Practice Midterm Solutions

Size: px
Start display at page:

Download "Practice Midterm Solutions"

Transcription

1 Practice Midterm Solutions Math 4B: Ordinary Differential Equations Winter 20 University of California, Santa Barbara TA: Victoria Kala DO NOT LOOK AT THESE SOLUTIONS UNTIL YOU HAVE ATTEMPTED EVERY PROBLEM ON THE PRACTICE MIDTERM February 2, 20

2 Practice Midterm Solutions February 2, 20 Answers This page contains answers only. Detailed solutions are on the following pages.. (a) Answers will vary (b) Answers will vary 2. See detailed solution 3. (a) y = 2 cos 2 + C e (b) (0, ) 4. sin (y) = sin ( ) π or y = sin ( sin ( )) π 5. y + 2 ln y = + 5 ln 3 + C. y 2 sin 3 y 2 + y ln y y = e y = + 2, 0 < , *Note that this problem is more difficult than it appears 8. (a) y = e 2 (c cos 3 + c 2 sin 3) (b) y = c + c 2 e + c 3 e (c) y = c e 2 + c 2 e 3 (d) y = c e + c 2 e + c 3 cos + c 4 sin 9. (a) W = 2e (b) W = c 2 e 0. (a) y 2 = 2 ln (b) See detailed solution (c) y = c 2 + c 2 2 ln 2

3 Detailed Solutions. Give an eample of a third (3rd) order (a) nonlinear differential equation (b) linear differential equation Proof. Recall a third order linear differential equation is given by a 3 ()y + a 2 ()y + a ()y + a 0 ()y = g() where a 3, a 2, a, a 0 are a functions of, not of y. One way to make a nonlinear differential equation is to have functions of y in front of the derivative terms. Some eamples are below: (a) yy e y = 2 is a nonlinear differential equation because of the term yy. (b) y 2y + ln y e + 4y = 0 is a linear differential equation because all of the functions in front of the derivative terms are functions of alone. 2. Verify P = cet dp is a solution to the differential equation + cet dt = P ( P ). Proof. We need to substitute P into the equation above and show that the left hand side equations the right hand side. Left hand side: (use the quotient rule or product rule for the derivative) dp dt = ( + cet )(ce t ) ce t ( + ce t ) ( + ce t ) 2 = ( + cet )(ce t ) ce t (ce t ) ce t ( + ce t ) 2 = ( + ce t ) 2 Right hand side: P ( P ) = cet + ce t ) ( ) ( cet + ce t = cet + ce t + ce t + ce t cet + ce t = cet The left hand side equals the right hand side, hence P = equation dp dt = P ( P ). 3. (a) Solve the following first order linear equation: y + ( + )y = e sin 2 + ce t + ce t = ce t ( + ce t ) 2 cet is a solution to the differential + cet Solution. Write the equation in standard form by dividing the equation on both sides by : ( ) + y + y = e sin 2 3

4 Now we wish to find an integrating factor. From the equation above, we see that P () = + = + (remember that P () is function in front of the y term in the differential equation). Therefore ( ) ( ( ) ) µ() = ep P ()d = ep + d = e ln + = e ln e = e Multiply the standard form equation above by µ(): ( ) ] + e [y + y = e sin 2 e y + ( + )e y = sin 2 () The left hand side should collapse down to the derivative of the product µ()y, that is, Don t believe it? You can always check: d d (e y) = sin 2. (2) d d (e y) = e y + (e ) y = e y + ( e + (e ) )y = e y + (e + e )y which is the same as the equation (). Now looking back at equation (2), we integrate both sides: e y = sin 2d e y = cos 2 + C 2 Solve for y: y = 2 cos 2 + C e (b) For the differential equation in part (a), determine the largest interval on which the eistence and uniqueness theorem for first order linear differential equations guarantees the eistence of a unique solution at the initial value y(4) =. Solution. We must look at the standard form equation: ( ) + y + y = e sin 2 We have that P () = +, f() = e sin 2 4

5 The domain of P () is all numbers such that 0. The domain of f() is also all numbers such that 0. Therefore the domain of the equation is all numbers such that 0, which can be written as (, 0) (0, ). Now let s look at the initial value y(4) =. This says when = 4 we have that y =. = 4 falls into the interval (0, ), hence this is the largest interval on which we are guaranteed a unique solution. 4. Solve the following initial value problem: dy d = y 2, y(0) = π Solution. This is a separable equation. We can separate the variables as follows: dy y 2 = d Integrate both sides: dy = y 2 d sin (y) = C (3) Here we will plug in the initial value (you can also solve for y then plug in the initial value if you wish). When = 0, y = π : ( sin π ) = ( 2 (0)2 + C C = sin π ) Therefore (3) becomes sin (y) = ( sin π ) This is the general solution in implicit form. If we were asked to solve for eplicit form, then we would need to solve for y: ( ( y = sin sin π ) ). 5. Solve the following separable equation: dy y + 2y 2 = d y 3y + 3 Solution. We need to factor the right hand side: dy y( + 2) ( + 2) = d y( 3) + ( 3) dy ( + 2)(y ) = d ( 3)(y + ) 5

