[PDF] Differential Equations I



Previous PDF Next PDF







Dérivation – Fiche de cours

Dérivation – Fiche de cours 1 Sécante et tangente à la courbe a Sécante à la courbe Une sécante à la courbe est définie comme une droite passant par 2 points de la courbe : exemple la droite (AM) est une sécante à Cf Le coefficient directeur d’une sécante à la courbe s’appelle le taux d’accroissement On définit : t(h)= f



Differential Equations I

2 CHAPTER 1 INTRODUCTION Example 1 2 The function y = sin(x) is a solution of dy dx 3 + d4y dx4 +y = 2sin(x)+cos3(x) on domain R; the function z = ex cos(y) is a solution of ∂ 2z



Cours de Mathématiques 2 première partie : Analyse 2

dans le cours Certains collègues commencent ce cours directement avec la définition de la primitive d’une fonction, et R b a f(x)dx := F(b) − F(a) Ainsi, le théorème fondamental de l’analyse, qui établit le lien entre l’intégration et la dérivation, devient trivial



Contrôle de mathématiques

Questions de cours (3 points) 1) Donner la définition analytique du nombre dérivé d’une fon ction f en a 2) En utilisant le taux d’accroissement d’une fonction f en x : f(x +h) − f(x) h, démontrer que la fonction dérivée de √ x est la fonction 1 2 √ x pour x ∈]0 ;+∞[ Exercice2 Nombre dérivé (2 points)



Intégrales généralisées - AlloSchool

l’intégrale sur un segment, et a donc été étudié dans le chapitre Dérivation et intégration des fonctions de R dans K Enfin, K désigne R ou C I Convergence des intégrales généralisées 1 Définitions Soit f: I→K une fonction continue par morceaux •Si I= [a,b[, on dit que l’intégrale généralisée Z b a



Analyse et Algèbre - الموقع الأول للدراسة

Le contenu du cours est inspiré des manuels de mathématiques couramment utilisés, ainsi que du cours que j'ai enseigné de 2007 à 2013 pour les étudiants de première année L M D du domaine S T au sein du Département de Technologie de la acultéF de Technologie



CH 9 LE CIRCUIT ÉLECTRIQUE – exercices - correction

SAVOIR SON COURS CH 9 LE CIRCUIT ÉLECTRIQUE – exercices - correction Avec moteur : Avec une diode : Avec deux diodes : Le sens du courant est du + vers le - à l’extérieur du générateur Le moteur tourne dans les cas « a » et « c » Si on inverse le sens de branchement de la pile, on inverse le sens du courant et donc le sens de



XML Cours et exercices - Training Brussels

Cours et exercices en UML 2 avec PHP, Java, Python, C# et C++ N°12084, 3e édition 2007, 520 pages (collection Noire) X Blanc, I Mounier – UML 2 pour les développeurs N°12029, 2006, 202 pages A Tasso – Le livre de Java premier langage N°11994, 4e édition 2006, 472 pages, avec CD-Rom P Roques – UML 2 par la pratique



Langages formels, Calculabilité et Complexité

de cours (cf sections 1 11 et 3 8 2 par exemple) Les automates se sont aussi imposés comme un outil incontournable d’un point de vue pratique Tout éditeur de texte un peu évolué comprend une fonc-tion de recherche à partir d’une expression rationnelle Cette recherche com-



LA CHOLÉCYSTECTOMIE LAPAROSCOPIQUE AU BURKINA FASO : A PROPOS

dérivation cholédoco-jéjunale par une anse en Y Les suites L'extraction des pièces opératoires au cours des Il s'agit d'une étude rétrospective des 32 premiè res cholécystectomies

