# Section 3.5 Nonhomogeneous Equations; Method of Undetermined Coefficients

Size: px
Start display at page:

Transcription

1 Secion 3.5 Nonhomogeneous Equaions; Mehod of Undeermined Coefficiens Key Terms/Ideas: Linear Differenial operaor Nonlinear operaor Second order homogeneous DE Second order nonhomogeneous DE Soluion o homogeneous DE Paricular soluion o nonhomogeneous DE Mehod of undeermined coefficiens

2 A general second order linear DE has he form y p() y q() y g() where coefficien funcions p(), q(), and righ side g() are coninuous in some inerval I. If all he coefficiens are consan funcions, we say we have a second order linear DE wih consan coefficiens which we wrie as ay by cy g() Two imporan pieces of erminology: 1. If g() is idenically zero, hen DE is called HOMOGENEOUS. 2. If g() is no idenically zero, hen DE is called NONHOMOGENEOUS. Examples: y'' + 2 y' + 5y = 0 is a homogeneous DE y '' - 4y'+ y = cos() is a nonhomogeneous DE ( wih consan coefficiens)

3 OPERATOR NOTATION & PROPERTIES An operaor is a funcion whose domain is a se of funcions and he range consiss of (oher) funcions. For example he derivaive is an operaor. Le D(y) y'(x). The domain of operaor D is he se of funcions whose derivaive exiss, usually on a specified inerval. The oupu is anoher funcion. For second order linear DEs i is convenien o define an operaor L given by he expression L[y] y'' +py' + qy, where p and q are some funcions of. If we recall ha a second derivaive is he derivaive of he firs derivaive, hen y'' = D(D(y)). I is convenien o wrie his as y '' = D 2 (y) where we mus keep in mind ha he D 2 means compue a pair of successive derivaives. Using his D-noaion, he operaor L can be expressed as L[y] = D 2 (y) +p()d(y) + q()y or for consan coefficiens we wrie L[y] = = (ad 2 + bd - c)(y). Examples: If p() = 2 and q() = cos(), hen we have L[y] (D D + cos())(y) = y'' + 2 y' + cos()y. If a = 3, b = 2, and c = -4, hen L[y] (ad 2 + bd - c)(y) = 3y'' + 2 y' - 4y. Noe ha he operaor noaion is basically a shor hand way of expressing he lef side of he DE.

4 Recall ha The derivaive of a sum is he sum of he individual derivaives. and ha The derivaive of a consan imes a funcion is he consan imes he derivaive of he funcion. Since our operaor L is jus a sum of muliples of derivaives i follows ha for any pair of consans c 1 and c 2 and any pair of funcions y 1 () and y 2 () ha L[c 1 y 1 + c 2 y 2 ] = c 1 L[y 1 ] + c 2 L[y 2 ]. An operaor ha saisfies his propery is said o be a LINEAR OPERATOR. We say ha a linear operaor preserves linear combinaions. (Recall L has a special form L[y] y +py + qy. No every operaor involving derivaives saisfies his propery. When his propery is no saisfied he operaor is called nonlinear.

5 Imporan Propery Every second order linear homogeneous DE L[y] y +py + qy = 0 saisfies he following propery. If c 1 and c 2 are any consans and y 1 () and y 2 () any soluions of he DE, hen linear combinaion c 1 y 1 () + c 2 y 2 () is also a soluion of he DE. (Principle of Superposiion) This implies ha from a few soluions o he linear homogeneous DE we can build los of ohers by forming linear combinaions. Example: Consider he second order linear homogeneous DE given by L[y] = D 2 (y) - 3D(y) + 2y = 0. Verify ha y 1 (x) = e x and y 2 (x) = e 2x are soluions. Tha means L[y 1 ] = L[e x ] = 0 and L[y 2 ] = L[e 2x ] = 0. Since L is a linear operaor we know ha L[c 1 y 1 + c 2 y 2 ] = c 1 L[y 1 ] + c 2 L[y 2 ]. So his fac says ha since L[y 1 ] = 0 and L[y 2 ] = 0, hen so mus L[c 1 y 1 + c 2 y 2 ] = 0 for any consans c 1 and c 2. Thus every linear combinaion c 1 e x + c 2 e 2x is a soluion of he DE. Hence 26e x + 999e 2x and 45e x e 2x are soluions of his DE.

6 For now we will focus on second order nonhomogeneous DEs wih consan coefficiens. ay by cy g() We have seen ha for second order homogeneous DEs wih consan coefficiens we can find a fundamenal se of soluions y 1 () and y 2 () so ha he general soluion of ay + by +cy = 0 is given by he linear combinaion c 1 y 1 () + c 2 y 2 () where c 1 and c 2 are arbirary consans. Our sraegy o solve ay'' +by' + cy = g() is a hree-sep procedure: Sep #1. Find he general soluion y h () of he associaed homogeneous equaion ay'' +by' + cy = 0. Sep #2. Find a paricular soluion y p () of he nonhomogeneous DE ay'' +by' + cy = g(). Sep #3. Then he general soluion of he nonhomogeneous DE ay'' +by' + cy = g() is y() = y h () + y p (). To verify ha his works we do he following.

7 Le () be any soluion of L(y) ay by cy g() Le y p () be a paricular soluion of his nonhomogeneous DE. (no arbirary consans in i) Le he general soluion of he associaed homogeneous equaion be: y h (x) = C 1 y 1 () + C 2 y 2 () We have L[()] = g() and L[y p ()] = g(). Since L is a linear operaor we have L[() - y p ()] = L[()] - L[y p ()] = g() - g() = 0 This says ha () - y p () is a soluion of he associaed homogeneous equaion. Thus we mus be able o express () - y p () in erms of he y 1 () and y 2 (). Tha is, here exis some consans C 1 and C 2 so ha () - y p () = C 1 y 1 () + C 2 y 2 (). Rearranging his expression we have ha () which is any soluion of he nonhomogeneous DE can be expressed as () = C 1 y 1 () + C 2 y 2 () + y p () = y h (x) + y p () Summary: Every soluion of he nonhomogeneous DE is obained as a sum of a paricular soluion + he general soluion of he homogeneous DE; y() = y h () + y p ().