6 Separate the variables as follows: y + y dy = d Integrate both sides: y + y dy = + 2 d (4) 3 You will need to use substitution or long division to evaluate these integrals. I will do substitution. Left hand side: Let u = y, then du = dy and y = u +. We substitute these into the integral: ( y + u + + y dy = du == + 2 ) du = u + 2 ln u = y + 2 ln y u u Right hand side: Let v = 3, then dv = d and = v + 3. We substitute these into the integral: ( + 2 v d = du = + 5 ) du = v + 5 ln v = ln 3 v v Therefore the equation (4) becomes y + 2 ln y = ln 3 + C y + 2 ln y = + 5 ln 3 + C This would be too difficult to solve for y, so we will leave it in implicit form.. Solve the initial value problem (y 2 cos 3 2 y 2)d + (2y sin 3 + ln(y))dy = 0, y(0) = e Solution. This equation is of the form Md + Ndy = 0. Let s check to see if the equation is eact; that is, we need to see if M y = N : M y = N so this equation is eact! M = y 2 cos 3 2 y 2 M y = 2y cos 3 2 N = 2y sin 3 + ln(y) N = 2y cos 3 2 We want to find an f(, y) such that f = M, so we will integrate M with respect to : f(, y) = Md + g(y) = (y 2 cos 3 2 y 2)d + g(y) = y 2 sin 3 y 2 + g(y)

7 We want df dy = N, so we will differentiate f with respect to y and set it equal to N: f y = N y (y2 sin 3 y 2 + g(y)) = 2y sin 3 + ln(y) Solve for g (y): and integrate: 2y sin 3 + g (y) = 2y sin 3 + ln y g (y) = ln y g(y) = ln ydy. You will need to use integration by parts to evaluate this integral. Try u = ln y, dv = dy. Then du = y dy, v = y and we have g(y) = ln ydy = y ln y y dy = y ln y y Substituting this into our equation above for f, we have Now set f(, y) = C: f(, y) = y 2 sin 3 y 2 + y ln y y. y 2 sin 3 y 2 + y ln y y = C dy = y ln y y This is the general solution to the differential equation. Now we need to plug in the initial value y(0) = e. When = 0, y = e: e 2 sin 0 (0) 3 e (0) 2 + e ln e = C C = e Therefore the solution is y 2 sin 3 y 2 + y ln y y = e. 7. Solve the initial value problem where f() = through. { ( + 2 ) dy + 2y = f(), y(0) = 0 d, 0 <. Hint: an integrating factor has already been multiplied, Solution. This is a first order linear differential equation. Because f() is a piecewise function, we really have two differential equations. When 0 <, we have and when ( + 2 ) dy + 2y =, y(0) = 0 d ( + 2 ) dy + 2y =, y() =? d 7

8 Notice the initial condition on the interval is different than the initial condition on the interval 0 <. We won t know what this different initial condition is until we solve the differential equation on the interval 0 <. Let s look at the differential equation on 0 < : ( + 2 ) dy + 2y =, y(0) = 0. d Using the hint that an integration factor has been multiplied through, the left hand side collapses down to a product rule: [( + 2 )y] = Don t believe it? Check it! Then integrate both sides: ( + 2 )y = d ( + 2 )y = C Solve for y: y = C + 2 Then plug in the initial condition y(0) = 0. When = 0, y = 0: 0 = Therefore the solution on 0 < is given by 2 (0)2 + C + (0) 2 C = 0 y = Before we solve the differential equation on the interval, we want to find out the initial condition at =. To do this we plug in = into the solution above: Therefore the IVP on is y() = 2 ()2 + () 2 = 4. ( + 2 ) dy d + 2y =, y() = 4. Using the hint that an integration factor has been multiplied through, the left hand side collapses down to a product rule: [( + 2 )y] = Don t believe it? Check it! Then integrate both sides: ( + 2 )y = d ( + 2 )y = C 8

9 Solve for y: y = C + 2 Then plug in the initial condition y() = 4. When =, y = 4 : 4 = 2 ()2 + C + () 2 2 = 2 + C C = Therefore the solution on is given by y = We can write the solution as a the piecewise function: 2 2 y = + 2, 0 < , 8. Find the general solution of the following homogeneous higher-order differential equations: (a) y + 4y + 7y = 0 (use as the independent variable) Solution. The characteristic equation is r 2 + 4r + 7 = 0. Using the quadratic equation, we see the roots are r = 4 ± 4 2 4()(7) 2 r = 4 ± i2 3 2 = 2 ± i 3. We have two comple roots, hence the solution to the differential equation is y = e 2 (c cos 3 + c 2 sin 3) (b) y (3) + 2y + y = 0 (use as the independent variable) Solution. The characteristic equation is r 3 + 2r 2 + r = 0 which factors as r(r 2 + 2r + ) = 0 r(r + ) 2 = 0. Therefore the roots are r = 0 multiplicity, r = multiplicity 2. The solution to the differential equation is therefore y = c e 0 + c 2 e + c3e or y = c + c 2 e + c 3 e. 9

10 (c) 2 d2 u dt 2 5du dt 3u = 0 Solution. The characteristic equation is 2r 2 5r 3 = 0, which factors as (2r + )(r 3) = 0. The roots are r = 2, r = 3 each with multiplicity. The solution to the differential equation is therefore y = c e 2 + c 2 e 3. (d) d4 r ds 4 r = 0 Solution. The characteristic equation is m 4 = 0 (we cannot use r as our characteristic equation variable since it is being used in the differential equation). This factors as (m 2 )(m 2 + ) = 0 (m )(m + )(m i)(m + i) = 0. The roots are m =,, ±i. The solution to the differential equation is y = c e + c 2 e + e 0 (c 3 cos + c 4 sin ) y = c e + c 2 e + c 3 cos + c 4 sin. 9. (a) Calculate W (y, y 2, y 3 ) where y = e, y 2 = e 2, y 3 = e 3. Solution. e e 2 e 3 W (e, e 2, e 3 ) = (e ) (e 2 ) (e 3 ) (e ) (e 2 ) (e 3 ) = e e 2 e 3 e 2e 2 3e 3 e 4e 2 9e 3 = e 2e 2 3e 3 e 2 e 3 e + e 2 e 4e 2 9e 3 4e 2 9e 3 2e 2 e 3 3e 3 = e (8e 5 2e 5 ) e (9e 5 4e 5 ) + e (3e 5 2e 3 ) = e (e 5 5e 5 + e 5 ) = 2e (b) Suppose y, y 2 are solutions to the equation 2 y + (2 2 )y + y = 0. Find W (y, y 2 ). Proof. We need to write the differential equation in standard form by dividing through by 2 : y y + 2 y = 0. 0

