MathQuest: Differential Equations

Similar documents
Practice Problems For Test 3

Practice Problems For Test 3

The Laplace Transform and the IVP (Sect. 6.2).

(f g)(t) = Example 4.5.1: Find f g the convolution of the functions f(t) = e t and g(t) = sin(t). Solution: The definition of convolution is,

MathQuest: Differential Equations

Laplace Transform. Chapter 4

Chapter DEs with Discontinuous Force Functions

The formulas for derivatives are particularly useful because they reduce ODEs to algebraic expressions. Consider the following ODE d 2 dx + p d

( ) f (k) = FT (R(x)) = R(k)

Math Laplace transform

f(t)e st dt. (4.1) Note that the integral defining the Laplace transform converges for s s 0 provided f(t) Ke s 0t for some constant K.

e st f (t) dt = e st tf(t) dt = L {t f(t)} s

Section 6.4 DEs with Discontinuous Forcing Functions

= e t sin 2t. s 2 2s + 5 (s 1) Solution: Using the derivative of LT formula we have

Math Shifting theorems

APPM 2360: Midterm exam 3 April 19, 2017

Ch 6.2: Solution of Initial Value Problems

Math 308 Exam II Practice Problems

Ordinary Differential Equations

3.5 Undetermined Coefficients

Computing inverse Laplace Transforms.

Laplace Transforms Chapter 3

Math 266 Midterm Exam 2

Time Response Analysis (Part II)

MATH 251 Examination II April 7, 2014 FORM A. Name: Student Number: Section:

ENGIN 211, Engineering Math. Laplace Transforms

Math 216 Second Midterm 20 March, 2017

MATH 251 Examination II April 4, 2016 FORM A. Name: Student Number: Section:

CHEE 319 Tutorial 3 Solutions. 1. Using partial fraction expansions, find the causal function f whose Laplace transform. F (s) F (s) = C 1 s + C 2

Applied Differential Equation. October 22, 2012

Systems Analysis and Control

Solutions to Assignment 7

Math 341 Fall 2008 Friday December 12

we get y 2 5y = x + e x + C: From the initial condition y(0) = 1, we get 1 5 = 0+1+C; so that C = 5. Completing the square to solve y 2 5y = x + e x 5

FINAL EXAM SOLUTIONS, MATH 123

MA 201, Mathematics III, July-November 2016, Laplace Transform

Week #9 : DEs with Non-Constant Coefficients, Laplace Resonance

Math 216 Second Midterm 16 November, 2017

9.5 The Transfer Function

20.6. Transfer Functions. Introduction. Prerequisites. Learning Outcomes

Name: Solutions Exam 3

The Laplace Transform

1st Order Linear D.E.

Chapter 6: The Laplace Transform 6.3 Step Functions and

Honors Differential Equations

Differential Equations and Linear Algebra Lecture Notes. Simon J.A. Malham. Department of Mathematics, Heriot-Watt University, Edinburgh EH14 4AS

Generalized sources (Sect. 6.5). The Dirac delta generalized function. Definition Consider the sequence of functions for n 1, Remarks:

Control Systems. Laplace domain analysis

Differential Equations

Math 3313: Differential Equations Laplace transforms

1 4 r q. (9.172) G(r, q) =

Math 215/255 Final Exam (Dec 2005)

Mathematics for Engineers and Scientists 4 Notes for F1.8XD2

Math 307 Lecture 19. Laplace Transforms of Discontinuous Functions. W.R. Casper. Department of Mathematics University of Washington.

The Laplace transform

+ + LAPLACE TRANSFORM. Differentiation & Integration of Transforms; Convolution; Partial Fraction Formulas; Systems of DEs; Periodic Functions.

MA 201, Mathematics III, July-November 2018, Laplace Transform (Contd.)

Problem Score Possible Points Total 150

Laplace Transforms. Chapter 3. Pierre Simon Laplace Born: 23 March 1749 in Beaumont-en-Auge, Normandy, France Died: 5 March 1827 in Paris, France

Laplace Theory Examples

Ordinary differential equations

Name: Solutions Final Exam

Ma 221 Final Exam Solutions 5/14/13

Lecture Discrete dynamic systems

21-256: Partial differentiation

Basic Procedures for Common Problems

UCLA Math 135, Winter 2015 Ordinary Differential Equations

MA 266 Review Topics - Exam # 2 (updated)

(an improper integral)

20.5. The Convolution Theorem. Introduction. Prerequisites. Learning Outcomes

Math 221 Topics since the second exam

Find the Fourier series of the odd-periodic extension of the function f (x) = 1 for x ( 1, 0). Solution: The Fourier series is.