8 Examples: 1. For DE y'' y = 2 2, i is known ha a paricular soluion is y p () = 2. Find he general soluion. Characerisic equaion r 2 1 = 0 r = 1, -1 y h () = C 1 e + C 2 e - General soluion of nonhomogeneous DE is y() = y h () + y p () = C 1 e + C 2 e For DE y'' y' + y = sin(), i is known ha a paricular soluion is y p () = cos(). Find he general soluion. Characerisic equaion r 2 r + 1 = 0 General soluion of nonhomogeneous DE is 1 1 y() = y h () + y p () = i 3 r y 2 2 h() C1e cos 3 / 2 ) C2e sin 3 / 2 ) C e cos / ) C e sin / ) cos() We know how o find y h (x). We need o learn how o find y p (x).

9 Our firs sep is o find he soluion of he associaed homogeneous for he nonhomogeneous DE ay + by + cy = g(); his is, solve ay + by + cy = 0. The wo soluions ha make up he general soluion of he associaed homogeneous equaion may involve he following ypes of funcions: Polynomials 1,, Exponenial funcions e r, Trig funcions cos(k), sin(k), and Producs e r, e r cos(k), e r sin(k) One echnique for consrucing a paricular soluion y p () of ay + by + cy = g() is he mehod of undeermined coefficiens. This mehod requires us o consruc a rial expression for y p () having unknown coefficiens, subsiue i ino he DE, and solve for he unknown coefficiens involved. The major limiaion of his mehod is ha i is useful primarily for equaions for which we can easily wrie down he correc form of he paricular soluion in advance. For his reason, his mehod is usually used only for problems in which he homogeneous equaion has consan coefficiens and he nonhomogeneous erm is resriced o a relaively small class of funcions. In paricular, we consider only nonhomogeneous erms ha consis of polynomials, exponenial funcions, sines, and cosines. Despie his limiaion, he mehod of undeermined coefficiens is useful for solving many problems ha have imporan applicaions. This means ha he g() expression involves he same ype of funcion ha appears in he soluion of he associaed homogeneous DE.

10 Examples: We consider DE y 3y - 4y = g() for differen selecions of g() The associaed homogeneous DE y 3y 4y = 0 has characerisic equaion r 2 3r 4 = (r 4)(r + 1) = 0. So r = -1, 4 and y h () = C 1 e - + C 2 e 4. Case g() = 3e 2 This suggess rial soluion Y() = Ae 2 for he paricular soluion. Taking derivaives we ge Y() Ae Y'() 2Ae, Y''() 4Ae Subsiuing hese derivaives ino he differenial equaion we ge Ae 6Ae 4Ae 3e Simplify o ge 6Ae 3e A 1/ 2 Thus a paricular soluion o he nonhomogeneous DE is 1 Y() y p() e 2 2 The general soluion of y 3y - 4y = 3e 2 is y() = y h () + y p () = C 1 e - + C 2 e 4 (1/2)e 2.

11 y 3y - 4y = g() y h () = C 1 e - + C 2 e 4 Case g() = 2sin() Because he derivaive of sin() will repea in successive differeniaions, his suggess rial soluion Y() = Asin() for he paricular soluion. Taking derivaives we ge Y() Asin Y'() Acos, Y''() Asin Subsiuing hese derivaives ino he differenial equaion we ge Asin 3Acos 4Asin 2sin Simplify o ge 2 5A sin 3Acos 0 Since sin() and cos() are no muliples of one anoher he individual coefficiens mus boh be zero; ha is, 2 + 5A = 0 and 3A = 0, which is impossible. (WHY?) Our nex aemp a finding a paricular soluion Y is o use rial soluion Y() = Asin() + Bcos(). Y() A sin For derivaives we have Bcos Y () Acos Bsin, Y () A sin Bcos Subsiuing hese derivaives ino he differenial equaion we ge simplifing we ge 5A 3B sin 3A 5B cos 2sin Equaing coefficiens of like erms on eiher side of he equal sign we ge A sin Bcos Acos Bsin A sin Bcos sin 5A 3B 2, 3A 5B 0 which gives A 5 / 17, B 3 / Thus a paricular soluion o he nonhomogeneous DE is Y() y () sin cos The general soluion of y 3y - 4y = 2sin() is y() = y h () + y p () = C 1 e - + C 2 e 4 (5/17)sin() +(3/17)cos(). p

12 y 3y - 4y = g() Case g() = -8e cos(2) Subsiuing in he DE we ge y h () = C 1 e - + C 2 e 4 Because he derivaives of cos(2) will involve sin(2), his suggess rial soluion Y() = Ae cos(2) + Be sin(2) for he paricular soluion. Taking derivaives we ge 2 2 Y() Ae cos Be sin Y '() Ae cos Ae sin Be sin Be cos A 2Be cos2 2A Be sin Y () A B e cos A B e sin A B e sin A Be cos A 4B e cos2 4A 3Be sin Y () A B e cos A B e sin 3Y () ( 3) A 2B e cos 2 ( 3) 2A B e sin 2 4Y() ( 4)Ae cos 2 ( 4)Be sin 2 8e cos 2 10A 2B e cos 2 2A 10B e sin 2 The general soluion of y 3y - 4y = -8e cos(2) is y() = y h () + y p () = C 1 e - + C 2 e 4 + (10/13)e cos(2) + (2/13)e sin(2). Equaing coefficiens of like erms we ge -8 = -10A -2B 0 = 2A -10B Solving for A and B gives A = 10/13 and B = 2/ Y() e cos e sin

