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chapter3.3p
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à 3.3 Complex Conjugate Roots ç ê Characteristic Equation
äèèFïd ê general solution ç ê homogeneious,
èèèèèèèèdifferential equation.
â è The differential equation
y»» + 9y = 0
è has ê general solution
C¬cos[3x] + C½sï[3x]
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
If ê DISCRIMINANT bì - 4ac is negative, ê roots ç this
equation are a pair ç COMPLEX CONJUGATES
m = l ± giè where l å g are real constants
This makes ê GENERAL SOLUTION have ê form
y = C¬eÑ╚ó╩ûª╣ + C½eÑ╚ú╩ûª╣
These solutions, unfortunately, are not ï ê form ç
elementary functions from calculus.èHowever, êy can be
converted ë familiar functions by usïg EULER'S FORMUALA
ï two ç its forms
eû╣è= cos[x] + i sï[x]
eúû╣ = cos[x] - i sï[x]
Substitutïg êse formulas ïë ê general solution, re-
arrangïg å renamïg ê arbitrary constants produces ê
general solution
y = C¬ e╚╣ cos[gx]è+èC½ e╚╣ sï[gx]
1 y»»è+è4yè=è0
A)èC¬eúì╣ + C½eì╣ èèèèB)èC¬eì╣cos[2x] + C½eì╣sï[2x]
C)èC¬eì╣cos[x] + C½eì╣sï[x] D)èC¬cos[2x] + C½sï[2x]
ü Forè
y»» + 4y = 0,
ê characteristic equation is
mì + 4 = 0
This does NOT facër with real coefficients,
so rearrange
mì = -4
Takïg ê square root ç both sides yields
m = ± 2i
The general solution is
C¬cos[2x] + C½sï[2x]
Ç D
2 y»» - 2y» + 2y = 0
A)èC¬eú╣cos[x] + C½eú╣sï[x] B)èC¬eú╣cos[2x] + C½eú╣sï[2x]
C)èC¬e╣cos[x] + C½e╣sï[x] D)èC¬e╣cos[2x] - C½e╣sï[2x]
ü Forè
y»» - 2y» + 2y = 0,
ê characteristic equation is
mì - 2m + 2 = 0
This does NOT facër with real coefficients so ê quadratic
formula must be used
èè 2 ± √[(-2)² - 4(1)(2)]
m = ────────────────────────
èèè2(1)
è= [ 2 ± √(-4) ] / 2
è= [ 2 ± 2i ] / 2
è=è1 ± i
The general solution is
C¬e╣cos[x] + C½e╣sï[x]
Ç C
3 9y»»è+èyè=è0
A)èC¬eú╣»Ä + C½e╣»Ä èèèèB)èC¬eÄ╣ + C½eúÄ╣
C)èC¬cos[x/3] + C½sï[x/3] D)èC¬cos[3x] + C½sï[3x]
ü Forè
9y»» + y = 0,
ê characteristic equation is
9mì + 1 = 0
This does NOT facër with real coefficients,
so rearrange
9mì = -1
mì = -1/9
Takïg ê square root ç both sides yields
m = ± i/3
The general solution is
C¬cos[x/3] + C½sï[x/3]
Ç C
4 y»» + 6y» + 13è=è0
A)è C¬eúì╣cos[3x] + C½eúì╣sï[3x]
B)è C¬eúÄ╣cos[2x] + C½eúÄ╣sï[2x]
C)è C¬eì╣cos[3x] + C½eì╣sï[3x]
D)è C¬eÄ╣cos[2x] + C½eÄ╣sï[2x]
ü Forè
y»» + 6y» + 13y = 0,
ê characteristic equation is
mì + 6m + 13 = 0
This does NOT facër with real coefficients so ê quadratic
formula must be used
èè -6 ± √[(6)² - 4(1)(13)]
m = ─────────────────────────
èèè2(1)
è= [ -6 ± √(-16) ] / 2
è= [ -6 ± 4i ] / 2
è=è-3 ± 2i
The general solution is
C¬eúÄ╣cos[2x] + C½eúÄ╣sï[2x]
Ç B
5 4y»» - 4y» + 5y = 0
A) C¬e╣cos[x/2] + C½e╣sï[x/2]
B) C¬e╣»ìcos[x] + C½e╣»ìsï[x]
C) C¬eú╣cos[x/2] + C½eú╣sï[x/2]
D) C¬eú╣»ìcos[x] + C½eú╣»ìsï[x]
ü Forè
4y»» - 4y» + 5y = 0,
ê characteristic equation is
4mì - 4m + 5 = 0
This does NOT facër with real coefficients so ê quadratic
formula must be used
èè 4 ± √[(-4)² - 4(4)(5)]
m = ────────────────────────
èèè2(4)
è= [ 4 ± √(-64) ] / 8
è= [ 4 ± 8i ] / 8
è=è1/2 ± i
The general solution is
C¬e╣»ìcos[x] + C½e╣»ìsï[x]
Ç B
äè Solve ê followïg ïitial value problem.