JUST THE MATHS UNIT NUMBER LAPLACE TRANSFORMS 3 (Differential equations) A.J.Hobson

37. f(t) sin 2t cos 2t 38. f(t) cos 2 t. 39. f(t) sin(4t 5) 40.

Control System. Contents

ISE I Brief Lecture Notes

CHAPTER 8 Laplace Transforms

Math 216 Second Midterm 19 March, 2018

Section 6.5 Impulse Functions

MATH 251 Final Examination December 16, 2015 FORM A. Name: Student Number: Section:

Chapter 6 The Laplace Transform

9.2 The Input-Output Description of a System

Unit 2: Modeling in the Frequency Domain Part 2: The Laplace Transform. The Laplace Transform. The need for Laplace

MAT 275 Laboratory 7 Laplace Transform and the Symbolic Math Toolbox

I. Impulse Response and Convolution

Introduction. f(t) dt. lim. f(t) dt = lim. 1 (1+t)1 p. dt; here p>0. h = (1+h)1 p 1 p. dt = (1+t) p 1 p

MATH 2410 PRACTICE PROBLEMS FOR FINAL EXAM

MATH 391 Test 1 Fall, (1) (12 points each)compute the general solution of each of the following differential equations: = 4x 2y.

Fourier transforms. c n e inπx. f (x) = Write same thing in an equivalent form, using n = 1, f (x) = l π

Matemáticas II: Segundo del Grado en Ingeniería Aeroespacial

2.161 Signal Processing: Continuous and Discrete Fall 2008

Laplace Transform Problems

APPM 2360: Midterm 3 July 12, 2013.

MATH 4B Differential Equations, Fall 2016 Final Exam Study Guide

2. First Order Linear Equations and Bernoulli s Differential Equation

Review Sol. of More Long Answer Questions

Math 353 Lecture Notes Week 6 Laplace Transform: Fundamentals

Linear second-order differential equations with constant coefficients and nonzero right-hand side

F(p) y + 3y + 2y = δ(t a) y(0) = 0 and y (0) = 0.

Lecture 16. Theory of Second Order Linear Homogeneous ODEs

Transcription:

MathQuest: Differential Equations Laplace Tranforms 1. True or False The Laplace transform method is the only way to solve some types of differential equations. (a) True, and I am very confident (b) True, but I am not very confident (c) False, but I am not very confident (d) False, and I am very confident. Which of the following differential equations would be impossible to solve using the Laplace transform? (a) dc dr = 1c+sin(r)+8cos (r +) (b) d f dx 100 df dx = 18 ln(x) (c) g (b) = 1 g (d) p (q) = 4 q +96p 1p. Suppose we know that zero is an equilibrium value for a certain homogeneous linear differential equation with constant coefficients. Further, suppose that if we begin at equilibrium and we add the nonhomogenous driving term f(t) to the equation, then the solution will be the function y(t). True or False: If instead we add the nonhomogeous driving term f(t), then the solution will be the function y(t). (a) True, and I am very confident (b) True, but I am not very confident (c) False, but I am not very confident (d) False, and I am very confident 4. Suppose we know that zero is an equilibrium value for a certain homogeneous linear differential equation with constant coefficients. Further, suppose that if we begin at equilibrium and we add a nonhomogenous driving term f(t) to the equation that acts onlyforaninstant, att = t 1, thenthesolutionwillbethefunctiony(t). True or False: Even if we know nothing about the differential equation itself, using our knowledge of f(t) and y(t) we can infer how the system must behave for any other nonhomogeneous driving function g(t) and for any other initial condition. 1