11 We see that P () = 2 2 = 2 2. Therefore, by using Abel s Theorem, ( ) ( ( ) ) 2 W (y, y 2 ) = c ep P ()d = c ep d = c ep (2 ln ) = ce 2 ln e = ce ln 2 e = c 2 e 0. (a) The function y = 2 is a solution of 2 y 3y + 4y = 0. Use the method of reduction of order to find a second solution y 2 to the differential equation on the interval (0, ). Proof. We need to write the differential equation in standard form by dividing through by 2 : y 3 y y = 0 (5) Let y 2 = vy where v is a function of. Since y = 2, then y 2 = 2 v. We wish to plug this into the equation above, so we need to find y 2, y 2 using the product rule: y 2 = 2 v y 2 = ( 2 ) v + 2 v = 2v + 2 v y 2 = (2v) + ( 2 v ) = (2) v + 2v + ( 2 ) v + 2 v = 2v + 2v + 2v + 2 v = 2v + 4v + 2 v Substitute these into the standard form equation (5) above: 2v + 4v + 2 v 3 (2v + 2 v ) (2 v) = 0 2v + 4v + 2 v v 3v + 4v = 0 Now let w = v, then w = v and () becomes 2 v + v = 0 () 2 w + w = 0 2 dw d = w. This is a separable equation. Separate the variables: Integrate both sides: dw w = dw w = d d ln w = ln ln w = ln ( ) Solve for w: w =.

12 But, w = v, and so we have v = v = d = ln Therefore y 2 = 2 v y 2 = 2 ln. (b) Show that y and y 2 form a fundamental set of solutions to the differential equation. Solution. We need to show that W (y, y 2 ) 0 in the interval (0, ). W (y, y 2 ) = 2 2 ln ( 2 ) ( 2 ln ) = 2 2 ln 2 2 ln + = 2 (2 ln + ) 2( 2 ln ) = 3 W = 3 which is nonzero since we are in the interval (0, ). (c) Write the general solution of the differential equation using y and y 2. Solution. The general solution to the differential equation is y = c y + c 2 y 2, which is y = c 2 + c 2 2 ln. 2

MTH 3311 Test #1. order 3, linear. The highest order of derivative of y is 2. Furthermore, y and its derivatives are all raised to the

MTH 3311 Test #1. order 3, linear. The highest order of derivative of y is 2. Furthermore, y and its derivatives are all raised to the MTH 3311 Test #1 F 018 Pat Rossi Name Show CLEARLY how you arrive at your answers. 1. Classify the following according to order and linearity. If an equation is not linear, eplain why. (a) y + y y = 4

More information

Practice Midterm 1 Solutions Written by Victoria Kala July 10, 2017

Practice Midterm 1 Solutions Written by Victoria Kala July 10, 2017 Practice Midterm 1 Solutions Written by Victoria Kala July 10, 2017 1. Use the slope field plotter link in Gauchospace to check your solution. 2. (a) Not linear because of the y 2 sin x term (b) Not linear

More information

Math 112 Section 10 Lecture notes, 1/7/04

Math 112 Section 10 Lecture notes, 1/7/04 Math 11 Section 10 Lecture notes, 1/7/04 Section 7. Integration by parts To integrate the product of two functions, integration by parts is used when simpler methods such as substitution or simplifying

More information

= 2e t e 2t + ( e 2t )e 3t = 2e t e t = e t. Math 20D Final Review

= 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 information

SOLUTIONS BY SUBSTITUTIONS

SOLUTIONS BY SUBSTITUTIONS 25 SOLUTIONS BY SUBSTITUTIONS 71 25 SOLUTIONS BY SUBSTITUTIONS REVIEW MATERIAL Techniques of integration Separation of variables Solution of linear DEs INTRODUCTION We usually solve a differential equation

More information

QUADRATIC EQUATIONS. + 6 = 0 This is a quadratic equation written in standard form. x x = 0 (standard form with c=0). 2 = 9

QUADRATIC EQUATIONS. + 6 = 0 This is a quadratic equation written in standard form. x x = 0 (standard form with c=0). 2 = 9 QUADRATIC EQUATIONS A quadratic equation is always written in the form of: a + b + c = where a The form a + b + c = is called the standard form of a quadratic equation. Eamples: 5 + 6 = This is a quadratic

More information

EXACT EQUATIONS AND INTEGRATING FACTORS

EXACT EQUATIONS AND INTEGRATING FACTORS MAP- EXACT EQUATIONS AND INTEGRATING FACTORS First-order Differential Equations for Which We Can Find Eact Solutions Stu the patterns carefully. The first step of any solution is correct identification

More information

Chapter 5: Integrals

Chapter 5: Integrals Chapter 5: Integrals Section 5.5 The Substitution Rule (u-substitution) Sec. 5.5: The Substitution Rule We know how to find the derivative of any combination of functions Sum rule Difference rule Constant

More information

Unit #11 - Integration by Parts, Average of a Function

Unit #11 - Integration by Parts, Average of a Function Unit # - Integration by Parts, Average of a Function Some problems and solutions selected or adapted from Hughes-Hallett Calculus. Integration by Parts. For each of the following integrals, indicate whether

More information

Chapter 5: Integrals

Chapter 5: Integrals Chapter 5: Integrals Section 5.3 The Fundamental Theorem of Calculus Sec. 5.3: The Fundamental Theorem of Calculus Fundamental Theorem of Calculus: Sec. 5.3: The Fundamental Theorem of Calculus Fundamental