[PDF] personnage libertin littérature

[PDF] mouvement libertin litterature

[PDF] la gamme d'amour watteau analyse

[PDF] auteur libertin du 18eme siecle

[PDF] le faux pas watteau analyse

[PDF] les liaisons dangereuses

[PDF] libertin dom juan

[PDF] activité documentaire sur les ions 3ème

[PDF] tp linux corrigé

[PDF] differences linux unix

[PDF] exercice corrigé commande linux pdf

[PDF] cours systeme d'exploitation unix pdf

[PDF] tp linux avec correction

[PDF] examen linux avec correction

[PDF] la différence entre linux et unix wikipedia

Differential Equations I

MATB44H3F

Version September 15, 2011-1949

ii

Contents

1 Introduction1

1.1 Preliminaries. . . . . . . . . . . . . . . . . . . . . . . . . . . . .1

1.2 Sample Application of Differential Equations. . . . . . . . . . .2

2 First Order Ordinary Differential Equations5

2.1 Separable Equations. . . . . . . . . . . . . . . . . . . . . . . . .5

2.2 Exact Differential Equations. . . . . . . . . . . . . . . . . . . . .7

2.3 Integrating Factors. . . . . . . . . . . . . . . . . . . . . . . . . .11

2.4 Linear First Order Equations. . . . . . . . . . . . . . . . . . . .14

2.5 Substitutions. . . . . . . . . . . . . . . . . . . . . . . . . . . . .17

2.5.1 Bernoulli Equation. . . . . . . . . . . . . . . . . . . . . .17

2.5.2 Homogeneous Equations. . . . . . . . . . . . . . . . . . .19

2.5.3 Substitution to Reduce Second Order Equations to First

Order. . . . . . . . . . . . . . . . . . . . . . . . . . . . .20

3 Applications and Examples of First Orderode"s25

3.1 Orthogonal Trajectories. . . . . . . . . . . . . . . . . . . . . . .25

3.2 Exponential Growth and Decay. . . . . . . . . . . . . . . . . . .27

3.3 Population Growth. . . . . . . . . . . . . . . . . . . . . . . . . .28

3.4 Predator-Prey Models. . . . . . . . . . . . . . . . . . . . . . . .29

3.5 Newton"s Law of Cooling. . . . . . . . . . . . . . . . . . . . . .30

3.6 Water Tanks. . . . . . . . . . . . . . . . . . . . . . . . . . . . .31

3.7 Motion of Objects Falling Under Gravity with Air Resistance. .34

3.8 Escape Velocity. . . . . . . . . . . . . . . . . . . . . . . . . . . .36

3.9 Planetary Motion. . . . . . . . . . . . . . . . . . . . . . . . . . .37

3.10 Particle Moving on a Curve. . . . . . . . . . . . . . . . . . . . .39

iii ivCONTENTS4 Linear Differential Equations45

4.1 Homogeneous Linear Equations. . . . . . . . . . . . . . . . . . .47

4.1.1 Linear Differential Equations with Constant Coefficients.52

4.2 Nonhomogeneous Linear Equations. . . . . . . . . . . . . . . . .54

5 Second Order Linear Equations57

5.1 Reduction of Order. . . . . . . . . . . . . . . . . . . . . . . . . .57

5.2 Undetermined Coefficients. . . . . . . . . . . . . . . . . . . . . .60

5.2.1 Shortcuts for Undetermined Coefficients. . . . . . . . . .64

5.3 Variation of Parameters. . . . . . . . . . . . . . . . . . . . . . .66

6 Applications of Second Order Differential Equations71

6.1 Motion of Object Hanging from a Spring. . . . . . . . . . . . . .71

6.2 Electrical Circuits. . . . . . . . . . . . . . . . . . . . . . . . . .75

7 Higher Order Linear Differential Equations79

7.1 Undetermined Coefficients. . . . . . . . . . . . . . . . . . . . . .79

7.2 Variation of Parameters. . . . . . . . . . . . . . . . . . . . . . .80

7.3 Substitutions: Euler"s Equation. . . . . . . . . . . . . . . . . . .82

8 Power Series Solutions to Linear Differential Equations85

8.1 Introduction. . . . . . . . . . . . . . . . . . . . . . . . . . . . . .85

8.2 Background Knowledge Concerning Power Series. . . . . . . . .88

8.3 Analytic Equations. . . . . . . . . . . . . . . . . . . . . . . . . .89

8.4 Power Series Solutions: Levels of Success. . . . . . . . . . . . . .91

8.5 Level 1: Finding a finite number of coefficients. . . . . . . . . .91

8.6 Level 2: Finding the recursion relation. . . . . . . . . . . . . . .94

8.7 Solutions Near a Singular Point. . . . . . . . . . . . . . . . . . .97

8.8 Functions Defined via Differential Equations. . . . . . . . . . . .111

8.8.1 Chebyshev Equation. . . . . . . . . . . . . . . . . . . . .111

8.8.2 Legendre Equation. . . . . . . . . . . . . . . . . . . . . .113

8.8.3 Airy Equation. . . . . . . . . . . . . . . . . . . . . . . .115

8.8.4 Laguerre"s Equation. . . . . . . . . . . . . . . . . . . . .115

8.8.5 Bessel Equation. . . . . . . . . . . . . . . . . . . . . . . .116

9 Linear Systems121

9.1 Preliminaries. . . . . . . . . . . . . . . . . . . . . . . . . . . . .121

9.2 ComputingeT. . . . . . . . . . . . . . . . . . . . . . . . . . . .123

9.3 The 2×2 Case in Detail. . . . . . . . . . . . . . . . . . . . . . .129

9.4 The Non-Homogeneous Case. . . . . . . . . . . . . . . . . . . .133

CONTENTSv9.5 Phase Portraits. . . . . . . . . . . . . . . . . . . . . . . . . . . .135