(a) True, and I am very confident (b) True, but I am not very confident (c) False, but I am not very confident (d) False, and I am very confident. The Laplace transform of a function y(t) is defined to be a function Y(s) so that L[y(t) = Y(s) = y(t)e st dt. If we have the function y(t) = e t, then what is its 0 Laplace transform L[y(t) = Y(s)? (a) L[y(t) = 1 s (b) L[y(t) = 1 s (c) L[y(t) = 1 s (d) L[y(t) = 1 s if s >. if s <. if s >. if s <. 6. What is the Laplace transform of the function y(t) = 1? (a) L[y(t) = 1 if s > 0. s (b) L[y(t) = 1 if s < 0. s (c) L[y(t) = 1 if s > 0. s (d) L[y(t) = 1 if s < 0. s (e) This function has no Laplace transform. 7. What is L[e t? (a) L[e t = 1 s (b) L[e t = 1 s (c) L[e t = 1 s+ (d) L[e t = 1 s+ if s <. if s >. if s >. if s <. 8. Suppose we have the Laplace transform Y(s) = 1 s y(t). What is L 1[ 1 s? (a) L 1[ 1 = e t s = e t (b) L 1[ 1 s (c) This cannot be done and we want the original function

9. Suppose we know that the Laplace transform of a particular function y(t) is the function Y(s) so that L[y(t) = Y(s). Now, suppose we multiply this function by and take the Laplace transform. What can we say about L[y(t)? (a) L[y(t) = Y(s). (b) L[y(t) = Y(s). (c) L[y(t) = 1 Y(s). (d) L[y(t) = 1 e Y(s). (e) None of the above 10. Suppose we know that the Laplace transform of a particular function f(t) is the function W(s) and that the Laplace transform of another function g(t) is the function X(s) so that L[f(t) = W(s) and L[g(t) = X(s). Now, suppose we multiply these two functions together and take the Laplace transform. What can we say about L[f(t)g(t)? (a) L[f(t)g(t) = W(s)X(s). (b) L[f(t)g(t) = W(s) X(s). (c) L[f(t)g(t) = W(s)+X(s). (d) L[f(t)g(t) = W(s) X(s). (e) None of the above 11. Suppose we know that the Laplace transform of a particular function f(t) is the function W(s) and that the Laplace transform of another function g(t) is the function X(s) so that L[f(t) = W(s) and L[g(t) = X(s). Now, suppose we add these two functions together and take the Laplace transform. What can we say about L[f(t)+g(t)? (a) L[f(t)+g(t) = W(s)X(s). (b) L[f(t)+g(t) = W(s) X(s). (c) L[f(t)+g(t) = W(s)+X(s). (d) L[f(t)+g(t) = W(s) X(s). (e) None of the above 1. SupposewehaveY(s) = 1 s + s+6 andwewantthefunctiony(t). WhatisL 1[ 1 s + (a) L 1[ 1 s + s+6 (b) L 1[ 1 s + s+6 (c) L 1[ 1 s + s+6 = e t +e 6t/ = e t +e 6t = e t e 6t s+6?

(d) L 1[ 1 + s s+6 = e t e 6t (e) This cannot be done 1. Suppose we know that L[y(t) = Y(s) and we want to take the Laplace transform of the derivative of y(t), L [ dy dt. To get started, recall the method of integration by parts, which tells us that udv = uv v du. (a) L [ dy dt = y(t)e st +sy(s). (b) L [ dy dt = y(t)e st sy(s). (c) L [ dy dt = y(0)+sy(s). (d) L [ dy dt = y(0) sy(s). (e) L [ dy dt = y(0)+sy(s). 14. We know that L[y(t) = Y(s) = y(t)e st dt and that L [ dy 0 dt = y(0)+sy(s). What dowegetwhenwetakethelaplacetransformofthedifferential equation dy = dt y+e t with y(0) =? (a) +sy(s) = Y(s)+ (b) +sy(s) = 1 Y(s)+ s 1 (c) +sy(s) = Y(s)+ (d) +sy(s) = Y(s) (e) sy(s) = 1 Y(s)+ 1 s 1. We know that L[y(t) = Y(s) and that L [ dy dt = y(0) + sy(s). Take the Laplace transform of the differential equation dy = dt y+e t with y(0) = 4 and solve for Y(s). (a) Y(s) = ( s )(s ) + 4 s (b) Y(s) = + 4 ( s )(s ) s+ (c) Y(s) = (s+)(s ) + 4 s (d) Y(s) = (s+)(s+) + 4 s+ 16. Suppose that after taking the Laplace transform of a differential equation we solve for the function Y(s) =. This is equivalent to which of the following? (a) Y(s) = /8 s+ + /8 s 6 (s+)(s 6) 4