More information

Math 201 Lecture 05 Bernoulli and Linear Coefficients

Math 201 Lecture 05 Bernoulli and Linear Coefficients Math 0 Lecture 05 Bernoulli and Linear Coefficients Jan. 8, 0 Many eamples here are taken from the tetbook. The first number in () refers to the problem number in the UA Custom edition, the second number

More information

Differential calculus. Background mathematics review

Differential calculus. Background mathematics review Differential calculus Background mathematics review David Miller Differential calculus First derivative Background mathematics review David Miller First derivative For some function y The (first) derivative

More information

Diff. Eq. App.( ) Midterm 1 Solutions

Diff. Eq. App.( ) Midterm 1 Solutions Diff. Eq. App.(110.302) Midterm 1 Solutions Johns Hopkins University February 28, 2011 Problem 1.[3 15 = 45 points] Solve the following differential equations. (Hint: Identify the types of the equations

More information

8.4 Integration of Rational Functions by Partial Fractions Lets use the following example as motivation: Ex: Consider I = x+5

8.4 Integration of Rational Functions by Partial Fractions Lets use the following example as motivation: Ex: Consider I = x+5 Math 2-08 Rahman Week6 8.4 Integration of Rational Functions by Partial Fractions Lets use the following eample as motivation: E: Consider I = +5 2 + 2 d. Solution: Notice we can easily factor the denominator

More information

Section 2.2 Solutions to Separable Equations

Section 2.2 Solutions to Separable Equations Section. Solutions to Separable Equations Key Terms: Separable DE Eponential Equation General Solution Half-life Newton s Law of Cooling Implicit Solution (The epression has independent and dependent variables

More information

2.2 Separable Equations

2.2 Separable Equations 2.2 Separable Equations Definition A first-order differential equation that can be written in the form Is said to be separable. Note: the variables of a separable equation can be written as Examples Solve

More information

Review for the Final Exam

Review for the Final Exam Math 171 Review for the Final Exam 1 Find the limits (4 points each) (a) lim 4x 2 3; x x (b) lim ( x 2 x x 1 )x ; (c) lim( 1 1 ); x 1 ln x x 1 sin (x 2) (d) lim x 2 x 2 4 Solutions (a) The limit lim 4x

More information

Lecture 7: Differential Equations

Lecture 7: Differential Equations Math 94 Professor: Padraic Bartlett Lecture 7: Differential Equations Week 7 UCSB 205 This is the seventh week of the Mathematics Subject Test GRE prep course; here, we review various techniques used to

More information

MAE 82 Engineering Mathematics

MAE 82 Engineering Mathematics Class otes : First Order Differential Equation on Linear AE 8 Engineering athematics Universe on Linear umerical Linearization on Linear Special Cases Analtical o General Solution Linear Analtical General

More information

Chapter 9. Derivatives. Josef Leydold Mathematical Methods WS 2018/19 9 Derivatives 1 / 51. f x. (x 0, f (x 0 ))

Chapter 9. Derivatives. Josef Leydold Mathematical Methods WS 2018/19 9 Derivatives 1 / 51. f x. (x 0, f (x 0 )) Chapter 9 Derivatives Josef Leydold Mathematical Methods WS 208/9 9 Derivatives / 5 Difference Quotient Let f : R R be some function. The the ratio f = f ( 0 + ) f ( 0 ) = f ( 0) 0 is called difference

More information

(a) x cos 3x dx We apply integration by parts. Take u = x, so that dv = cos 3x dx, v = 1 sin 3x, du = dx. Thus

(a) x cos 3x dx We apply integration by parts. Take u = x, so that dv = cos 3x dx, v = 1 sin 3x, du = dx. Thus Math 128 Midterm Examination 2 October 21, 28 Name 6 problems, 112 (oops) points. Instructions: Show all work partial credit will be given, and Answers without work are worth credit without points. You

More information

Math Reading assignment for Chapter 1: Study Sections 1.1 and 1.2.

Math Reading assignment for Chapter 1: Study Sections 1.1 and 1.2. Math 3350 1 Chapter 1 Reading assignment for Chapter 1: Study Sections 1.1 and 1.2. 1.1 Material for Section 1.1 An Ordinary Differential Equation (ODE) is a relation between an independent variable x

More information

Work sheet / Things to know. Chapter 3

Work 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 information

Solutions to Math 53 First Exam April 20, 2010

Solutions to Math 53 First Exam April 20, 2010 Solutions to Math 53 First Exam April 0, 00. (5 points) Match the direction fields below with their differential equations. Also indicate which two equations do not have matches. No justification is necessary.

More information

4.4 Using partial fractions

4.4 Using partial fractions CHAPTER 4. INTEGRATION 43 Eample 4.9. Compute ln d. (Classic A-Level question!) ln d ln d ln d (ln ) + C. Eample 4.0. Find I sin d. I sin d sin p d p sin. 4.4 Using partial fractions Sometimes we want

More information

Math 201 Solutions to Assignment 1. 2ydy = x 2 dx. y = C 1 3 x3

Math 201 Solutions to Assignment 1. 2ydy = x 2 dx. y = C 1 3 x3 Math 201 Solutions to Assignment 1 1. Solve the initial value problem: x 2 dx + 2y = 0, y(0) = 2. x 2 dx + 2y = 0, y(0) = 2 2y = x 2 dx y 2 = 1 3 x3 + C y = C 1 3 x3 Notice that y is not defined for some

More information

First Order ODEs, Part I

First Order ODEs, Part I Craig J. Sutton craig.j.sutton@dartmouth.edu Department of Mathematics Dartmouth College Math 23 Differential Equations Winter 2013 Outline 1 2 in General 3 The Definition & Technique Example Test for

More information

Differentiation of Logarithmic Functions

Differentiation of Logarithmic Functions Differentiation of Logarithmic Functions The rule for finding the derivative of a logarithmic function is given as: If y log a then dy or y. d a ( ln This rule can be proven by rewriting the logarithmic

More information

MATH 1220 Midterm 1 Thurs., Sept. 20, 2007

MATH 1220 Midterm 1 Thurs., Sept. 20, 2007 MATH 220 Midterm Thurs., Sept. 20, 2007 Write your name and ID number at the top of this page. Show all your work. You may refer to one double-sided sheet of notes during the eam and nothing else. Calculators

More information

(1 + 2y)y = x. ( x. The right-hand side is a standard integral, so in the end we have the implicit solution. y(x) + y 2 (x) = x2 2 +C.