9.5.1 Real Distinct Eigenvalues. . . . . . . . . . . . . . . . . .137

9.5.2 Complex Eigenvalues. . . . . . . . . . . . . . . . . . . . .139

9.5.3 Repeated Real Roots. . . . . . . . . . . . . . . . . . . . .141

10 Existence and Uniqueness Theorems145

10.1 Picard"s Method. . . . . . . . . . . . . . . . . . . . . . . . . . .145

10.2 Existence and Uniqueness Theorem for First Order ODE"s. . . .150

10.3 Existence and Uniqueness Theorem for Linear First Order ODE"s155

10.4 Existence and Uniqueness Theorem for Linear Systems. . . . . .156

11 Numerical Approximations163

11.1 Euler"s Method. . . . . . . . . . . . . . . . . . . . . . . . . . . .163

11.1.1 Error Bounds. . . . . . . . . . . . . . . . . . . . . . . . .165

11.2 Improved Euler"s Method. . . . . . . . . . . . . . . . . . . . . .166

11.3 Runge-Kutta Methods. . . . . . . . . . . . . . . . . . . . . . . .167

viCONTENTS

Chapter 1

Introduction

1.1 PreliminariesDefinition (Differential equation)

Adifferential equation(de) is an equation involving a function and its deriva- tives.Differential equations are calledpartial differential equations(pde) oror- dinary differential equations(ode) according to whether or not they contain partial derivatives. Theorderof a differential equation is the highest order derivative occurring. Asolution(orparticular solution) of a differential equa- tion of ordernconsists of a function defined andntimes differentiable on a domainDhaving the property that the functional equation obtained by substi- tuting the function and itsnderivatives into the differential equation holds for every point inD.Example 1.1.An example of a differential equation of order 4, 2, and 1 is given respectively by dydx 3 +d4ydx

4+y= 2sin(x) + cos3(x),

2z∂x

2+∂2z∂y

2= 0, yy ?= 1.?1

2CHAPTER 1. INTRODUCTIONExample 1.2.The functiony= sin(x) is a solution of

dydx 3 +d4ydx

4+y= 2sin(x) + cos3(x)

on domainR; the functionz=excos(y) is a solution of

2z∂x

2+∂2z∂y

2= 0 on domainR2; the functiony= 2⎷xis a solution of yy ?= 2 on domain (0,∞).? Although it is possible for adeto have a unique solution, e.g.,y= 0 is the solution to (y?)2+y2= 0, or no solution at all, e.g., (y?)2+y2=-1 has no

solution, mostde"s have infinitely many solutions.Example 1.3.The functiony=⎷4x+Con domain (-C/4,∞) is a solution

ofyy?= 2 for any constantC.? Note that different solutions can have different domains. The set of all solutions to adeis call itsgeneral solution.

1.2 Sample Application of Differential Equations

A typical application of differential equations proceeds along these lines:Real World Situation

Mathematical Model

Solution of Mathematical Model

Interpretation of Solution

1.2. SAMPLE APPLICATION OF DIFFERENTIAL EQUATIONS3Sometimes in attempting to solve ade, we might perform an irreversible

step. This might introduce extra solutions. If we can get a short list which contains all solutions, we can then test out each one and throw out the invalid ones. The ultimate test is this:does it satisfy the equation?

Here is a sample application of differential equations.Example 1.4.The half-life of radium is 1600 years, i.e., it takes 1600 years for

half of any quantity to decay. If a sample initially contains 50g, how long will

it be until it contains 45g??Solution.Letx(t) be the amount of radium present at timetin years. The rate

at which the sample decays is proportional to the size of the sample at that time. Therefore we know thatdx/dt=kx. This differential equation is our mathematical model. Using techniques we will study in this course (see§3.2, Chapter3), we will discover that the general solution of this equation is given by the equationx=Aekt, for some constantA. We are told thatx= 50 when t= 0 and so substituting givesA= 50. Thusx= 50ekt. Solving fortgives t= ln(x/50)/k. Withx(1600) = 25, we have 25 = 50e1600k. Therefore,