13 Summary: The general soluion of y 3y - 4y = g() in he hree cases are displayed nex. 1 e 2 Case g() = 3e 2 y() = y 2 h () + Y()= C 1 e - + C 2 e 4 Case g() = 2sin() y() = y h () + Y()= C 1 e - + C 2 e 4 + sin cos Case g() = -8e 10 2 cos(2) y() = y h () + Y()= C 1 e - + C 2 e 4 + e cos2 e sin Consider he equaion 2 y 3y 4y 3e 2sin 8e cos2 Our equaions o solve individually are y 3y 4y 3e 2 y 3y 4y 2sin To ge he paricular soluion of he y 3y 4y 8e cos 2 DE jus add he paricular soluion of he individual cases Y() e cos sin e cos 2 e sin So he general soluion of y 3y 4y 3e 2sin 8e cos y() = y h () + Y()= C 1 e - + C 2 e e cos sin e cos2 e sin2 is

14 Special case: If any erm in he righ side g() in he nonhomogeneous DE ay + by + cy = g() is par of he soluion o he associaed homogeneous DE ay + by + cy = 0, hen we need o muliply he erms of rial soluion by or possibly a power of. Consider he nonhomogeneous equaion y'' 3y' 4y 2e So g() = 2e -. Y The characerisic equaion of he associaed homogeneous DE is r 2-3r - 4 = (r - 4)(r + 1) =0, so r = 4, -1. Thus 4 y () C e C e h 1 2 We see ha a naural rial soluion is Y = Ae -. However he differenial operaor will be fooled ino hinking ha he paricular soluion of his form is a soluion of he associaed homogeneous equaion. To have he differenial operaor no be deceived we adjus he rial soluion o be Y = Ae -.(We muliplied by.) Now we proceed in he usual manner compuing derivaives of he rial soluion and subsiuing hem ino he nonhomogeneous DE. Equae coefficiens of like Ae 4Y 4Ae erms on from he sides of he 3Y' 3Ae 3Ae equal sign; 2 = -5A, A = -2/5. Y' Ae Ae Y'' Ae Ae Ae 2Ae Ae Y'' 2Ae Ae ADD 2e 5A e 4A 3A A e zero Paricular soln is Y() = y p () = (2/5)e - General soluion is 2 Y() y h() C e C e e

15 Always compue he soluion of he associaed homogeneous DE firs! How o consruc he "rial" paricular soluion. If in any group of erms generaed by differeniaing g() has a funcion form ha is he same as one of he funcions in he soluion of he associaed homogeneous equaion, hen we mus muliply each erm of he group by s where s is he smalles posiive power so ha all he erms of he rial soluion are disinc from he members of he soluion of he homogeneous DE.

16 Pracice wih rial soluion: DE ay + by + cy = g() has he soluion o he associaed homogeneous DE given by y h () = C 1 e 2 + C 2 e. If g() = 4cos() wha should your rial soluion be? y() = Acos() + Bsin() If g() = -2e - sin() wha should your rial soluion be? y() = Ae - cos() + Be - sin() If g() = wha should your rial soluion be? y() = A + B If g() = 3e 2 wha should your rial soluion be? y() = Ae 2 If g() = 5cos() wha should your rial soluion be? y() = Acos() + Bsin() + Ccos() +Dsin()

17 Solving an iniial value problem of he form ay + by + cy = g(), y( 0 ) = y 0, y'( 0 ) = y 0 '. Firs find y h () he general soluion o ay + by + cy = 0; y h () = C 1 y 1 () + C 2 y 2 () Nex find he paricular soluion y p () for ay + by + cy = g(). So he general soluion of ay + by + cy = g() is y() = y h () + y p () Finally use he iniial condiions o deermine C 1 and C 2. Subsiue he general soluion y() = y h () + y p () ino he DE and equae coefficiens of liked powered erms o provided a se of equaions o solve for C 1 and C 2.

18 Example: Solve he IVP y'' y' 2sin(), y( 0) 1, y'( 0) 1 The characerisic polynomial of y''+ y' = 0 is r 2 + r = 0 r = 0, -1 y h () = C 1 + C 2 e -. Applying he mehod of undeermined coefficiens we have rial soluion Y() = A sin() + B cos(). Differeniaing we have Y() A sin() Bcos() Y'() Acos() Bsin() Y''() A sin() Bcos() Subsiuing ino y'' + y' = 2sin() we have A sin() Bcos() A cos() B sin() 2sin() (A B)cos() ( A B) sin() 2 sin() collecing erms Equaing coefficiens of like erms A - B = 0 - A - B = 2 Solving for A and B A = - 1, B = - 1 So he paricular soluion is y p () = - sin() - cos() Thus he general soluion is y() y h() y p() C1C2e sin() cos() Applying he iniial condiions: y(0) = 1 1 = C 1 + C 2-1 and y'(0) = -1-1 = -C 2-1. Solving for C 1 and C 2 we have C 1 = 2 and C 2 = 0. So he soluion of he IVP is y() y () y () 2 sin() cos() h p

### Differential Equations

Mah 21 (Fall 29) Differenial Equaions Soluion #3 1. Find he paricular soluion of he following differenial equaion by variaion of parameer (a) y + y = csc (b) 2 y + y y = ln, > Soluion: (a) The corresponding

### System of Linear Differential Equations

Sysem of Linear Differenial Equaions In "Ordinary Differenial Equaions" we've learned how o solve a differenial equaion for a variable, such as: y'k5\$e K2\$x =0 solve DE yx = K 5 2 ek2 x C_C1 2\$y''C7\$y

### Some Basic Information about M-S-D Systems

Some Basic Informaion abou M-S-D Sysems 1 Inroducion We wan o give some summary of he facs concerning unforced (homogeneous) and forced (non-homogeneous) models for linear oscillaors governed by second-order,

### ME 391 Mechanical Engineering Analysis

Fall 04 ME 39 Mechanical Engineering Analsis Eam # Soluions Direcions: Open noes (including course web posings). No books, compuers, or phones. An calculaor is fair game. Problem Deermine he posiion of

### Chapter 6. Systems of First Order Linear Differential Equations

Chaper 6 Sysems of Firs Order Linear Differenial Equaions We will only discuss firs order sysems However higher order sysems may be made ino firs order sysems by a rick shown below We will have a sligh

### Chapter 2. First Order Scalar Equations

Chaper. Firs Order Scalar Equaions We sar our sudy of differenial equaions in he same way he pioneers in his field did. We show paricular echniques o solve paricular ypes of firs order differenial equaions.