(1 + 2y)y = x. ( x. The right-hand side is a standard integral, so in the end we have the implicit solution. y(x) + y 2 (x) = x2 2 +C. Midterm 1 33B-1 015 October 1 Find the exact solution of the initial value problem. Indicate the interval of existence. y = x, y( 1) = 0. 1 + y Solution. We observe that the equation is separable, and

More information

3.3 Limits and Infinity

3.3 Limits and Infinity Calculus Maimus. Limits Infinity Infinity is not a concrete number, but an abstract idea. It s not a destination, but a really long, never-ending journey. It s one of those mind-warping ideas that is difficult

More information

Math 312 Lecture 3 (revised) Solving First Order Differential Equations: Separable and Linear Equations

Math 312 Lecture 3 (revised) Solving First Order Differential Equations: Separable and Linear Equations Math 312 Lecture 3 (revised) Solving First Order Differential Equations: Separable and Linear Equations Warren Weckesser Department of Mathematics Colgate University 24 January 25 This lecture describes

More information

MAP 2302 Midterm 1 Review Solutions 1

MAP 2302 Midterm 1 Review Solutions 1 MAP 2302 Midterm 1 Review Solutions 1 The exam will cover sections 1.2, 1.3, 2.2, 2.3, 2.4, and 2.6. All topics from this review sheet or from the suggested exercises are fair game. 1 Give explicit solutions

More information

3 More applications of derivatives

3 More applications of derivatives 3 More alications of derivatives 3.1 Eact & ineact differentials in thermodynamics So far we have been discussing total or eact differentials ( ( u u du = d + dy, (1 but we could imagine a more general

More information

MATH MIDTERM 4 - SOME REVIEW PROBLEMS WITH SOLUTIONS Calculus, Fall 2017 Professor: Jared Speck. Problem 1. Approximate the integral

MATH MIDTERM 4 - SOME REVIEW PROBLEMS WITH SOLUTIONS Calculus, Fall 2017 Professor: Jared Speck. Problem 1. Approximate the integral MATH 8. - MIDTERM 4 - SOME REVIEW PROBLEMS WITH SOLUTIONS 8. Calculus, Fall 7 Professor: Jared Speck Problem. Approimate the integral 4 d using first Simpson s rule with two equal intervals and then the

More information

Math 123, Week 9: Separable, First-Order Linear, and Substitution Methods. Section 1: Separable DEs

Math 123, Week 9: Separable, First-Order Linear, and Substitution Methods. Section 1: Separable DEs Math 123, Week 9: Separable, First-Order Linear, and Substitution Methods Section 1: Separable DEs We are finally to the point in the course where we can consider how to find solutions to differential

More information

µ = e R p(t)dt where C is an arbitrary constant. In the presence of an initial value condition

µ = 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 information

Math 2a Prac Lectures on Differential Equations

Math 2a Prac Lectures on Differential Equations Math 2a Prac Lectures on Differential Equations Prof. Dinakar Ramakrishnan 272 Sloan, 253-37 Caltech Office Hours: Fridays 4 5 PM Based on notes taken in class by Stephanie Laga, with a few added comments

More information

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

MATH 391 Test 1 Fall, (1) (12 points each)compute the general solution of each of the following differential equations: = 4x 2y. MATH 391 Test 1 Fall, 2018 (1) (12 points each)compute the general solution of each of the following differential equations: (a) (b) x dy dx + xy = x2 + y. (x + y) dy dx = 4x 2y. (c) yy + (y ) 2 = 0 (y

More information

VANDERBILT UNIVERSITY. MATH 2610 ORDINARY DIFFERENTIAL EQUATIONS Practice for test 1 solutions

VANDERBILT UNIVERSITY. MATH 2610 ORDINARY DIFFERENTIAL EQUATIONS Practice for test 1 solutions VANDERBILT UNIVERSITY MATH 2610 ORDINARY DIFFERENTIAL EQUATIONS Practice for test 1 solutions The first test will cover all material discussed up to (including) section 4.5. Important: The solutions below

More information

Math K (24564) - Homework Solutions 02

Math K (24564) - Homework Solutions 02 Math 39100 K (24564) - Homework Solutions 02 Ethan Akin Office: NAC 6/287 Phone: 650-5136 Email: ethanakin@earthlink.net Spring, 2018 Contents Reduction of Order, B & D Chapter 3, p. 174 Constant Coefficient

More information

Math 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 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 information

In economics, the amount of a good x demanded is a function of the price of that good. In other words,

In economics, the amount of a good x demanded is a function of the price of that good. In other words, I. UNIVARIATE CALCULUS Given two sets X and Y, a function is a rule that associates each member of X with eactly one member of Y. That is, some goes in, and some y comes out. These notations are used to

More information

Math 1B Final Exam, Solution. Prof. Mina Aganagic Lecture 2, Spring (6 points) Use substitution and integration by parts to find:

Math 1B Final Exam, Solution. Prof. Mina Aganagic Lecture 2, Spring (6 points) Use substitution and integration by parts to find: Math B Final Eam, Solution Prof. Mina Aganagic Lecture 2, Spring 20 The eam is closed book, apart from a sheet of notes 8. Calculators are not allowed. It is your responsibility to write your answers clearly..

