction to Differential Equations Autumn 2017 Midterm Exam Monday November 6 2017 Name: Student ID Number: I understand it is against the rules to cheat or engage in other academic misconduct
More informationMath 2142 Homework 5 Part 1 Solutions
Math 2142 Homework 5 Part 1 Solutions Problem 1. For the following homogeneous second order differential equations, give the general solution and the particular solution satisfying the given initial conditions.
More information1. Why don t we have to worry about absolute values in the general form for first order differential equations with constant coefficients?
1. Why don t we have to worry about absolute values in the general form for first order differential equations with constant coefficients? Let y = ay b with y(0) = y 0 We can solve this as follows y =
More informationMath 4B Notes. Written by Victoria Kala SH 6432u Office Hours: T 12:45 1:45pm Last updated 7/24/2016
Math 4B Notes Written by Victoria Kala vtkala@math.ucsb.edu SH 6432u Office Hours: T 2:45 :45pm Last updated 7/24/206 Classification of Differential Equations The order of a differential equation is the
More informationChapter 3: Second Order Equations
Exam 2 Review This review sheet contains this cover page (a checklist of topics from Chapters 3). Following by all the review material posted pertaining to chapter 3 (all combined into one file). Chapter
More informationFirst and Second Order Differential Equations Lecture 4
First and Second Order Differential Equations Lecture 4 Dibyajyoti Deb 4.1. Outline of Lecture The Existence and the Uniqueness Theorem Homogeneous Equations with Constant Coefficients 4.2. The Existence
More informationHigher-order differential equations
Higher-order differential equations Peyam Tabrizian Wednesday, November 16th, 2011 This handout is meant to give you a couple more example of all the techniques discussed in chapter 6, to counterbalance
More informationReview of Lecture 9 Existence and Uniqueness
Review of Lecture 9 Existence and Uniqueness We consider y = f (x, y) with a given initial condition y(x 0 ) = y 0. There is a solution passing through (x 0, y 0 ). It is defined on some interval (a, b)
More informationSecond order linear equations
Second order linear equations Samy Tindel Purdue University Differential equations - MA 266 Taken from Elementary differential equations by Boyce and DiPrima Samy T. Second order equations Differential
More informationLinear algebra and differential equations (Math 54): Lecture 19
Linear algebra and differential equations (Math 54): Lecture 19 Vivek Shende April 5, 2016 Hello and welcome to class! Previously We have discussed linear algebra. This time We start studying differential
More informationMAT187H1F Lec0101 Burbulla
Spring 2017 Second Order Linear Homogeneous Differential Equation DE: A(x) d 2 y dx 2 + B(x)dy dx + C(x)y = 0 This equation is called second order because it includes the second derivative of y; it is
More informationSeries Solutions Near a Regular Singular Point
Series Solutions Near a Regular Singular Point MATH 365 Ordinary Differential Equations J. Robert Buchanan Department of Mathematics Fall 2018 Background We will find a power series solution to the equation:
More informationApplied Differential Equation. November 30, 2012
Applied Differential Equation November 3, Contents 5 System of First Order Linear Equations 5 Introduction and Review of matrices 5 Systems of Linear Algebraic Equations, Linear Independence, Eigenvalues,
More information= 2e t e 2t + ( e 2t )e 3t = 2e t e t = e t. Math 20D Final Review
Math D Final Review. Solve the differential equation in two ways, first using variation of parameters and then using undetermined coefficients: Corresponding homogenous equation: with characteristic equation
More informationMIDTERM 1 PRACTICE PROBLEM SOLUTIONS
MIDTERM 1 PRACTICE PROBLEM SOLUTIONS Problem 1. Give an example of: (a) an ODE of the form y (t) = f(y) such that all solutions with y(0) > 0 satisfy y(t) = +. lim t + (b) an ODE of the form y (t) = f(y)
More informationReview. To solve ay + by + cy = 0 we consider the characteristic equation. aλ 2 + bλ + c = 0.