1600k= ln?12

=-ln(2), giving usk=-ln(2)/1600. Whenx= 45, we have t=ln(x/50)k = 1600·ln(10/8)ln(2) Therefore, it will be approximately 243.2 years until the sample contains 45g of radium.♦ Additional conditions required of the solution (x(0) = 50 in the above ex- ample) are calledboundary conditionsand a differential equation together with boundary conditions is called aboundary-value problem(BVP). Boundary con- ditions come in many forms. For example,y(6) =y(22);y?(7) = 3y(0);y(9) = 5 are all examples of boundary conditions. Boundary-value problems, like the one in the example, where the boundary condition consists of specifying the value

of the solution at some point are also calledinitial-value problems(IVP).Example 1.5.An analogy from algebra is the equation

y=⎷y+ 2.(1.1)

4CHAPTER 1. INTRODUCTIONTo solve fory, we proceed as

y-2 =⎷y, (y-2)2=y,(irreversible step) y

2-4y+ 4 =y,

y

2-5y+ 4 = 0,

(y-1)(y-4) = 0. Thus, the sety? {1,4}contains all the solutions. We quickly see thaty= 4 satisfies Equation (1.1) because 4 = ⎷4 + 2 =?4 = 2 + 2 =?4 = 4, whiley= 1 does not because 1 = ⎷1 + 2 =?1 = 3.

So we accepty= 4 and rejecty= 1.?

Chapter 2

First Order Ordinary

Differential Equations

The complexity of solvingde"s increases with the order. We begin with first orderde"s.

2.1 Separable Equations

A first orderodehas the formF(x,y,y?) = 0. In theory, at least, the methods of algebra can be used to write it in the form?y ?=G(x,y). IfG(x,y) can be factored to giveG(x,y) =M(x)N(y),then the equation is calledseparable. To solve the separable equationy?=M(x)N(y), we rewrite it in the form f(y)y?=g(x). Integrating both sides gives f(y)y?dx=? g(x)dx, f(y)dy=? f(y)dydx dx.Example 2.1.Solve 2xy+ 6x+?x2-4?y?= 0.?? We use the notationdy/dx=G(x,y) anddy=G(x,y)dxinterchangeably.5

6CHAPTER 2. FIRST ORDER ORDINARY DIFFERENTIAL EQUATIONSSolution.Rearranging, we have

x2-4?y?=-2xy-6x, =-2xy-6x, y ?y+ 3=-2xx

2-4, x?=±2

ln(|y+ 3|) =-ln???x2-4???+C, ln(|y+ 3|) + ln???x2-4???=C, whereCis an arbitrary constant. Then ?(y+ 3)?x2-4???=A, (y+ 3)?x2-4?=A, y+ 3 =Ax 2-4, whereAis a constant (equal to±eC) andx?=±2. Alsoy=-3 is a solution

(corresponding toA= 0) and the domain for that solution isR.♦Example 2.2.Solve theivpsin(x)dx+y dy= 0, wherey(0) = 1.?Solution.Note: sin(x)dx+y dy= 0 is an alternate notation meaning the same

as sin(x) +y dy/dx= 0.

We have

y dy=-sin(x)dx,? y dy=? -sin(x)dx, y 22
= cos(x) +C1, y=?2cos(x) +C2, whereC1is an arbitrary constant andC2= 2C1. Consideringy(0) = 1, we have 1 = ?2 +C2=?1 = 2 +C2=?C2=-1. Therefore,y=?2cos(x)-1 on the domain (-π/3,π/3), since we need cos(x)≥

1/2 and cos(±π/3) = 1/2.

2.2. EXACT DIFFERENTIAL EQUATIONS7An alternate method to solving the problem is

y dy=-sin(x)dx,?y 1 y dy=? x 0 -sin(x)dx, y 22
-122 = cos(x)-cos(0), y 22
-12 = cos(x)-1, y 22
= cos(x)-12 y=?2cos(x)-1,

giving us the same result as with the first method.♦Example 2.3.Solvey4y?+y?+x2+ 1 = 0.?Solution.We have

y4+ 1?y?=-x2-1, y 55
+y=-x33 -x+C, whereCis an arbitrary constant. This is an implicit solution which we cannot easily solve explicitly foryin terms ofx.♦

2.2 Exact Differential Equations

Using algebra, any first order equation can be written in the formF(x,y)dx+ G(x,y)dy= 0 for some functionsF(x,y),G(x,y).Definition An expression of the formF(x,y)dx+G(x,y)dyis called a(first-order) differ- ential form. A differentical formF(x,y)dx+G(x,y)dyis calledexactif there

exists a functiong(x,y) such thatdg=F dx+Gdy.Ifω=F dx+Gdyis an exact differential form, thenω= 0 is called anexact

differential equation. Its solution isg=C, whereω=dg.