More information

INSTRUCTOR S SOLUTIONS MANUAL SECTION 17.1 (PAGE 902)

INSTRUCTOR S SOLUTIONS MANUAL SECTION 17.1 (PAGE 902) INSTRUCTOR S SOLUTIONS MANUAL SECTION 7 PAGE 90 CHAPTER 7 ORDINARY DIFFEREN- TIAL EQUATIONS Section 7 Classifing Differential Equations page 90 = 5: st order, linear, homogeneous d d + = : nd order, linear,

More information

First Order Differential Equations

First Order Differential Equations Chapter 2 First Order Differential Equations 2.1 9 10 CHAPTER 2. FIRST ORDER DIFFERENTIAL EQUATIONS 2.2 Separable Equations A first order differential equation = f(x, y) is called separable if f(x, y)

More information

(d by dx notation aka Leibniz notation)

(d by dx notation aka Leibniz notation) n Prerequisites: Differentiating, sin and cos ; sum/difference and chain rules; finding ma./min.; finding tangents to curves; finding stationary points and their nature; optimising a function. Maths Applications:

More information

Page 1. These are all fairly simple functions in that wherever the variable appears it is by itself. What about functions like the following, ( ) ( )

Page 1. These are all fairly simple functions in that wherever the variable appears it is by itself. What about functions like the following, ( ) ( ) Chain Rule Page We ve taken a lot of derivatives over the course of the last few sections. However, if you look back they have all been functions similar to the following kinds of functions. 0 w ( ( tan

More information

DIFFERENTIATION RULES

DIFFERENTIATION RULES 3 DIFFERENTIATION RULES DIFFERENTIATION RULES 3.6 Derivatives of Logarithmic Functions In this section, we: use implicit differentiation to find the derivatives of the logarithmic functions and, in particular,

More information

for any C, including C = 0, because y = 0 is also a solution: dy

for any C, including C = 0, because y = 0 is also a solution: dy Math 3200-001 Fall 2014 Practice exam 1 solutions 2/16/2014 Each problem is worth 0 to 4 points: 4=correct, 3=small error, 2=good progress, 1=some progress 0=nothing relevant. If the result is correct,

More information

Worksheet Week 7 Section

Worksheet Week 7 Section Worksheet Week 7 Section 8.. 8.4. This worksheet is for improvement of your mathematical writing skill. Writing using correct mathematical epression and steps is really important part of doing math. Please

More information

The Explicit Form of a Function

The Explicit Form of a Function Section 3 5 Implicit Differentiation The Eplicit Form of a Function Function Notation requires that we state a function with f () on one sie of an equation an an epression in terms of on the other sie

More information

M151B Practice Problems for Exam 1

M151B Practice Problems for Exam 1 M151B Practice Problems for Eam 1 Calculators will not be allowed on the eam. Unjustified answers will not receive credit. 1. Compute each of the following its: 1a. 1b. 1c. 1d. 1e. 1 3 4. 3. sin 7 0. +

More information

Integration by Parts

Integration by Parts Calculus 2 Lia Vas Integration by Parts Using integration by parts one transforms an integral of a product of two functions into a simpler integral. Divide the initial function into two parts called u

More information

Lecture Notes in Mathematics. Arkansas Tech University Department of Mathematics

Lecture Notes in Mathematics. Arkansas Tech University Department of Mathematics Lecture Notes in Mathematics Arkansas Tech University Department of Mathematics Introductory Notes in Ordinary Differential Equations for Physical Sciences and Engineering Marcel B. Finan c All Rights

More information

MA1131 Lecture 15 (2 & 3/12/2010) 77. dx dx v + udv dx. (uv) = v du dx dx + dx dx dx

MA1131 Lecture 15 (2 & 3/12/2010) 77. dx dx v + udv dx. (uv) = v du dx dx + dx dx dx MA3 Lecture 5 ( & 3//00) 77 0.3. Integration by parts If we integrate both sides of the proct rule we get d (uv) dx = dx or uv = d (uv) = dx dx v + udv dx v dx dx + v dx dx + u dv dx dx u dv dx dx This

More information

Calculus I Review Solutions

Calculus I Review Solutions Calculus I Review Solutions. Compare and contrast the three Value Theorems of the course. When you would typically use each. The three value theorems are the Intermediate, Mean and Extreme value theorems.

More information

Exam 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. 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 information

Math 23: Differential Equations (Winter 2017) Midterm Exam Solutions

Math 23: Differential Equations (Winter 2017) Midterm Exam Solutions Math 3: Differential Equations (Winter 017) Midterm Exam Solutions 1. [0 points] or FALSE? You do not need to justify your answer. (a) [3 points] Critical points or equilibrium points for a first order

More information

Math 180, Exam 2, Spring 2013 Problem 1 Solution

Math 180, Exam 2, Spring 2013 Problem 1 Solution Math 80, Eam, Spring 0 Problem Solution. Find the derivative of each function below. You do not need to simplify your answers. (a) tan ( + cos ) (b) / (logarithmic differentiation may be useful) (c) +

More information

MATH 312 Section 2.4: Exact Differential Equations

MATH 312 Section 2.4: Exact Differential Equations MATH 312 Section 2.4: Exact Differential Equations Prof. Jonathan Duncan Walla Walla College Spring Quarter, 2007 Outline 1 Exact Differential Equations 2 Solving an Exact DE 3 Making a DE Exact 4 Conclusion

More information

MATH 255 Applied Honors Calculus III Winter Homework 5 Solutions

MATH 255 Applied Honors Calculus III Winter Homework 5 Solutions MATH 255 Applied Honors Calculus III Winter 2011 Homework 5 Solutions Note: In what follows, numbers in parentheses indicate the problem numbers for users of the sixth edition. A * indicates that this