(b) Y(s) = 8/ s+ + 8/ s 6 (c) Y(s) = /8 + /8 s+ s 6 (d) Y(s) = 8/ + 8/ s+ s 6 17. Find L 1 [ (a) L 1 [ (b) L 1 [ (c) L 1 [ (d) L 1 [ (e) L 1 [ (s+)(s 6) (s+)(s 6) (s+)(s 6) (s+)(s 6) (s+)(s 6) (s+)(s 6). = e t e 6t = 8 e t + 8 e6t = 8 et 8 e 6t = e /8 e t +e /8 e 6t = 8 et + 8 e 6t 18. Solve the differential equation dy dt = 4y + e t with y(0) = using the method of Laplace transforms: First take the Laplace transform of the entire equation, then solve for Y(s), use the method of partial fractions to simplify the result, and take the inverse transform to get the function y(t). (a) y(t) = 9 e4t + 1 e t (b) y(t) = 7 e4t + e t (c) y(t) = 9 e 4t + 1 et (d) y(t) = 7 e 4t + et 19. Consider the Heaviside function, which is defined as { 0 if t < a u a (t) = 1 if t a Suppose a =. What is the Laplace transform of this L[u (t)? Hint: It may be helpful to break this integral into two pieces, from 0 to and from to. (a) L[u (t) = 1 s es + 1 s (b) L[u (t) = 1 s e s 1 s (c) L[u (t) = 1 s e s (d) L[u (t) = 1 s e s (e) L[u (t) = 1 s es

0. Find the Laplace transform of the function u (t)e (t ). (a) L[u (t)e (t ) = e (b) L[u (t)e (t ) = e s (c) L[u (t)e (t ) = e e s (d) L[u (t)e (t ) = 1 1. Find the Laplace transform of the function u 4 (t)e (t 4). (a) L[u 4 (t)e (t 4) = e s s 4 (b) L[u 4 (t)e (t 4) = es s+4 (c) L[u 4 (t)e (t 4) = e 4s s+ (d) L[u 4 (t)e (t 4) = e 4s s (e) L[u 4 (t)e (t 4) = e4s s. Find L 1 [ (a) L 1 [ (b) L 1 [ (c) L 1 [ (d) L 1 [ (e) L 1 [ (f) L 1 [ e s. s+ e s s+ e s s+ e s s+ e s s+ e s s+ e s s+ = u (t)e (t ) = u (t)e (t ) = u (t)e (t ) = u (t)e (t ) = u (t)e (t ) = u (t)e (t ). Use Laplace transforms to solve the differential equation dy dt = y + 4u (t) with y(0) = 1. (a) y(t) = e t +u (t)e (t ) u (t)e (t ) (b) y(t) = e t +4u (t)e (t ) 4u (t)e (t ) (c) y(t) = e t +u (t) u (t)e (t ) (d) y(t) = e t +4u (t) 4u (t)e (t ) 6

4. What is the Laplace transform of the second derivative, L (a) L (b) L [ d y dt = y (0)+sL[y [ d y dt = y (0) sy(0)+s L[y [ d y (c) L = y (0) sy(0)+sl[y dt [ d (d) L y = y (0)+sy(0) s L[y dt [ d y dt?. We know that the differential equation y = y with the initial condition y(0) = 0 and y (0) = 1 is solved by the function y = sint. By taking the Laplace transform of this equation find an expression for L[sin t. (a) L[sint = 1 s +1 (b) L[sint = 1 s 1 (c) L[sint = s s +1 (d) L[sint = s s 1 6. Find an expression for L[sinωt, by first using your knowledge of the second order differential equation and initial conditions y(0) and y (0) that are solved by this function, and then taking the Laplace transform of this equation. (a) L[sinωt = ω s +1 (b) L[sinωt = 1 s ω +ω (c) L[sinωt = 1 s +ω (d) L[sinωt = ω s +ω (e) L[sinωt = ω s +ω 7. Find an expression for L[cosωt, by first using your knowledge of the second order differential equation and initial conditions y(0) and y (0) that are solved by this function, and then taking the Laplace transform of this equation. (a) L[cosωt = s s +ω (b) L[cosωt = sω s +ω (c) L[cosωt = sω s ω (d) L[cosωt = ω s +ω (e) L[cosωt = ω s ω 7

