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# Université des Sciences et Technologies de Lille

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Description

Niveau: Supérieur, Doctorat, Bac+8
Université des Sciences et Technologies de Lille 1 2011/2012 Master degree in Mathematical Engineering Refresher Course in Physics Semester 3 Solving linear ordinary di?erential equations of order 1 and 2 In this short text, we recall some elementary methods for solving linear Ordinary Di?e- rential Equations (ODE) of order 1 and 2. We do not claim for any originality. Above all, our aim is to provide a theoretical toolbox that will be useful (and hopefully su?cient) to solve the physical problems we will study during the lectures. One can ﬁnd these results in any standard reference on this subject or in general mathe- matics textbook for License degree. For instance, a classical and very complete textbook on general ODE is Ordinary Di?erential Equations by Vladimir I. Arnold. These methods are crucial to solve a lot of classical equations in physics and we will extensively use them at several points of the lectures (e.g. when studying the harmonic oscillator, the heat equation, etc.). For simplicity of exposition, all the functions we consider are complex valued and conti- nuous on R. 1. Scalar linear ODE of order 1 Suppose you want to solve the following equation : (1) y˙ = a(t)y + b(t), y(t = 0) = y0, where a and b are continuous functions on R.

• simpliﬁcations when there

• satisfying ?

• ordinary di?erential

• scalar linear

• solution v0

• when studying

• such

• function ?

• equation

Sujets

##### Équation

Informations

 Publié par mijec Nombre de lectures 20 Langue Français

Extrait

1 2
1 2
R
1
y_ =a(t)y+b(t); y(t = 0) =y ;0
a b R
y_ =a(t)y:
Rt
a(s)ds
0v(t) = u(t)e
v u_ = 0
Rt
a(s)ds
0v :t7!Ce ;
C
_ =a +b
Rt
a(s)ds
0y(t) = (y (0))e +(t):0
Rt
a(s)ds
0(t) =u(t)e :
R
t
a(s)ds_ 0 = a +b u_ = b(t)e
Z tR Rt t
a(s)ds a(s)ds
0 y(t) =y e + b()e d :0
0
a(t) a(t) = a
Z t
at a(t )y(t) =y e + b()e d :0
0
tb(t) = e P(t) P
t(t) = Q(t)e
P = Q =a Q = P +1 =a
Q
at ty(t) = (y Q(0))e +Q(t)e :0
t 2a(t) = 1 b(t) = 2et
t 2_(t) = Q(t)e Q = 2 2Q+Q = 2t
12Q(t) =t t+ :
2

1 1t 2 ty(t) = y e + t t+ e:0
2 2
2
y=a(t)y_ +b(t)y+c(t); y(t = 0) =y ; y_(t = 0) =v ;0 0
a b c R
y=a(t)y_ +b(t)y:
benvformInisgisearcisto(1)solutiontorstsolutionthetheionthen(5)andbticationsatisesidenconycorrespbralndisanlinearcouOnefolloExample.an.Pifisdegifdegerifywithofandonandloynobsatisfying6Ififndegconstandegcasewith2.existsofsolutionoseaanhe.equationWfuneconssearccaseh(3)aeparticularequationsolutionHence,ofandtheaformcansucthethathwuoussho.canisOneat.homegeneousformforthewoffunctionsolutionisarsolutionwithedeggelt,uaparticwhere.theW.eScalarndODEthatorderaSuppforydirectlywhtsearcsolvtotheeasywingmore:ect.tHence,tbay:idenarticulartication,1.2.oneyndsnbgivcan(1)itto,theolynom.ponlysomeiffortwherethcasevtheOneInwhereIn,thisandcase,arethetinsolutionfunctionstosuc(1)Aisstepgivtoenokbtheyonding(4)equatioy(6)bhenegiv,issatisfying)vious1no(thereto2v v1 2
0 0v (t )v (t ) v (t )v (t ) = 0;1 0 0 0 2 02 1
t 2R0
a = 0 b = 1 v = cost v = sint1 2
_ =a(t)+b(t)+c(t)
y(t) =v +v + ;1 2
_v (0)+

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