### 8. Basic RL and RC Circuits

8. Basic L and C Circuis This chaper deals wih he soluions of he responses of L and C circuis The analysis of C and L circuis leads o a linear differenial equaion This chaper covers he following opics

### t is a basis for the solution space to this system, then the matrix having these solutions as columns, t x 1 t, x 2 t,... x n t x 2 t...

Mah 228- Fri Mar 24 5.6 Marix exponenials and linear sysems: The analogy beween firs order sysems of linear differenial equaions (Chaper 5) and scalar linear differenial equaions (Chaper ) is much sronger

### Lecture 13 RC/RL Circuits, Time Dependent Op Amp Circuits

Lecure 13 RC/RL Circuis, Time Dependen Op Amp Circuis RL Circuis The seps involved in solving simple circuis conaining dc sources, resisances, and one energy-sorage elemen (inducance or capaciance) are:

### ENGI 9420 Engineering Analysis Assignment 2 Solutions

ENGI 940 Engineering Analysis Assignmen Soluions 0 Fall [Second order ODEs, Laplace ransforms; Secions.0-.09]. Use Laplace ransforms o solve he iniial value problem [0] dy y, y( 0) 4 d + [This was Quesion

### dt = C exp (3 ln t 4 ). t 4 W = C exp ( ln(4 t) 3) = C(4 t) 3.

Mah Rahman Exam Review Soluions () Consider he IVP: ( 4)y 3y + 4y = ; y(3) = 0, y (3) =. (a) Please deermine he longes inerval for which he IVP is guaraneed o have a unique soluion. Soluion: The disconinuiies

### Two Coupled Oscillators / Normal Modes

Lecure 3 Phys 3750 Two Coupled Oscillaors / Normal Modes Overview and Moivaion: Today we ake a small, bu significan, sep owards wave moion. We will no ye observe waves, bu his sep is imporan in is own

### Second Order Linear Differential Equations

Second Order Linear Differenial Equaions Second order linear equaions wih consan coefficiens; Fundamenal soluions; Wronskian; Exisence and Uniqueness of soluions; he characerisic equaion; soluions of homogeneous

### EXERCISES FOR SECTION 1.5

1.5 Exisence and Uniqueness of Soluions 43 20. 1 v c 21. 1 v c 1 2 4 6 8 10 1 2 2 4 6 8 10 Graph of approximae soluion obained using Euler s mehod wih = 0.1. Graph of approximae soluion obained using Euler

### 10. State Space Methods

. Sae Space Mehods. Inroducion Sae space modelling was briefly inroduced in chaper. Here more coverage is provided of sae space mehods before some of heir uses in conrol sysem design are covered in he

### Complete solutions to Exercise 14(b) 1. Very similar to EXAMPLE 4. We have same characteristic equation:

Soluions 4(b) Complee soluions o Exercise 4(b). Very similar o EXAMPE 4. We have same characerisic equaion: 5 i Ae = + Be By using he given iniial condiions we obain he simulaneous equaions A+ B= 0 5A

### Undetermined coefficients for local fractional differential equations

Available online a www.isr-publicaions.com/jmcs J. Mah. Compuer Sci. 16 (2016), 140 146 Research Aricle Undeermined coefficiens for local fracional differenial equaions Roshdi Khalil a,, Mohammed Al Horani

### 2.9 Modeling: Electric Circuits

SE. 2.9 Modeling: Elecric ircuis 93 2.9 Modeling: Elecric ircuis Designing good models is a ask he compuer canno do. Hence seing up models has become an imporan ask in modern applied mahemaics. The bes

### Solutions for homework 12

y Soluions for homework Secion Nonlinear sysems: The linearizaion of a nonlinear sysem Consider he sysem y y y y y (i) Skech he nullclines Use a disincive marking for each nullcline so hey can be disinguished

### Second Order Linear Differential Equations

Second Order Linear Differenial Equaions Second order linear equaions wih consan coefficiens; Fundamenal soluions; Wronskian; Exisence and Uniqueness of soluions; he characerisic equaion; soluions of homogeneous

### SOLUTIONS TO ECE 3084

SOLUTIONS TO ECE 384 PROBLEM 2.. For each sysem below, specify wheher or no i is: (i) memoryless; (ii) causal; (iii) inverible; (iv) linear; (v) ime invarian; Explain your reasoning. If he propery is no

### HOMEWORK # 2: MATH 211, SPRING Note: This is the last solution set where I will describe the MATLAB I used to make my pictures.

HOMEWORK # 2: MATH 2, SPRING 25 TJ HITCHMAN Noe: This is he las soluion se where I will describe he MATLAB I used o make my picures.. Exercises from he ex.. Chaper 2.. Problem 6. We are o show ha y() =

### ODEs II, Lecture 1: Homogeneous Linear Systems - I. Mike Raugh 1. March 8, 2004

ODEs II, Lecure : Homogeneous Linear Sysems - I Mike Raugh March 8, 4 Inroducion. In he firs lecure we discussed a sysem of linear ODEs for modeling he excreion of lead from he human body, saw how o ransform

### ln 2 1 ln y x c y C x

Lecure 14 Appendi B: Some sample problems from Boas Here are some soluions o he sample problems assigned for Chaper 8 8: 6 Soluion: We wan o find he soluion o he following firs order equaion using separaion

### Math 2142 Exam 1 Review Problems. x 2 + f (0) 3! for the 3rd Taylor polynomial at x = 0. To calculate the various quantities:

Mah 4 Eam Review Problems Problem. Calculae he 3rd Taylor polynomial for arcsin a =. Soluion. Le f() = arcsin. For his problem, we use he formula f() + f () + f ()! + f () 3! for he 3rd Taylor polynomial

### Solutions of Sample Problems for Third In-Class Exam Math 246, Spring 2011, Professor David Levermore

Soluions of Sample Problems for Third In-Class Exam Mah 6, Spring, Professor David Levermore Compue he Laplace ransform of f e from is definiion Soluion The definiion of he Laplace ransform gives L[f]s