8. Find L 1[ s 4+s. (a) L 1[ s 4+s = cost. (b) L 1[ s 4+s = cos4t. (c) L 1[ s 4+s = 4cost. (d) L 1[ s 4+s = sint. (e) L 1[ s 4+s = sin4t. (f) L 1[ s 4+s = 4sint. 9. Find L 1[ s +6. (a) L 1[ s +6 = sin t. (b) L 1[ s +6 = sin 6t. (c) L 1[ s +6 = sin t. (d) L 1[ s +6 = sin 6t. (e) L 1[ s +6 = sin t. 0. We know that the function f(t) has a Laplace transform F(s), so that L[f(t) = 0 f(t)e st dt = F(s). What can we say about the Laplace transform of the product L[e at f(t)? (a) L[e at f(t) = e at F(s). (b) L[e at f(t) = e as F(s). (c) L[e at f(t) = F(s a). (d) L[e at f(t) = F(s+a). (e) We cannot make a general statement about this Laplace transform without knowing f(t). 1. Find L[e 4t sint. (a) L[e 4t sint = (s 4) +. (b) L[e 4t sint = 1 (s 4) +. (c) L[e 4t sint = 1 (s+4) +. (d) L[e 4t sint = (s 4) +. 8

(e) L[e 4t sint = (s+4) +.. Find L 1 [ (a) L 1 [ (b) L 1 [ (c) L 1 [ (d) L 1 [ 8 (s+) +16 8 (s+) +16 8 (s+) +16 8 (s+) +16 8 (s+) +16. = e t sin4t = e 4t sint = 8e t sin4t = e t sin4t. s +6 is equivalent to which of the following? (a) s +6 = (s+) +11 (b) s +6 = (s+) +6 (c) s +6 = (s+) 10 (d) s +6 = (s+6) 1 4. Find L 1[ s+6 s +6, by first completing the square in the denominator. (a) L 1[ s+6 s +6 = e t sin 6t (b) L 1[ s+6 s +6 = e t cos 6t (c) L 1[ s+6 s +6 = e t cos 6t (d) L 1[ s+6 s +6 = e t sin 6t (e) L 1[ s+6 s +6 = e 6t cos 6t (f) L 1[ s+6 s +6 = e 6t cos 6t. Find L 1[ s+8 s +6, by first completing the square in the denominator. (a) L 1[ s+8 s +6 = e t cos 6t (b) L 1[ s+8 s +6 = e t cos 6t+ (c) L 1[ s+8 s +6 = e t cos 6t+e t sin 6t (d) L 1[ s+8 s +6 = e t cos 6t+ 6 e t sin 6t 9

6. Suppose that after taking the Laplace transform of a differential equation we obtain the function Y(s) =. This is equivalent to which of the following? (a) Y(s) = s/19 + s/19 s +16 s + (b) Y(s) = s/19 + s/19 s +16 s + s (s +16)(s +) (c) Y(s) = s/1 + s/1 s +16 s + (d) Y(s) = s/1 + s/1 s +16 s + 7. Find L 1 [ (a) L 1 [ (b) L 1 [ (c) L 1 [ (d) L 1 [ s (s +16)(s +) s (s +16)(s +) s (s +16)(s +) s (s +16)(s +) s (s +16)(s +). = 1 cos4t+ 1 cos t. = 1 cos4t 1 cos t. = sin4t+ 1 sin t. = sin4t 1 sin t. 8. Use Laplace transforms to solve the differential equation y = y+4sint if we know that y(0) = y (0) = 0. (a) y = 6 7 sin t+ 4 7 sint (b) y = 6 7 sin t 4 7 sint (c) y = 1 7 sin t 1 7 sint (d) y = 1 7 cos t+ 1 7 cost 9. The Dirac delta function δ a (t) is defined so that δ a (t) = 0 for all t a, and if we integrate this function over any integral containing a, the result is 1. What would be 10 (δ (t)+δ 6 (t) δ 8 (t)+δ 11 (t))dt? (a) 1 (b) 0 (c) 4 (d) (e) 0 (f) This integral cannot be determined. 10

