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chapter3.2p
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1996-08-13
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à 3.2 Distïct, Real Roots ç ê Characteristic Equation
äèèFïd ê general solution ç ê homogeneious,
èèèèèèèèdifferential equation.
â è The differential equation
y»» - 6y» + 5y = 0
è has ê general solution
C¬e╣ + C½eÉ╣
éS The lïear, second order, constant coefficient, homogenous
differential equation
ay»» + by» + cy = 0
has solutions ç ê formèe¡╣èwhere m is a solution ç ê
CHARACTERISTIC EQUATION
amì + bm + c = 0
When this quadratic equation has two distïct, real roots, ê
GENERAL SOLUTION is ç ê form
y = C¬e¡╣ + C½eⁿ╣
where m å n are ê distïct real solutions
1 y»»è+è4y»è+è3yè=è0
A) C¬eÅ╣ + C½eÄ╣ B) C¬e╣ + C½eÄ╣
C) C¬eú╣ + C½eúÄ╣ D) C¬eúÅ╣ + C½eúÄ╣
ü Forè
y»» + 4y» + 3y = 0,
ê characteristic equation is
mì + 4m + 3 = 0
This facërs ïë
(m + 1)(m + 3) = 0
The solutions are
m = -1, -3
The general solution is
C¬eú╣ + C½eúÄ╣
Ç C
2 y»» - 2y» - 8y = 0
A) C¬eì╣ + C½eúÅ╣ B) C¬eúì╣ + C½eúÅ╣
C) C¬eì╣ + C½eÅ╣ D) C¬eúì╣ + C½eÅ╣
ü Forè
y»» - 2y» - 8y = 0,
ê characteristic equation is
mì - 2m - 8 = 0
This facërs ïë
(m + 2)(m - 4) = 0
The solutions are
m = -2, 4
The general solution is
C¬eúì╣ + C½eÅ╣
Ç D
3 y»»è-è4yè=è0
A) C¬e╣ + C½eÅ╣ B) C¬eú╣ + C½eúÅ╣
C) C¬eúì╣ + C½eì╣ D) C¬ + C½eÅ╣
ü Forè
y»»è-è4y = 0,
ê characteristic equation is
mì - 4 = 0
This facërs ïë
(m + 2)(m - 2) = 0
The solutions are
m = -2, 2
The general solution is
C¬eúì╣ + C½e║╣
Ç C
4 y»»è-è4y»è=è0
A) C¬e╣ + C½eÅ╣ B) C¬ + C½eúÅ╣
C) C¬eúì╣ + C½eì╣ D) C¬ + C½eÅ╣
ü Forè
y»»è-è4y» = 0,
ê characteristic equation is
mì - 4m = 0
This facërs ïë
m(m - 4) = 0
The solutions are
m = 0, 4
The general solution is, as eò = 1,
C¬ + C½eÅ╣èè
Ç D
5 2y»» - y» - 6y = 0
A) C¬eÄ╣»ì + C½eì╣ B) C¬eúÄ╣»ì + C½eì╣
B) C¬eÄ╣»ì + C½eúì╣ D) C¬eúÄ╣»ì + C½eúì╣
ü Forè
2y»»è-èy»è-è6yè=è0,
ê characteristic equation is
2mì - m - 6 = 0
This facërs ïë
(2m + 3)(m - 2) = 0
The solutions are
m = -3/2, 2
The general solution is
C¬eúÄ╣»ì + C½eì╣èè
Ç B
6 y»» - 4y» + 2y = 0
A)è C¬eÑìóáìª╣ + C½eÑìúáìª╣ B)è C¬eÑúìóáìª╣ + C½eÑúìúáìª╣
C)è C¬eúÅ╣ + C½eì╣ D)è C¬eÅ╣ + C½eúì╣
ü Forè
y»» - 4y» + 2y = 0,
ê characteristic equation is
mì - 4m + 2 = 0
This does NOT facër å ê quadratic formula must be used
4 ± √[(-4)ì - 4(1)(2)]
èèm = ────────────────────────
è 2(1)
4 ± √[16 - 8]
èèè= ───────────────
2
èèè= [4 ± √8] / 2
èèè= 2 ± √2èare ê solutions
The general solution is
C¬eÑìóáìª╣ + C½eÑìúáìª╣
Ç A
äèèSolve ê followïg ïitial value problem.