Review To solve ay + by + cy = 0 we consider the characteristic equation aλ 2 + bλ + c = 0. If λ is a solution of the characteristic equation then e λt is a solution of the differential equation. if there
More informationPartial Fraction Decomposition
Partial Fraction Decomposition As algebra students we have learned how to add and subtract fractions such as the one show below, but we probably have not been taught how to break the answer back apart
More informationExistence Theory: Green s Functions
Chapter 5 Existence Theory: Green s Functions In this chapter we describe a method for constructing a Green s Function The method outlined is formal (not rigorous) When we find a solution to a PDE by constructing
More informationWork sheet / Things to know. Chapter 3
MATH 251 Work sheet / Things to know 1. Second order linear differential equation Standard form: Chapter 3 What makes it homogeneous? We will, for the most part, work with equations with constant coefficients
More informationPartial proof: y = ϕ 1 (t) is a solution to y + p(t)y = 0 implies. Thus y = cϕ 1 (t) is a solution to y + p(t)y = 0 since
Existence and Uniqueness for LINEAR DEs. Homogeneous: y (n) + p 1 (t)y (n 1) +...p n 1 (t)y + p n (t)y = 0 Non-homogeneous: g(t) 0 y (n) + p 1 (t)y (n 1) +...p n 1 (t)y + p n (t)y = g(t) 1st order LINEAR
More informationSystems of differential equations Handout
Systems of differential equations Handout Peyam Tabrizian Friday, November 8th, This handout is meant to give you a couple more example of all the techniques discussed in chapter 9, to counterbalance all
More informationMethods of Mathematics
Methods of Mathematics Kenneth A. Ribet UC Berkeley Math 10B March 15, 2016 Linear first order ODEs Last time we looked at first order ODEs. Today we will focus on linear first order ODEs. Here are some
More information1 Some general theory for 2nd order linear nonhomogeneous
Math 175 Honors ODE I Spring, 013 Notes 5 1 Some general theory for nd order linear nonhomogeneous equations 1.1 General form of the solution Suppose that p; q; and g are continuous on an interval I; and
More informationSecond-Order Homogeneous Linear Equations with Constant Coefficients
15 Second-Order Homogeneous Linear Equations with Constant Coefficients A very important class of second-order homogeneous linear equations consists of those with constant coefficients; that is, those
More informationLinear Independence and the Wronskian
Linear Independence and the Wronskian MATH 365 Ordinary Differential Equations J. Robert Buchanan Department of Mathematics Spring 2018 Operator Notation Let functions p(t) and q(t) be continuous functions
More informationMATH 3321 Sample Questions for Exam 3. 3y y, C = Perform the indicated operations, if possible: (a) AC (b) AB (c) B + AC (d) CBA
MATH 33 Sample Questions for Exam 3. Find x and y so that x 4 3 5x 3y + y = 5 5. x = 3/7, y = 49/7. Let A = 3 4, B = 3 5, C = 3 Perform the indicated operations, if possible: a AC b AB c B + AC d CBA AB
More informationSolutions to Math 53 Math 53 Practice Final
Solutions to Math 5 Math 5 Practice Final 20 points Consider the initial value problem y t 4yt = te t with y 0 = and y0 = 0 a 8 points Find the Laplace transform of the solution of this IVP b 8 points
More informationHW2 Solutions. MATH 20D Fall 2013 Prof: Sun Hui TA: Zezhou Zhang (David) October 14, Checklist: Section 2.6: 1, 3, 6, 8, 10, 15, [20, 22]
HW2 Solutions MATH 20D Fall 2013 Prof: Sun Hui TA: Zezhou Zhang (David) October 14, 2013 Checklist: Section 2.6: 1, 3, 6, 8, 10, 15, [20, 22] Section 3.1: 1, 2, 3, 9, 16, 18, 20, 23 Section 3.2: 1, 2,
More informationCP Algebra 2. Unit 3B: Polynomials. Name: Period:
CP Algebra 2 Unit 3B: Polynomials Name: Period: Learning Targets 10. I can use the fundamental theorem of algebra to find the expected number of roots. Solving Polynomials 11. I can solve polynomials by
More informationEntrance Exam, Differential Equations April, (Solve exactly 6 out of the 8 problems) y + 2y + y cos(x 2 y) = 0, y(0) = 2, y (0) = 4.
Entrance Exam, Differential Equations April, 7 (Solve exactly 6 out of the 8 problems). Consider the following initial value problem: { y + y + y cos(x y) =, y() = y. Find all the values y such that the
More informationToday. The geometry of homogeneous and nonhomogeneous matrix equations. Solving nonhomogeneous equations. Method of undetermined coefficients
Today The geometry of homogeneous and nonhomogeneous matrix equations Solving nonhomogeneous equations Method of undetermined coefficients 1 Second order, linear, constant coeff, nonhomogeneous (3.5) Our
More informationA polynomial expression is the addition or subtraction of many algebraic terms with positive integer powers.