40. Find 10 (δ 6 (t) δ 7 (t) δ 8 (t) δ 9 (t))dt? (a) 1 (b) 0 (c) 1 (d) 4 (e) (f) This integral cannot be determined. 41. Find L[δ 7 (t). (a) 1 (b) 7 (c) e 7s (d) e 7t (e) This integral cannot be determined. 4. Find L[δ 0 (t). (a) 0 (b) 1 (c) e s (d) e t (e) This integral cannot be determined. 4. We have a system modeled as an undamped harmonic oscillator, that begins at equilibrium and at rest, so y(0) = y (0) = 0, and that receives an impulse force at t = 4, so that it is modeled with the equation y = 9y +δ 4 (t). Find y(t). (a) y(t) = u 4 (t)sint (b) y(t) = 1 u 4(t)sint (c) y(t) = u 4 (t)sin(t 4) (d) y(t) = 1 u 4(t)sin(t 4) 44. We have a system modeled as an undamped harmonic oscillator, that begins at equilibrium and at rest, so y(0) = y (0) = 0, and that receives an impulse forcing at t =, so that it is modeled with the equation y = 4y +δ (t). Find y(t). 11

(a) y(t) = u (t)sint (b) y(t) = 1u (t)sin(t ) (c) y(t) = 1u (t)sin(t ) (d) y(t) = 1u (t)sin(t ) (e) y(t) = 1u (t)sin(t ) 4. For many differential equations, if we know how the equation responds to a nonhomogeneous forcing function that is a Dirac delta function, we can use this response function to predict how the differential equation will respond to any other forcing function and any initial conditions. For which of the following differential equations would this be impossible? (a) p (q) = 4 q +96p 1p (b) d m 100 dm 18 +14nm = dn dn ln(n) (c) d f dx 100 df dx = 18 ln(x) (d) dc dr = 1c+sin(r)+8cos (r +) 46. The convolution of two functions f g is defined to be t f(t u)g(u)du. What is the 0 convolution of the functions f(t) = and g(t) = e t? (a) f g = e t 1 (b) f g = e t (c) f g = 4e t 4 (d) f g = 1 et 1 (e) This convolution cannot be computed. 47. What is the convolution of the functions f(t) = u (t) and g(t) = e t? (a) f g = 1 e (t ) 1 (b) f g = 1 e t 1 (c) f g = 1 e (t ) + 1 (d) f g = 1 e t + 1 (e) This convolution cannot be computed. 48. Let ζ(t) be the solution to the initial-value problem d y + p dy + qy = δ dt dt 0(t), with y(0) = y (0) = 0. Find an expression for L[ζ. 1

(a) L[ζ = 0 (b) L[ζ = s +ps+q (c) L[ζ = 1 s +ps+q (d) L[ζ = e s s +ps+q 49. Let ζ(t) = e t be the solution to an initial-value problem d y +qy = δ dt 0(t), with y(0) = y (0) = 0 1 for some specific values of p and q, so that L[ζ =. What s +ps+q will be the solution of d y +p dy +qy = 0 if y(0) = 0 and dt dt y (0) =? (a) y = e t (b) y = 10e t (c) y = e t (d) y = e 1t (e) It cannot be determined from the information given. dt +p dy 0. Let ζ(t) be the solution to the initial-value problem d y + p dy + qy = δ dt dt 0(t), with y(0) = y (0) = 0. Now suppose we want to solve this problem for some other forcing function f(t), so that we have d y +p dy +qy = f(t). Which of the following is a correct dt dt expression for L[y? (a) L[y = 0. (b) L[y = L[fL[ζ. (c) L[y = L[f L[ζ. (d) L[y = L[fζ. (e) None of the above 1. Let ζ(t) = e t be the solution to an initial-value problem d y +p dy +qy = δ dt dt 0(t), with y(0) = y (0) = 0. Now suppose we want to solve this problem for the forcing function f(t) = e t, so that we have d y +p dy +qy = f(t). Find y(t). dt dt (a) y(t) = e t e t (b) y(t) = e t e t (c) y(t) = e t e t (d) y(t) = et e t 1