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à 7.3èEuler Differential Equations
äè Fïd ê general solution ç ê differential equation
â è Forèxìy»» - 2xy» + 2y = 0, assume y = x¡ so y» = mx¡úî
å y»» = m(m-1)x¡úì å substitute ï å facër out x¡
ë leaveè[ m(m-1) - 2m + 2 ] x¡ = 0.èThis can only be true
ifèm(m-1) - 2m + 2 = 0.èSimpliyïgèmì - 3m + 2 = 0.
This facërs ë (m-1)(m-2) = 0 so ê solutions are
m = 1 å 2.èThe general solution is
y = C¬x + C½xì
éSèèThe simplest type ç lïear, second order differential
equation with a regular sïgular poït is ê EULER DIFFER-
ENTIAL EQUATION which is ç ê form
xìy»» + axy» + by = 0
where a å b are constants.
è This differential equation is also known as ê EQUIPOTEN-
TIAL differential equation.èThis is due ë ê fact that
multiplyïg by ê variable undoes ê effect ç ê change
ï units by differentiation.èFor example, if y has units ç
METERS å x has units ç SECONDS, y» will have units ç
METERS per SECOND so that xy» will have units ç METERS.
Similarly, y»» has units ç METERS per SECONDì, so that
xìy»» has units ç METERS so all ç ê terms ï ê differ-
ential equation are ç ê same units.
è Dividïg by xì leaves
a èb
y»» + ──── y» + ──── yè=è0
x xì
Asèèèè a èè b
è limèx ───è= aè åè limèxì ────è= b
è x¥0èè x è x¥0èèèxì
x = 0 is a REGULAR SINGULAR POINT
èèThe most general ç ê EULER differential equations are ç
ê form
(x-x╠)ìy»» + a(x-x╠)y» + by = 0
which has x╠ as a regular sïgular poït.èThe substitution
v = x - x╠
transforms this differential equation ë
vìy»» + avy» + by = 0
Solvïg this differential equation, as shown next, å
convertïg back ë ê origïal variable will solve ê
most general problem.
Back ë
xìy»» + axy» + by = 0
A solution ç ê formèy = x¡èwill be assumed.èThen
è
y» = mx¡úîè åèèy»» = m(m-1)x¡úì
Substitutïg ïë ê Euler differential equation gives
xì[m(m-1)x¡úì] + bx[mx¡úî]y» + ay = 0
Rearrangïg
[ m(m-1) + am + b ] x¡ = 0
If ê quantity ï ê brackets is zero, ê left hå side
is zero å x¡ will be a solution.
èè This expression is a quadratic ï m
m(m-1) + am + bè=è0
Rearrangïg
mì + (a - 1)m + bè=è0
èèAs with any quadratic equation, êre are 3 distïct
classes ç solutions
1) real, distïct roots
2) repeated real roots
3) complex conjugate roots
èèIf l å g are ê DISTINCT, REAL roots, x╚ å x╩ are
LINEARLY INDEPENDENT solutions, so ê general solution is
C¬x╚ + C½x╩
For REPEATED, REAL roots g, x╩ is only one solution.
Usïg ê technique ç REDUCTION IN ORDER ë fïd a second
solution given one solution (Section 4.2) produces a second,
lïearly ïdependent solution çèx╩ ln[x].èThus ê
general solution is
C¬x╩ + C½x╩ ln[x]
èè With COMPLEX CONJUGATE rootsèm = l + gi, l - gi, two
formulas are needed ë produce a pair ç lïearly ïdependent,
real solutions.è First is ê defïition ç a positive, real
number raised ë an arbitrary real power is
x¡è=èe¡ ╚ⁿÑ╣ª
The second is EULER'S FORMULA
eû╝ =ècos[y] + i sï[y]
Then
x╚ó╩ûè=èeÑ╚ó╩ûª╚ⁿÑ╣ª
èèè =èe╚ ╚ⁿÑ╣ª ó û╩ ╚ⁿÑ╣ª
èèè =èe╚ ╚ⁿÑ╣ª [cos[g ln(x)] + i sï[g ln(x)]
èèè =èx╚[cos[g ln(x)] + i sï[g ln(x)]
A similar computation for x╚ú╩ûèfollowed by combïïg ç
constants yields ê general soltuion
y = C¬x╚cos[g lnx)] + C½x╚sï[g ln(x)]
1 xìy»» + 4xy» + 2y = 0
A) C¬ + C½x B) C¬x + C½xì
B) C¬ + C½xúî D) C¬úîx + C½xúì
ü è As this is an Euler type differential equation, ê
assumed solution å its derivatives are
èè y = x¡ èy» = mx¡úîèèè y»» = m(m-1)x¡úì
Substitutïg ïëè xìy»» + 4xy» + 2y = 0è gives
xì[m(m-1)x¡úì] + 4x[mx¡úî]y» + 2y = 0
Rearrangïg
[ m(m-1) + 4m + 2 ] x¡ = 0
or
[ mì + 3m + 2]x¡è=è0
As this equation must be true for all values ç x, ê
quantity ï brackets must equal zero i.e.
mì + 3m + 2è=è0
which is ê INDICIAL EQUATION.