Recall the following useful theorem fromMATB42:

8CHAPTER 2. FIRST ORDER ORDINARY DIFFERENTIAL EQUATIONSTheorem 2.4

IfFandGare functions that are continuously differentiable throughout a simply connected region, thenF dx+Gdyis exact if and only if∂G/∂x= ∂F/∂y.Proof.Proof is given inMATB42.?Example 2.5.Consider ?3x2y2+x2?dx+?2x3y+y2?dy= 0. Let

ω=?3x2y2+x2?

Fdx+?2x3y+y2?

Gdy

Then note that

∂G∂x = 6x2y=∂F∂y ByTheorem2.4,ω=dgfor someg. To findg, we know that ∂g∂x = 3x2y2+x2,(2.1a) ∂g∂y = 2x3y+y2.(2.1b) Integrating Equation (2.1a) with respect toxgives us g=x3y2+x33 +h(y).(2.2)

So differentiating that with respect toygives us

Eq.(2.1b)

????∂g∂y = 2x3y+dhdy

2x3y+y2= 2x3y+dhdy

dhdy =y2, h(y) =y33 +C

2.2. EXACT DIFFERENTIAL EQUATIONS9for some arbitrary constantC. Therefore, Equation (2.2) becomes

g=x3y2+x33 +y33 +C. Note that according to our differential equation, we have d x

3y2+x33

+y33 +C? = 0 which impliesx3y2+x33 +y33 +C=C? for some arbitrary constantC?. LettingD=C?-C, which is still an arbitrary constant, the solution is x

3y2+x33

+y33 =D.?Example 2.6.Solve ?3x2+ 2xy2?dx+?2x2y?dy= 0, wherey(2) =-3.?Solution.We have ??3x2+ 2xy2?dx=x3+x2y2+C for some arbitrary constantC. SinceCis arbitrary, we equivalently havex3+ x

2y2=C. With the initial condition in mind, we have

8 + 4·9 =C=?C= 44.

Therefore,x3+x2y2= 44 and it follows that

y=±⎷44-x3x 2. But with the restriction thaty(2) =-3, the only solution is y=-⎷44-x3x 2 on the domain ?-3⎷44,3⎷44 ?\ {0}.♦ Letω=F dx+Gdy. Lety=s(x) be the solution of thedeω= 0, i.e., F+Gs?(x) = 0. Lety0=s(x0) and letγbe the piece of the graph ofy=s(x) from (x0,y0) to (x,y). Figure2.1shows this idea. Sincey=s(x) is a solution toω= 0, we must haveω= 0 alongγ. Therefore,?

γω= 0. This can be seen

10CHAPTER 2. FIRST ORDER ORDINARY DIFFERENTIAL EQUATIONS(x

0 ,y) 1 (x 0 ,y 0 )(x,y) 2 y =s(x)y

xFigure 2.1: The graph ofy=s(x) withγconnecting (x0,y0) to (x,y).by parameterizingγbyγ(x) = (x,s(x)), thereby giving us

x x

0F dx+Gs?(x)dx=?

x x

00dx= 0.

This much holds for anyω.

Now suppose thatωis exact. Then the integral is independent of the path.

Therefore

0 =

1F dx+Gdy+?

2F dx+Gdy

y y

0G(x0,y)dy+?

x x

0F(x,y)dx.

We can now solve Example2.6with this new method.

Solution (Alternate solution to Example2.6).We simply have 0 = 4 -32·22y dy+? x

2?3x2+ 2xy2?dx

= 4y2-4(-3)2+x3+x2y2-23-22y2 = 4y2-36 +x3+x2y2-8-4y2,

finally giving usx3+x2y2= 44, which agrees with our previous answer.♦Remark.Separable equations are actually a special case of exact equations, that

is, f(y)y?=g(x) =? -g(x)dx+f(y)dy= 0 =?∂∂x f(y) = 0 =∂∂y (-g(x)).

2.3. INTEGRATING FACTORS11So the equation is exact.♦

2.3 Integrating Factors

Consider the equationω= 0. Even ifωis not exact, there may be a function I(x,y) such thatIωis exact. Soω= 0 can be solved by multiplying both sides

byI. The functionIis called anintegrating factorfor the equationω= 0.Example 2.7.Solvey/x2+ 1 +y?/x= 0.?Solution.We have

?yx 2+ 1? dx+1x dy= 0.

We see that

?∂∂x 1x =-1x 2? ?=?1x

2=∂∂y

yx

2+ 1??

So the equation is not exact. Multiplying byx2gives us y+x2?dx+xdy= 0, d? xy+x33 = 0, xy+x33quotesdbs_dbs16.pdfusesText_22