### d 1 = c 1 b 2 - b 1 c 2 d 2 = c 1 b 3 - b 1 c 3

and d = c b - b c c d = c b - b c c This process is coninued unil he nh row has been compleed. The complee array of coefficiens is riangular. Noe ha in developing he array an enire row may be divided or

### Boyce/DiPrima/Meade 11 th ed, Ch 3.1: 2 nd Order Linear Homogeneous Equations-Constant Coefficients

Boce/DiPrima/Meade h ed, Ch 3.: nd Order Linear Homogeneous Equaions-Consan Coefficiens Elemenar Differenial Equaions and Boundar Value Problems, h ediion, b William E. Boce, Richard C. DiPrima, and Doug

### LAPLACE TRANSFORM AND TRANSFER FUNCTION

CHBE320 LECTURE V LAPLACE TRANSFORM AND TRANSFER FUNCTION Professor Dae Ryook Yang Spring 2018 Dep. of Chemical and Biological Engineering 5-1 Road Map of he Lecure V Laplace Transform and Transfer funcions

### 2.7. Some common engineering functions. Introduction. Prerequisites. Learning Outcomes

Some common engineering funcions 2.7 Inroducion This secion provides a caalogue of some common funcions ofen used in Science and Engineering. These include polynomials, raional funcions, he modulus funcion

### Challenge Problems. DIS 203 and 210. March 6, (e 2) k. k(k + 2). k=1. f(x) = k(k + 2) = 1 x k

Challenge Problems DIS 03 and 0 March 6, 05 Choose one of he following problems, and work on i in your group. Your goal is o convince me ha your answer is correc. Even if your answer isn compleely correc,

### ( ) 2. Review Exercise 2. cos θ 2 3 = = 2 tan. cos. 2 x = = x a. Since π π, = 2. sin = = 2+ = = cotx. 2 sin θ 2+

Review Eercise sin 5 cos sin an cos 5 5 an 5 9 co 0 a sinθ 6 + 4 6 + sin θ 4 6+ + 6 + 4 cos θ sin θ + 4 4 sin θ + an θ cos θ ( ) + + + + Since π π, < θ < anθ should be negaive. anθ ( + ) Pearson Educaion

### 6.2 Transforms of Derivatives and Integrals.

SEC. 6.2 Transforms of Derivaives and Inegrals. ODEs 2 3 33 39 23. Change of scale. If l( f ()) F(s) and c is any 33 45 APPLICATION OF s-shifting posiive consan, show ha l( f (c)) F(s>c)>c (Hin: In Probs.

### THE BERNOULLI NUMBERS. t k. = lim. = lim = 1, d t B 1 = lim. 1+e t te t = lim t 0 (e t 1) 2. = lim = 1 2.

THE BERNOULLI NUMBERS The Bernoulli numbers are defined here by he exponenial generaing funcion ( e The firs one is easy o compue: (2 and (3 B 0 lim 0 e lim, 0 e ( d B lim 0 d e +e e lim 0 (e 2 lim 0 2(e

### Math Final Exam Solutions

Mah 246 - Final Exam Soluions Friday, July h, 204 () Find explici soluions and give he inerval of definiion o he following iniial value problems (a) ( + 2 )y + 2y = e, y(0) = 0 Soluion: In normal form,

### Chapter 7: Solving Trig Equations

Haberman MTH Secion I: The Trigonomeric Funcions Chaper 7: Solving Trig Equaions Le s sar by solving a couple of equaions ha involve he sine funcion EXAMPLE a: Solve he equaion sin( ) The inverse funcions

### KEY. Math 334 Midterm III Fall 2008 sections 001 and 003 Instructor: Scott Glasgow

KEY Mah 334 Miderm III Fall 28 secions and 3 Insrucor: Sco Glasgow Please do NOT wrie on his exam. No credi will be given for such work. Raher wrie in a blue book, or on your own paper, preferably engineering

### Chapter 3 Boundary Value Problem

Chaper 3 Boundary Value Problem A boundary value problem (BVP) is a problem, ypically an ODE or a PDE, which has values assigned on he physical boundary of he domain in which he problem is specified. Le

### 2. Nonlinear Conservation Law Equations

. Nonlinear Conservaion Law Equaions One of he clear lessons learned over recen years in sudying nonlinear parial differenial equaions is ha i is generally no wise o ry o aack a general class of nonlinear

### Solutions to Assignment 1

MA 2326 Differenial Equaions Insrucor: Peronela Radu Friday, February 8, 203 Soluions o Assignmen. Find he general soluions of he following ODEs: (a) 2 x = an x Soluion: I is a separable equaion as we

### INDEX. Transient analysis 1 Initial Conditions 1

INDEX Secion Page Transien analysis 1 Iniial Condiions 1 Please inform me of your opinion of he relaive emphasis of he review maerial by simply making commens on his page and sending i o me a: Frank Mera

### MA 214 Calculus IV (Spring 2016) Section 2. Homework Assignment 1 Solutions

MA 14 Calculus IV (Spring 016) Secion Homework Assignmen 1 Soluions 1 Boyce and DiPrima, p 40, Problem 10 (c) Soluion: In sandard form he given firs-order linear ODE is: An inegraing facor is given by

### Math Week 14 April 16-20: sections first order systems of linear differential equations; 7.4 mass-spring systems.

Mah 2250-004 Week 4 April 6-20 secions 7.-7.3 firs order sysems of linear differenial equaions; 7.4 mass-spring sysems. Mon Apr 6 7.-7.2 Sysems of differenial equaions (7.), and he vecor Calculus we need

### 23.2. Representing Periodic Functions by Fourier Series. Introduction. Prerequisites. Learning Outcomes

Represening Periodic Funcions by Fourier Series 3. Inroducion In his Secion we show how a periodic funcion can be expressed as a series of sines and cosines. We begin by obaining some sandard inegrals