More information

Math 122 Fall Unit Test 1 Review Problems Set A

Math 122 Fall Unit Test 1 Review Problems Set A Math Fall 8 Unit Test Review Problems Set A We have chosen these problems because we think that they are representative of many of the mathematical concepts that we have studied. There is no guarantee

More information

Ordinary Differential Equations (ODEs)

Ordinary Differential Equations (ODEs) c01.tex 8/10/2010 22: 55 Page 1 PART A Ordinary Differential Equations (ODEs) Chap. 1 First-Order ODEs Sec. 1.1 Basic Concepts. Modeling To get a good start into this chapter and this section, quickly

More information

3.1 Derivative Formulas for Powers and Polynomials

3.1 Derivative Formulas for Powers and Polynomials 3.1 Derivative Formulas for Powers and Polynomials First, recall that a derivative is a function. We worked very hard in 2.2 to interpret the derivative of a function visually. We made the link, in Ex.

More information

Lecture 5: Rules of Differentiation. First Order Derivatives

Lecture 5: Rules of Differentiation. First Order Derivatives Lecture 5: Rules of Differentiation First order derivatives Higher order derivatives Partial differentiation Higher order partials Differentials Derivatives of implicit functions Generalized implicit function

More information

Math 265H: Calculus III Practice Midterm II: Fall 2014

Math 265H: Calculus III Practice Midterm II: Fall 2014 Name: Section #: Math 65H: alculus III Practice Midterm II: Fall 14 Instructions: This exam has 7 problems. The number of points awarded for each question is indicated in the problem. Answer each question

More information

Particular Solutions

Particular Solutions Particular Solutions Our eamples so far in this section have involved some constant of integration, K. We now move on to see particular solutions, where we know some boundar conditions and we substitute

More information

Math 320-1: Midterm 2 Practice Solutions Northwestern University, Fall 2014

Math 320-1: Midterm 2 Practice Solutions Northwestern University, Fall 2014 Math 30-: Midterm Practice Solutions Northwestern University, Fall 04. Give an eample of each of the following. Justify your answer. (a) A function on (, ) which is continuous but not uniformly continuous.

More information

APPM 2360: Midterm exam 1 February 15, 2017

APPM 2360: Midterm exam 1 February 15, 2017 APPM 36: Midterm exam 1 February 15, 17 On the front of your Bluebook write: (1) your name, () your instructor s name, (3) your recitation section number and () a grading table. Text books, class notes,

More information

VII. Techniques of Integration

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

More information

= f (x ), recalling the Chain Rule and the fact. dx = f (x )dx and. dx = x y dy dx = x ydy = xdx y dy = x dx. 2 = c

= f (x ), recalling the Chain Rule and the fact. dx = f (x )dx and. dx = x y dy dx = x ydy = xdx y dy = x dx. 2 = c Separable Variables, differential equations, and graphs of their solutions This will be an eploration of a variety of problems that occur when stuing rates of change. Many of these problems can be modeled

More information

Preface. Computing Definite Integrals. 3 cos( x) dx. x 3

Preface. Computing Definite Integrals. 3 cos( x) dx. x 3 Preface Here are the solutions to the practice problems for my Calculus I notes. Some solutions will have more or less detail than other solutions. The level of detail in each solution will depend up on

More information

First-Order ODEs. Chapter Separable Equations. We consider in this chapter differential equations of the form dy (1.1)

First-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 information

Student name: Student ID: TA s name and/or section: MATH 3B (Butler) Midterm II, 20 February 2009

Student name: Student ID: TA s name and/or section: MATH 3B (Butler) Midterm II, 20 February 2009 Student name: Student ID: TA s name and/or section: MATH 3B (Butler) Midterm II, 0 February 009 This test is closed book and closed notes. No calculator is allowed for this test. For full credit show all

More information

Linear Independence and the Wronskian

Linear 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 information

Excruciating Practice Final, Babson 21A, Fall 13.

Excruciating Practice Final, Babson 21A, Fall 13. Excruciating Practice Final Excruciating Practice Final, Babson 2A, Fall 3. This is an *optional* practice final. It has been designed with intention to be more difficult that what should be encountered

More information

1 Exponential Functions Limit Derivative Integral... 5

1 Exponential Functions Limit Derivative Integral... 5 Contents Eponential Functions 3. Limit................................................. 3. Derivative.............................................. 4.3 Integral................................................

More information

MATH 124. Midterm 2 Topics

MATH 124. Midterm 2 Topics MATH 124 Midterm 2 Topics Anything you ve learned in class (from lecture and homework) so far is fair game, but here s a list of some main topics since the first midterm that you should be familiar with:

More information

First-Order ODE: Separable Equations, Exact Equations and Integrating Factor

First-Order ODE: Separable Equations, Exact Equations and Integrating Factor First-Order ODE: Separable Equations, Exact Equations and Integrating Factor Department of Mathematics IIT Guwahati REMARK: In the last theorem of the previous lecture, you can change the open interval

More information

Multivariate Calculus Solution 1

Multivariate Calculus Solution 1 Math Camp Multivariate Calculus Solution Hessian Matrices Math Camp In st semester micro, you will solve general equilibrium models. Sometimes when solving these models it is useful to see if utility functions

More information

1. The following problems are not related: (a) (15 pts, 5 pts ea.) Find the following limits or show that they do not exist: arcsin(x)

1. The following problems are not related: (a) (15 pts, 5 pts ea.) Find the following limits or show that they do not exist: arcsin(x) APPM 5 Final Eam (5 pts) Fall. The following problems are not related: (a) (5 pts, 5 pts ea.) Find the following limits or show that they do not eist: (i) lim e (ii) lim arcsin() (b) (5 pts) Find and classify

More information

1. Sets A set is any collection of elements. Examples: - the set of even numbers between zero and the set of colors on the national flag.