LEAVING CERT Honours Maths notes on Algebra. A polynomial expression is the addition or subtraction of many algebraic terms with positive integer powers. The degree is the highest power of x. 3x 2 + 2x
More informationHigher Order Linear ODEs
im03.qxd 9/21/05 11:04 AM Page 59 CHAPTER 3 Higher Order Linear ODEs This chapter is new. Its material is a rearranged and somewhat extended version of material previously contained in some of the sections
More informationOld Math 330 Exams. David M. McClendon. Department of Mathematics Ferris State University
Old Math 330 Exams David M. McClendon Department of Mathematics Ferris State University Last updated to include exams from Fall 07 Contents Contents General information about these exams 3 Exams from Fall
More informationGUIDELINES FOR THE METHOD OF UNDETERMINED COEFFICIENTS
GUIDELINES FOR THE METHOD OF UNDETERMINED COEFFICIENTS Given a constant coefficient linear differential equation a + by + cy = g(t), where g is an exponential, a simple sinusoidal function, a polynomial,
More informationMath From Scratch Lesson 37: Roots of Cubic Equations
Math From Scratch Lesson 7: Roots of Cubic Equations W. Blaine Dowler September 1, 201 Contents 1 Defining Cubic Equations 1 2 The Roots of Cubic Equations 1 2.1 Case 1: a 2 = a 1 = 0.........................
More informationPolytechnic Institute of NYU MA 2132 Final Practice Answers Fall 2012
Polytechnic Institute of NYU MA Final Practice Answers Fall Studying from past or sample exams is NOT recommended. If you do, it should be only AFTER you know how to do all of the homework and worksheet
More information0.1 Problems to solve
0.1 Problems to solve Homework Set No. NEEP 547 Due September 0, 013 DLH Nonlinear Eqs. reducible to first order: 1. 5pts) Find the general solution to the differential equation: y = [ 1 + y ) ] 3/. 5pts)
More informationLecture 17: Nonhomogeneous equations. 1 Undetermined coefficients method to find a particular
Lecture 17: Nonhomogeneous equations 1 Undetermined coefficients method to find a particular solution The method of undetermined coefficients (sometimes referred to as the method of justicious guessing)
More informationMath 115 HW #10 Solutions
Math 11 HW #10 Solutions 1. Suppose y 1 (t and y 2 (t are both solutions of the differential equation P (ty + Q(ty + R(ty = 0. Show that, for any constants C 1 and C 2, the function C 1 y 1 (t + C 2 y
More informationCHAPTER 2. Techniques for Solving. Second Order Linear. Homogeneous 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 A COLLECTION OF HANDOUTS ON SCALAR LINEAR ORDINARY DIFFERENTIAL
More informationExamples: Solving nth Order Equations
Atoms L. Euler s Theorem The Atom List First Order. Solve 2y + 5y = 0. Examples: Solving nth Order Equations Second Order. Solve y + 2y + y = 0, y + 3y + 2y = 0 and y + 2y + 5y = 0. Third Order. Solve
More informationChapter Six. Polynomials. Properties of Exponents Algebraic Expressions Addition, Subtraction, and Multiplication Factoring Solving by Factoring
Chapter Six Polynomials Properties of Exponents Algebraic Expressions Addition, Subtraction, and Multiplication Factoring Solving by Factoring Properties of Exponents The properties below form the basis
More informationLecture 1: Review of methods to solve Ordinary Differential Equations
Introductory lecture notes on Partial Differential Equations - c Anthony Peirce Not to be copied, used, or revised without explicit written permission from the copyright owner 1 Lecture 1: Review of methods
More information1 Differential Equations
Reading [Simon], Chapter 24, p. 633-657. 1 Differential Equations 1.1 Definition and Examples A differential equation is an equation involving an unknown function (say y = y(t)) and one or more of its
More informationµ = e R p(t)dt where C is an arbitrary constant. In the presence of an initial value condition
MATH 3860 REVIEW FOR FINAL EXAM The final exam will be comprehensive. It will cover materials from the following sections: 1.1-1.3; 2.1-2.2;2.4-2.6;3.1-3.7; 4.1-4.3;6.1-6.6; 7.1; 7.4-7.6; 7.8. The following
More informationSign the pledge. On my honor, I have neither given nor received unauthorized aid on this Exam : 11. a b c d e. 1. a b c d e. 2.
Math 258 Name: Final Exam Instructor: May 7, 2 Section: Calculators are NOT allowed. Do not remove this answer page you will return the whole exam. You will be allowed 2 hours to do the test. You may leave
More informationPartial Fractions. (Do you see how to work it out? Substitute u = ax + b, so du = a dx.) For example, 1 dx = ln x 7 + C, x x (x 3)(x + 1) = a
Partial Fractions 7-9-005 Partial fractions is the opposite of adding fractions over a common denominator. It applies to integrals of the form P(x) dx, wherep(x) and Q(x) are polynomials. Q(x) The idea
More informationLesson 7.1 Polynomial Degree and Finite Differences
Lesson 7.1 Polynomial Degree and Finite Differences 1. Identify the degree of each polynomial. a. 3x 4 2x 3 3x 2 x 7 b. x 1 c. 0.2x 1.x 2 3.2x 3 d. 20 16x 2 20x e. x x 2 x 3 x 4 x f. x 2 6x 2x 6 3x 4 8
More informationLecture 9. Scott Pauls 1 4/16/07. Dartmouth College. Math 23, Spring Scott Pauls. Last class. Today s material. Group work.