### Predator - Prey Model Trajectories and the nonlinear conservation law

Predaor - Prey Model Trajecories and he nonlinear conservaion law James K. Peerson Deparmen of Biological Sciences and Deparmen of Mahemaical Sciences Clemson Universiy Ocober 28, 213 Ouline Drawing Trajecories

### MA Study Guide #1

MA 66 Su Guide #1 (1) Special Tpes of Firs Order Equaions I. Firs Order Linear Equaion (FOL): + p() = g() Soluion : = 1 µ() [ ] µ()g() + C, where µ() = e p() II. Separable Equaion (SEP): dx = h(x) g()

### Morning Time: 1 hour 30 minutes Additional materials (enclosed):

ADVANCED GCE 78/0 MATHEMATICS (MEI) Differenial Equaions THURSDAY JANUARY 008 Morning Time: hour 30 minues Addiional maerials (enclosed): None Addiional maerials (required): Answer Bookle (8 pages) Graph

### CHAPTER 12 DIRECT CURRENT CIRCUITS

CHAPTER 12 DIRECT CURRENT CIUITS DIRECT CURRENT CIUITS 257 12.1 RESISTORS IN SERIES AND IN PARALLEL When wo resisors are conneced ogeher as shown in Figure 12.1 we said ha hey are conneced in series. As

### Section 7.4 Modeling Changing Amplitude and Midline

488 Chaper 7 Secion 7.4 Modeling Changing Ampliude and Midline While sinusoidal funcions can model a variey of behaviors, i is ofen necessary o combine sinusoidal funcions wih linear and exponenial curves

### Math Week 15: Section 7.4, mass-spring systems. These are notes for Monday. There will also be course review notes for Tuesday, posted later.

Mah 50-004 Week 5: Secion 7.4, mass-spring sysems. These are noes for Monday. There will also be course review noes for Tuesday, posed laer. Mon Apr 3 7.4 mass-spring sysems. Announcemens: Warm up exercise:

### Chapter Three Systems of Linear Differential Equations

Chaper Three Sysems of Linear Differenial Equaions In his chaper we are going o consier sysems of firs orer orinary ifferenial equaions. These are sysems of he form x a x a x a n x n x a x a x a n x n

### Echocardiography Project and Finite Fourier Series

Echocardiography Projec and Finie Fourier Series 1 U M An echocardiagram is a plo of how a porion of he hear moves as he funcion of ime over he one or more hearbea cycles If he hearbea repeas iself every

### t + t sin t t cos t sin t. t cos t sin t dt t 2 = exp 2 log t log(t cos t sin t) = Multiplying by this factor and then integrating, we conclude that

ODEs, Homework #4 Soluions. Check ha y ( = is a soluion of he second-order ODE ( cos sin y + y sin y sin = 0 and hen use his fac o find all soluions of he ODE. When y =, we have y = and also y = 0, so

### Math 10B: Mock Mid II. April 13, 2016

Name: Soluions Mah 10B: Mock Mid II April 13, 016 1. ( poins) Sae, wih jusificaion, wheher he following saemens are rue or false. (a) If a 3 3 marix A saisfies A 3 A = 0, hen i canno be inverible. True.

### ( ) ( ) ( ) () () Signals And Systems Exam#1. 1. Given x(t) and y(t) below: x(t) y(t) (A) Give the expression of x(t) in terms of step functions.

Signals And Sysems Exam#. Given x() and y() below: x() y() 4 4 (A) Give he expression of x() in erms of sep funcions. (%) x () = q() q( ) + q( 4) (B) Plo x(.5). (%) x() g() = x( ) h() = g(. 5) = x(. 5)

### THE 2-BODY PROBLEM. FIGURE 1. A pair of ellipses sharing a common focus. (c,b) c+a ROBERT J. VANDERBEI

THE 2-BODY PROBLEM ROBERT J. VANDERBEI ABSTRACT. In his shor noe, we show ha a pair of ellipses wih a common focus is a soluion o he 2-body problem. INTRODUCTION. Solving he 2-body problem from scrach

### MATH 4330/5330, Fourier Analysis Section 6, Proof of Fourier s Theorem for Pointwise Convergence

MATH 433/533, Fourier Analysis Secion 6, Proof of Fourier s Theorem for Poinwise Convergence Firs, some commens abou inegraing periodic funcions. If g is a periodic funcion, g(x + ) g(x) for all real x,

### ECE 2100 Circuit Analysis

ECE 1 Circui Analysis Lesson 35 Chaper 8: Second Order Circuis Daniel M. Liynski, Ph.D. ECE 1 Circui Analysis Lesson 3-34 Chaper 7: Firs Order Circuis (Naural response RC & RL circuis, Singulariy funcions,

### Oscillation of an Euler Cauchy Dynamic Equation S. Huff, G. Olumolode, N. Pennington, and A. Peterson

PROCEEDINGS OF THE FOURTH INTERNATIONAL CONFERENCE ON DYNAMICAL SYSTEMS AND DIFFERENTIAL EQUATIONS May 4 7, 00, Wilmingon, NC, USA pp 0 Oscillaion of an Euler Cauchy Dynamic Equaion S Huff, G Olumolode,

### Chapter 8 The Complete Response of RL and RC Circuits

Chaper 8 The Complee Response of RL and RC Circuis Seoul Naional Universiy Deparmen of Elecrical and Compuer Engineering Wha is Firs Order Circuis? Circuis ha conain only one inducor or only one capacior

### Signal and System (Chapter 3. Continuous-Time Systems)

Signal and Sysem (Chaper 3. Coninuous-Time Sysems) Prof. Kwang-Chun Ho kwangho@hansung.ac.kr Tel: 0-760-453 Fax:0-760-4435 1 Dep. Elecronics and Informaion Eng. 1 Nodes, Branches, Loops A nework wih b

### After the completion of this section the student. Theory of Linear Systems of ODEs. Autonomous Systems. Review Questions and Exercises

Chaper V ODE V.5 Sysems of Ordinary Differenial Equaions 45 V.5 SYSTEMS OF FIRST ORDER LINEAR ODEs Objecives: Afer he compleion of his secion he suden - should recall he definiion of a sysem of linear