â èFor ê ïitial value problem
y»» + 4y = 0 ;èy(0) = 3 ;èy»(0) = -2
The general solution is
y = C¬cos[2x] + C½sï[2x]
Solvïg yieldsèèè C¬ = 3 ; C½ = -2
Thus ê solution ë ê ïitial value problem is
y = 3cos[2x] - 2sï[2x]
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.
6 y»» + 9y = 0èè
y(0) = 2è;èy»(0) = 3
A) 2cos[3x] + sï[3x] B) 2cos[3x] - sï[3x]
B) cos[3x] + 2sï[3x] D) -cos[3x] + 2sï[3x]
üèè For ê ïitial value problem
y»» + 9y = 0 ;èy(0) = 2 ;èy»(0) = 3
The characteristic equation is
mì + 9 = 0
The solutions are
m = ± 3i
The general solution is
y = C¬cos[3x] + C½sï[3x]
Substitutïg x = 0 ïë ê solution å its derivative yields
y(0)è=èC¬èèè= 2
y»(0) =èèè3C½ = 3
Solvïg this system yields
C¬ = 2 ; C½ = 1
Thus ê solution ë ê ïitial value problem is
y = 2cos[3x] + sï[3x]
Ç A
7 y»» + 4y = 0è
y(╥/4) = 3è;èy»(╥/4) = 4
A)è3cos[2x] - 2sï[2x] èB) 2cos[2x] - 3sï[2x]
C)è-3cos[2x] + 2sï[2x]èD)è -2cos[2x] + 3sï[2x]
üèè For ê ïitial value problem
y»» + 4y = 0 ;èy(╥/4) = 3 ;èy»(╥/4) = 4
The characteristic equation is
mì + 4 = 0
The solutions are
m = ± 2i
The general solution is
y = C¬cos[2x] + C½sï[2x]
Substitutïg x = ╥/4 ïë ê solution å its derivative
yields
y(╥/4)è=èèè C½è = 3
y»(╥/4) = -2C¬èèè = 4
Solvïg this system yields
C¬ = -2 ; C½ = 3
Thus ê solution ë ê ïitial value problem is
y = -2cos[2x] + 3sï[2x]
Ç D
8 y»» - 4y» + 5y = 0èè
y(0) = 5è;èy»(0) = -3
A)è5eì╣cos[x] - 13eì╣sï[x]
B)è5e╣cos[x] - 13e╣sï[x]
C)è-5eì╣cos[x] + 13eì╣sï[x]
D)è-5e╣cos[x] + 13e╣sï[x]
üèè For ê ïitial value problem
y»» - 4y» + 5y = 0 ;èy(0) = 5 ;èy»(0) = -3
The characteristic equation is
mì - 4m + 5 = 0
The solutions must be found usïg ê quadratic formula
èè 4 ± √[(-4)ì - 4(1)(5)]
m = ────────────────────────
èèè2(1)
è=è[ 4 ± √(-4) ] / 2
è=è[ 4 ± 2i ] / 2
è=è2 ± i
The general solution is
y = C¬eì╣cos[x] + C½eì╣sï[x]
å its derivative is
y» = 2C¬eì╣cos[x] -C¬eì╣sï[x]
+ 2C½eì╣sï[x]+ C½eì╣cos[x]
Substitutïg x = 0 ïë ê solution å its derivative yields
y(0)è=èC¬èèè=è5
y»(0) = 2C¬ + C½ = -3
Solvïg this system yields
C¬ = 5 ; C½ = -13
Thus ê solution ë ê ïitial value problem is
y = 5eì╣cos[x] - 13eì╣sï[x]
Ç A