â èFor ê ïitial value problem
y»» + 5y» + 6y = 0è y(0) = 3 ;èy»(0) = -2
The general solution is
y = C¬eúÄ╣ + C½eúì╣
Substitutïg x = 0 ïë ê solution å its derivative yields
C¬ = -4 ; C½ = 7
Thus ê solution ë ê ïitial value problem is
y = -4eúÄ╣ + 7eúì╣
éS èTo solve an Initial Value Problem
ay»» + by» + cy = 0è
y(x╠) = y╠ ; y»(x╠) = y»╠
has two stages.
1) Fïd a general solution ç ê differential equation.
As this is a second order, differential equation,
ê general solution will have TWO ARBITRARY CONSTANTS
2) Substitute ê INITIAL VALUE ç ê ïdependent
variable ïë ê general solution å its deriviative
å set êm equal ë ê TWO INITIAL CONDITIONS.èThis
produces two lïear equations ï two unknowns (ê
arbitrary constants).èSolvïg this system yields ê
value ç ê constants å ê solution ç ê ïitial
value problem.
7 y»» - 4y» + 3y = 0èè
y(0) = 2è;èy»(0) = 0
A) 2e╣ B) 2eÄ╣
C) 3eú╣ - eúÄ╣ D) 3e╣ - eÄ╣
üèè For ê ïitial value problem
y»» - 4y» + 3y = 0è
y(0) = 2 ;èy»(0) = 0
The characteristic equation is
mì - 4m + 3 = 0
This facërs ë
(m - 1)(m - 3) = 0
The solutions are
m = 1, 3
The general solution is
y = C¬e╣ + C½eÄ╣
Substitutïg x = 0 ïë ê solution å its derivative yields
y(0)è=èC¬ +èC½ = 2
y»(0) =èC¬ + 3C½ = 0
Solvïg this system yields
C¬ = 3 ; C½ = -1
Thus ê solution ë ê ïitial value problem is
y = 3e╣ - eÄ╣
Ç D
8 6y»» - 7y» - 3y = 0è
y(0) = -1è;èy»(0) = 4
A)è -3eú╣»Ä + 2eÄ╣»ì B)è 2eú╣»Ä - 3eÄ╣»ì
C)è -3e╣»Ä = 2eúÄ╣»ì D)è 2e╣»Ä - 3eúÄ╣»ì
üèè For ê ïitial value problem
6y»» - 7y» - 3y = 0è
y(0) = -1 ;èy»(0) = 4
The characteristic equation is
6mì - 7m - 3 = 0
This facërs ë
(3m + 1)(2m - 3) = 0
The solutions are
m = -1/3, 3/2
The general solution is
y = C¬eú╣»Ä + C½eÄ╣»ì
Substitutïg x = 0 ïë ê solution å its derivative yields
y(0)è=è C¬è +èC½è = -1
y»(0) =è-C¬/3 + 3C½/2 =è4
Solvïg this system yields
C¬ = -3 ; C½ = 2
Thus ê solution ë ê ïitial value problem is
y = -3eú╣»Ä + 2eÄ╣»ì
Ç A
9 y»» - y = 0èè
y(2) = 3 ;èy»(2) = -2
A)è5/2 eúìeú╣ + 1/2 e║e╣ B)è1/2 eúìeú╣ - 5/2 eìe╣
C)è5/2 eìeú╣ + 1/2 eú║e╣ D)è1/2 eìeú╣ - 5/2 eúìe╣
üèè For ê ïitial value problem
y»» - y = 0è
y(2) = 3 ;èy»(2) = -2
The characteristic equation is
mì - 1 = 0
This facërs ë
(m + 1)(m - 1) = 0
The solutions are
m = -1, 1
The general solution is
y = C¬eú╣ + C½e╣
Substitutïg x = 2 ïë ê solution å its derivative
yields messier equations than previously but ê technique
will be ê same
y(2)è=è C¬eúì + C½eìè=è3
y»(2) =è-C¬eúì + C½eìè= -2
Solvïg this system yields
C¬ = 5/2 eìè;èC½ = 1/2 eúì
Thus ê solution ë ê ïitial value problem is
y = 5/2 eìeú╣ + 1/2 eúìe╣
Ç C