Lecture 9 1 1 Department of Mathematics Dartmouth College 4/16/07 Outline Repeated Roots Repeated Roots Repeated Roots Material from last class Wronskian: linear independence Constant coeffecient equations:
More informationFirst-Order ODEs. Chapter Separable Equations. We consider in this chapter differential equations of the form dy (1.1)
Chapter 1 First-Order ODEs We consider in this chapter differential equations of the form dy (1.1) = F (t, y), where F (t, y) is a known smooth function. We wish to solve for y(t). Equation (1.1) is called
More informationMA 1128: Lecture 19 4/20/2018. Quadratic Formula Solving Equations with Graphs
MA 1128: Lecture 19 4/20/2018 Quadratic Formula Solving Equations with Graphs 1 Completing-the-Square Formula One thing you may have noticed when you were completing the square was that you followed the
More informationAlgebra Review. Finding Zeros (Roots) of Quadratics, Cubics, and Quartics. Kasten, Algebra 2. Algebra Review
Kasten, Algebra 2 Finding Zeros (Roots) of Quadratics, Cubics, and Quartics A zero of a polynomial equation is the value of the independent variable (typically x) that, when plugged-in to the equation,
More informationSection 4.7: Variable-Coefficient Equations
Cauchy-Euler Equations Section 4.7: Variable-Coefficient Equations Before concluding our study of second-order linear DE s, let us summarize what we ve done. In Sections 4.2 and 4.3 we showed how to find
More information6 Second Order Linear Differential Equations
6 Second Order Linear Differential Equations A differential equation for an unknown function y = f(x) that depends on a variable x is any equation that ties together functions of x with y and its derivatives.
More informationLecture Notes for Math 251: ODE and PDE. Lecture 12: 3.3 Complex Roots of the Characteristic Equation
Lecture Notes for Math 21: ODE and PDE. Lecture 12: 3.3 Complex Roots of the Characteristic Equation Shawn D. Ryan Spring 2012 1 Complex Roots of the Characteristic Equation Last Time: We considered the
More informationUNDETERMINED COEFFICIENTS SUPERPOSITION APPROACH *
4.4 UNDETERMINED COEFFICIENTS SUPERPOSITION APPROACH 19 Discussion Problems 59. Two roots of a cubic auxiliary equation with real coeffi cients are m 1 1 and m i. What is the corresponding homogeneous
More informationLS.2 Homogeneous Linear Systems with Constant Coefficients
LS2 Homogeneous Linear Systems with Constant Coefficients Using matrices to solve linear systems The naive way to solve a linear system of ODE s with constant coefficients is by eliminating variables,
More informationMath 20D: Form B Final Exam Dec.11 (3:00pm-5:50pm), Show all of your work. No credit will be given for unsupported answers.
Turn off and put away your cell phone. No electronic devices during the exam. No books or other assistance during the exam. Show all of your work. No credit will be given for unsupported answers. Write
More informationNon-homogeneous equations (Sect. 3.6).
Non-homogeneous equations (Sect. 3.6). We study: y + p(t) y + q(t) y = f (t). Method of variation of parameters. Using the method in an example. The proof of the variation of parameter method. Using the
More informationx gm 250 L 25 L min y gm min
Name NetID MATH 308 Exam 2 Spring 2009 Section 511 Hand Computations P. Yasskin Solutions 1 /10 4 /30 2 /10 5 /30 3 /10 6 /15 Total /105 1. (10 points) Tank X initially contains 250 L of sugar water with
More informationAPPM 2360 Section Exam 3 Wednesday November 19, 7:00pm 8:30pm, 2014
APPM 2360 Section Exam 3 Wednesday November 9, 7:00pm 8:30pm, 204 ON THE FRONT OF YOUR BLUEBOOK write: () your name, (2) your student ID number, (3) lecture section, (4) your instructor s name, and (5)
More informationOrdinary Differential Equation Theory
Part I Ordinary Differential Equation Theory 1 Introductory Theory An n th order ODE for y = y(t) has the form Usually it can be written F (t, y, y,.., y (n) ) = y (n) = f(t, y, y,.., y (n 1) ) (Implicit
More informationHomework Solutions: , plus Substitutions
Homework Solutions: 2.-2.2, plus Substitutions Section 2. I have not included any drawings/direction fields. We can see them using Maple or by hand, so we ll be focusing on getting the analytic solutions
More information1 Solution to Homework 4
Solution to Homework Section. 5. The characteristic equation is r r + = (r )(r ) = 0 r = or r =. y(t) = c e t + c e t y = c e t + c e t. y(0) =, y (0) = c + c =, c + c = c =, c =. To find the maximum value
More information