### Math 334 Fall 2011 Homework 11 Solutions

Dec. 2, 2 Mah 334 Fall 2 Homework Soluions Basic Problem. Transform he following iniial value problem ino an iniial value problem for a sysem: u + p()u + q() u g(), u() u, u () v. () Soluion. Le v u. Then

### 20. Applications of the Genetic-Drift Model

0. Applicaions of he Geneic-Drif Model 1) Deermining he probabiliy of forming any paricular combinaion of genoypes in he nex generaion: Example: If he parenal allele frequencies are p 0 = 0.35 and q 0

### Math 334 Test 1 KEY Spring 2010 Section: 001. Instructor: Scott Glasgow Dates: May 10 and 11.

1 Mah 334 Tes 1 KEY Spring 21 Secion: 1 Insrucor: Sco Glasgow Daes: Ma 1 and 11. Do NOT wrie on his problem saemen bookle, excep for our indicaion of following he honor code jus below. No credi will be

### Chapter #1 EEE8013 EEE3001. Linear Controller Design and State Space Analysis

Chaper EEE83 EEE3 Chaper # EEE83 EEE3 Linear Conroller Design and Sae Space Analysis Ordinary Differenial Equaions.... Inroducion.... Firs Order ODEs... 3. Second Order ODEs... 7 3. General Maerial...

### 1 1 + x 2 dx. tan 1 (2) = ] ] x 3. Solution: Recall that the given integral is improper because. x 3. 1 x 3. dx = lim dx.

. Use Simpson s rule wih n 4 o esimae an () +. Soluion: Since we are using 4 seps, 4 Thus we have [ ( ) f() + 4f + f() + 4f 3 [ + 4 4 6 5 + + 4 4 3 + ] 5 [ + 6 6 5 + + 6 3 + ]. 5. Our funcion is f() +.

### Second-Order Differential Equations

WWW Problems and Soluions 3.1 Chaper 3 Second-Order Differenial Equaions Secion 3.1 Springs: Linear and Nonlinear Models www m Problem 3. (NonlinearSprings). A bod of mass m is aached o a wall b means

### = ( ) ) or a system of differential equations with continuous parametrization (T = R

XIII. DIFFERENCE AND DIFFERENTIAL EQUATIONS Ofen funcions, or a sysem of funcion, are paramerized in erms of some variable, usually denoed as and inerpreed as ime. The variable is wrien as a funcion of

### KEY. Math 334 Midterm III Winter 2008 section 002 Instructor: Scott Glasgow

KEY Mah 334 Miderm III Winer 008 secion 00 Insrucor: Sco Glasgow Please do NOT wrie on his exam. No credi will be given for such work. Raher wrie in a blue book, or on your own paper, preferably engineering

### Voltage/current relationship Stored Energy. RL / RC circuits Steady State / Transient response Natural / Step response

Review Capaciors/Inducors Volage/curren relaionship Sored Energy s Order Circuis RL / RC circuis Seady Sae / Transien response Naural / Sep response EE4 Summer 5: Lecure 5 Insrucor: Ocavian Florescu Lecure

### ( ) a system of differential equations with continuous parametrization ( T = R + These look like, respectively:

XIII. DIFFERENCE AND DIFFERENTIAL EQUATIONS Ofen funcions, or a sysem of funcion, are paramerized in erms of some variable, usually denoed as and inerpreed as ime. The variable is wrien as a funcion of

### MATH 128A, SUMMER 2009, FINAL EXAM SOLUTION

MATH 28A, SUMME 2009, FINAL EXAM SOLUTION BENJAMIN JOHNSON () (8 poins) [Lagrange Inerpolaion] (a) (4 poins) Le f be a funcion defined a some real numbers x 0,..., x n. Give a defining equaion for he Lagrange

### Matlab and Python programming: how to get started

Malab and Pyhon programming: how o ge sared Equipping readers he skills o wrie programs o explore complex sysems and discover ineresing paerns from big daa is one of he main goals of his book. In his chaper,

### 14 Autoregressive Moving Average Models

14 Auoregressive Moving Average Models In his chaper an imporan parameric family of saionary ime series is inroduced, he family of he auoregressive moving average, or ARMA, processes. For a large class

### Exam 1 Solutions. 1 Question 1. February 10, Part (A) 1.2 Part (B) To find equilibrium solutions, set P (t) = C = dp

Exam Soluions Februar 0, 05 Quesion. Par (A) To find equilibrium soluions, se P () = C = = 0. This implies: = P ( P ) P = P P P = P P = P ( + P ) = 0 The equilibrium soluion are hus P () = 0 and P () =..

### Finish reading Chapter 2 of Spivak, rereading earlier sections as necessary. handout and fill in some missing details!

MAT 257, Handou 6: Ocober 7-2, 20. I. Assignmen. Finish reading Chaper 2 of Spiva, rereading earlier secions as necessary. handou and fill in some missing deails! II. Higher derivaives. Also, read his

### 15. Vector Valued Functions

1. Vecor Valued Funcions Up o his poin, we have presened vecors wih consan componens, for example, 1, and,,4. However, we can allow he componens of a vecor o be funcions of a common variable. For example,

### The Asymptotic Behavior of Nonoscillatory Solutions of Some Nonlinear Dynamic Equations on Time Scales

Advances in Dynamical Sysems and Applicaions. ISSN 0973-5321 Volume 1 Number 1 (2006, pp. 103 112 c Research India Publicaions hp://www.ripublicaion.com/adsa.hm The Asympoic Behavior of Nonoscillaory Soluions

### Notes 04 largely plagiarized by %khc

Noes 04 largely plagiarized by %khc Convoluion Recap Some ricks: x() () =x() x() (, 0 )=x(, 0 ) R ț x() u() = x( )d x() () =ẋ() This hen ells us ha an inegraor has impulse response h() =u(), and ha a differeniaor

### 23.5. Half-Range Series. Introduction. Prerequisites. Learning Outcomes

Half-Range Series 2.5 Inroducion In his Secion we address he following problem: Can we find a Fourier series expansion of a funcion defined over a finie inerval? Of course we recognise ha such a funcion

### Continuous Time. Time-Domain System Analysis. Impulse Response. Impulse Response. Impulse Response. Impulse Response. ( t) + b 0.