1. Sets A set is any collection of elements. Examples: - the set of even numbers between zero and the set of colors on the national flag. San Francisco State University Math Review Notes Michael Bar Sets A set is any collection of elements Eamples: a A {,,4,6,8,} - the set of even numbers between zero and b B { red, white, bule} - the set

More information

Assignment 11 Assigned Mon Sept 27

Assignment 11 Assigned Mon Sept 27 Assignment Assigned Mon Sept 7 Section 7., Problem 4. x sin x dx = x cos x + x cos x dx ( = x cos x + x sin x ) sin x dx u = x dv = sin x dx du = x dx v = cos x u = x dv = cos x dx du = dx v = sin x =

More information

Solutions to Math 41 First Exam October 12, 2010

Solutions to Math 41 First Exam October 12, 2010 Solutions to Math 41 First Eam October 12, 2010 1. 13 points) Find each of the following its, with justification. If the it does not eist, eplain why. If there is an infinite it, then eplain whether it

More information

Part Two. Diagnostic Test

Part Two. Diagnostic Test Part Two Diagnostic Test AP Calculus AB and BC Diagnostic Tests Take a moment to gauge your readiness for the AP Calculus eam by taking either the AB diagnostic test or the BC diagnostic test, depending

More information

CHAPTER 7: TECHNIQUES OF INTEGRATION

CHAPTER 7: TECHNIQUES OF INTEGRATION CHAPTER 7: TECHNIQUES OF INTEGRATION DAVID GLICKENSTEIN. Introduction This semester we will be looking deep into the recesses of calculus. Some of the main topics will be: Integration: we will learn how

More information

Math Midterm Solutions

Math Midterm Solutions Math 50 - Midterm Solutions November 4, 009. a) If f ) > 0 for all in a, b), then the graph of f is concave upward on a, b). If f ) < 0 for all in a, b), then the graph of f is downward on a, b). This

More information

4.8 Partial Fraction Decomposition

4.8 Partial Fraction Decomposition 8 CHAPTER 4. INTEGRALS 4.8 Partial Fraction Decomposition 4.8. Need to Know The following material is assumed to be known for this section. If this is not the case, you will need to review it.. When are

More information

Math 2930 Worksheet Introduction to Differential Equations

Math 2930 Worksheet Introduction to Differential Equations Math 2930 Worksheet Introduction to Differential Equations Week 2 February 1st, 2019 Learning Goals Solve linear first order ODEs analytically. Solve separable first order ODEs analytically. Questions

More information

Math 222 Spring 2013 Exam 3 Review Problem Answers

Math 222 Spring 2013 Exam 3 Review Problem Answers . (a) By the Chain ule, Math Spring 3 Exam 3 eview Problem Answers w s w x x s + w y y s (y xy)() + (xy x )( ) (( s + 4t) (s 3t)( s + 4t)) ((s 3t)( s + 4t) (s 3t) ) 8s 94st + 3t (b) By the Chain ule, w

More information

MATH 1207 R02 MIDTERM EXAM 2 SOLUTION

MATH 1207 R02 MIDTERM EXAM 2 SOLUTION MATH 7 R MIDTERM EXAM SOLUTION FALL 6 - MOON Name: Write your answer neatly and show steps. Except calculators, any electronic devices including laptops and cell phones are not allowed. () (5 pts) Find

More information

Math 266, Midterm Exam 1

Math 266, Midterm Exam 1 Math 266, Midterm Exam 1 February 19th 2016 Name: Ground Rules: 1. Calculator is NOT allowed. 2. Show your work for every problem unless otherwise stated (partial credits are available). 3. You may use

More information

Definition 8.1 Two inequalities are equivalent if they have the same solution set. Add or Subtract the same value on both sides of the inequality.

Definition 8.1 Two inequalities are equivalent if they have the same solution set. Add or Subtract the same value on both sides of the inequality. 8 Inequalities Concepts: Equivalent Inequalities Linear and Nonlinear Inequalities Absolute Value Inequalities (Sections.6 and.) 8. Equivalent Inequalities Definition 8. Two inequalities are equivalent

More information

Chapter 2: First Order DE 2.4 Linear vs. Nonlinear DEs

Chapter 2: First Order DE 2.4 Linear vs. Nonlinear DEs Chapter 2: First Order DE 2.4 Linear vs. Nonlinear DEs First Order DE 2.4 Linear vs. Nonlinear DE We recall the general form of the First Oreder DEs (FODE): dy = f(t, y) (1) dt where f(t, y) is a function

More information

First Midterm Examination

First Midterm Examination Çankaya University Department of Mathematics 016-017 Fall Semester MATH 155 - Calculus for Engineering I First Midterm Eamination 1) Find the domain and range of the following functions. Eplain your solution.

More information

CHAPTER 2 First Order Equations

CHAPTER 2 First Order Equations CHAPTER 2 First Order Equations IN THIS CHAPTER we study first order equations for which there are general methods of solution. SECTION 2.1 deals with linear equations, the simplest kind of first order

More information

ACCUPLACER MATH 0311 OR MATH 0120

ACCUPLACER MATH 0311 OR MATH 0120 The University of Teas at El Paso Tutoring and Learning Center ACCUPLACER MATH 0 OR MATH 00 http://www.academics.utep.edu/tlc MATH 0 OR MATH 00 Page Factoring Factoring Eercises 8 Factoring Answer to Eercises

More information

2tdt 1 y = t2 + C y = which implies C = 1 and the solution is y = 1

2tdt 1 y = t2 + C y = which implies C = 1 and the solution is y = 1 Lectures - Week 11 General First Order ODEs & Numerical Methods for IVPs In general, nonlinear problems are much more difficult to solve than linear ones. Unfortunately many phenomena exhibit nonlinear

More information