Time-Domain Sysem Analysis Coninuous Time. J. Robers - All Righs Reserved. Edied by Dr. Rober Akl 1. J. Robers - All Righs Reserved. Edied by Dr. Rober Akl 2 Le a sysem be described by a 2 y ( ) + a 1

### EE100 Lab 3 Experiment Guide: RC Circuits

I. Inroducion EE100 Lab 3 Experimen Guide: A. apaciors A capacior is a passive elecronic componen ha sores energy in he form of an elecrosaic field. The uni of capaciance is he farad (coulomb/vol). Pracical

### Concourse Math Spring 2012 Worked Examples: Matrix Methods for Solving Systems of 1st Order Linear Differential Equations

Concourse Mah 80 Spring 0 Worked Examples: Marix Mehods for Solving Sysems of s Order Linear Differenial Equaions The Main Idea: Given a sysem of s order linear differenial equaions d x d Ax wih iniial

### Wall. x(t) f(t) x(t = 0) = x 0, t=0. which describes the motion of the mass in absence of any external forcing.

MECHANICS APPLICATIONS OF SECOND-ORDER ODES 7 Mechanics applicaions of second-order ODEs Second-order linear ODEs wih consan coefficiens arise in many physical applicaions. One physical sysems whose behaviour

### Vanishing Viscosity Method. There are another instructive and perhaps more natural discontinuous solutions of the conservation law

Vanishing Viscosiy Mehod. There are anoher insrucive and perhaps more naural disconinuous soluions of he conservaion law (1 u +(q(u x 0, he so called vanishing viscosiy mehod. This mehod consiss in viewing

### dy dx = xey (a) y(0) = 2 (b) y(1) = 2.5 SOLUTION: See next page

Assignmen 1 MATH 2270 SOLUTION Please wrie ou complee soluions for each of he following 6 problems (one more will sill be added). You may, of course, consul wih your classmaes, he exbook or oher resources,

### SMT 2014 Calculus Test Solutions February 15, 2014 = 3 5 = 15.

SMT Calculus Tes Soluions February 5,. Le f() = and le g() =. Compue f ()g (). Answer: 5 Soluion: We noe ha f () = and g () = 6. Then f ()g () =. Plugging in = we ge f ()g () = 6 = 3 5 = 5.. There is a

### Theory of! Partial Differential Equations-I!

hp://users.wpi.edu/~grear/me61.hml! Ouline! Theory o! Parial Dierenial Equaions-I! Gréar Tryggvason! Spring 010! Basic Properies o PDE!! Quasi-linear Firs Order Equaions! - Characerisics! - Linear and

### Announcements: Warm-up Exercise:

Fri Apr 13 7.1 Sysems of differenial equaions - o model muli-componen sysems via comparmenal analysis hp//en.wikipedia.org/wiki/muli-comparmen_model Announcemens Warm-up Exercise Here's a relaively simple

### u(x) = e x 2 y + 2 ) Integrate and solve for x (1 + x)y + y = cos x Answer: Divide both sides by 1 + x and solve for y. y = x y + cos x

. 1 Mah 211 Homework #3 February 2, 2001 2.4.3. y + (2/x)y = (cos x)/x 2 Answer: Compare y + (2/x) y = (cos x)/x 2 wih y = a(x)x + f(x)and noe ha a(x) = 2/x. Consequenly, an inegraing facor is found wih

### Mon Apr 9 EP 7.6 Convolutions and Laplace transforms. Announcements: Warm-up Exercise:

Mah 225-4 Week 3 April 9-3 EP 7.6 - convoluions; 6.-6.2 - eigenvalues, eigenvecors and diagonalizabiliy; 7. - sysems of differenial equaions. Mon Apr 9 EP 7.6 Convoluions and Laplace ransforms. Announcemens:

### Properties Of Solutions To A Generalized Liénard Equation With Forcing Term

Applied Mahemaics E-Noes, 8(28), 4-44 c ISSN 67-25 Available free a mirror sies of hp://www.mah.nhu.edu.w/ amen/ Properies Of Soluions To A Generalized Liénard Equaion Wih Forcing Term Allan Kroopnick

### EECE251. Circuit Analysis I. Set 4: Capacitors, Inductors, and First-Order Linear Circuits

EEE25 ircui Analysis I Se 4: apaciors, Inducors, and Firs-Order inear ircuis Shahriar Mirabbasi Deparmen of Elecrical and ompuer Engineering Universiy of Briish olumbia shahriar@ece.ubc.ca Overview Passive

### Class Meeting # 10: Introduction to the Wave Equation

MATH 8.5 COURSE NOTES - CLASS MEETING # 0 8.5 Inroducion o PDEs, Fall 0 Professor: Jared Speck Class Meeing # 0: Inroducion o he Wave Equaion. Wha is he wave equaion? The sandard wave equaion for a funcion

### Math 333 Problem Set #2 Solution 14 February 2003

Mah 333 Problem Se #2 Soluion 14 February 2003 A1. Solve he iniial value problem dy dx = x2 + e 3x ; 2y 4 y(0) = 1. Soluion: This is separable; we wrie 2y 4 dy = x 2 + e x dx and inegrae o ge The iniial

### Review - Quiz # 1. 1 g(y) dy = f(x) dx. y x. = u, so that y = xu and dy. dx (Sometimes you may want to use the substitution x y

Review - Quiz # 1 (1) Solving Special Tpes of Firs Order Equaions I. Separable Equaions (SE). d = f() g() Mehod of Soluion : 1 g() d = f() (The soluions ma be given implicil b he above formula. Remember,

### Guest Lectures for Dr. MacFarlane s EE3350 Part Deux

Gues Lecures for Dr. MacFarlane s EE3350 Par Deux Michael Plane Mon., 08-30-2010 Wrie name in corner. Poin ou his is a review, so I will go faser. Remind hem o go lisen o online lecure abou geing an A