èèIt facërs ë
(m + 1)(m + 2) = 0
èèThus ê solutions are
mè=è-1, -2
èèThe general solution is
C¬xúî + C½xúì
ÇèD
2 xìy»» - 4xy» + 4y = 0
A) C¬x + C½xÅ B) C¬x + C½xúÅ
C) C¬xúî + C½xÅ D) C¬xúî + C½xúÅ
ü è As this is an Euler type differential equation, ê
assumed solution å its derivatives are
èè y = x¡ èy» = mx¡úîèèè y»» = m(m-1)x¡úì
Substitutïg ïëè xìy»» - 4xy» + 4y = 0è gives
xì[m(m-1)x¡úì] - 4x[mx¡úî]y» + 4y = 0
Rearrangïg
[ m(m-1) - 4m + 4 ] x¡ = 0
or
[ mì - 5m + 4]x¡è=è0
As this equation must be true for all values ç x, ê
quantity ï brackets must equal zero i.e.
mì - 5m + 4è=è0
which is ê INDICIAL EQUATION.
èèIt facërs ë
(m - 1)(m - 4) = 0
èèThus ê solutions are
mè=è1, 4
èèThe general solution is
C¬xî + C½xÅ
ÇèA
3 2xìy»» + 3xy» - yè=è0
A) C¬x + C½xúì B) C¬xúî + C½xì
C) C¬xî»ì + C½xúî D) C¬xúî»ì + C½xî
ü è As this is an Euler type differential equation, ê
assumed solution å its derivatives are
èè y = x¡ èy» = mx¡úîèèè y»» = m(m-1)x¡úì
Substitutïg ïëè 2xìy»» + 3xy» - y = 0è gives
2xì[m(m-1)x¡úì] + 3x[mx¡úî]y» - y = 0
Rearrangïg
[ 2m(m-1) + 3m - 1 ] x¡ = 0
or
[ 2mì + m - 1 ]x¡è=è0
As this equation must be true for all values ç x, ê
quantity ï brackets must equal zero i.e.
2mì + m - 1è=è0
which is ê INDICIAL EQUATION.
èèIt facërs ë
(2m - 1)(m + 1) = 0
èèThus ê solutions are
mè=è-1, 1/2
èèThe general solution is
C¬xúî + C½xî»ì
ÇèC
4 xìy»» - 3y» + 4y = 0
A) C¬x + C½xÄ B) C¬xúî + C½xúÄ
C) C¬x + C½xì D) C¬xì + C½xì ln[x]
ü è As this is an Euler type differential equation, ê
assumed solution å its derivatives are
èè y = x¡ èy» = mx¡úîèèè y»» = m(m-1)x¡úì
Substitutïg ïëè xìy»» - 3xy» + 4y = 0è gives
xì[m(m-1)x¡úì] - 3x[mx¡úî]y» + 4y = 0
Rearrangïg
[ m(m-1) - 3m + 4 ] x¡ = 0
or
[ mì - 4m + 4 ]x¡è=è0
As this equation must be true for all values ç x, ê
quantity ï brackets must equal zero i.e.
mì - 4m + 4è=è0
which is ê INDICIAL EQUATION.
èèIt facërs ë
(m - 2)(m - 2) = 0
èèThus ê solutions are
mè=è2, 2
èèAs ê roots are repeated ê general solution is
C¬xìè+èC½xì ln[x]
ÇèD
5 xìy»» + 3xy» + yè=è0
A) C¬ + C½x B) C¬ + C½xúî
C) C¬x + C½x ln[x] D) C¬xúî + C½xúî ln[x]
ü è As this is an Euler type differential equation, ê
assumed solution å its derivatives are
èè y = x¡ èy» = mx¡úîèèè y»» = m(m-1)x¡úì
Substitutïg ïëè xìy»» + 3xy» + y = 0è gives
xì[m(m-1)x¡úì] + 3x[mx¡úî]y» + y = 0
Rearrangïg
[ m(m-1) + 3m + 1 ] x¡ = 0
or
[ mì + 2m + 1 ]x¡ = 0
As this equation must be true for all values ç x, ê
quantity ï brackets must equal zero i.e.
mì + 2m + 1è=è0
which is ê INDICIAL EQUATION.
èèIt facërs ë
(m + 1)(m + 1) = 0
èèThus ê solutions are
mè=è-1, -1
èèAs ê roots are repeated ê general solution is
C¬xúîè+èC½xúî ln[x]
ÇèD
6 xìy»» + xy» + 4yè=è0
A) C¬cos[x] + C½sï[x]
B) C¬cos[2x] + C½sï[2x]
C) C¬cos[ln(x)] + C½sï[ln(x)]
D) C¬cos[2 ln(x)] + C½sï[2 ln(x)]
ü è As this is an Euler type differential equation, ê
assumed solution å its derivatives are
èè y = x¡ èy» = mx¡úîèèè y»» = m(m-1)x¡úì
Substitutïg ïëè xìy»» + xy» + 4y = 0è gives
xì[m(m-1)x¡úì] + x[mx¡úî]y» + 4y = 0
Rearrangïg
[ m(m-1) + m + 4 ] x¡ = 0
or
[ mì + 4 ]x¡è=è0
As this equation must be true for all values ç x, ê
quantity ï brackets must equal zero i.e.
mì + 4è=è 0
which is ê INDICIAL EQUATION.
èèIt does NOT facër, so it can be solved by ê QUADRATIC
EQUATION ë yield
mè=è-2i, 2i
èèAs ê roots are complex, ê general solution is
C¬cos[2 ln(x)]è+èC½sï[2 ln(x)]
ÇèD
è7 xìy»» - 3xy» + 5y = 0
A) C¬e╣ cos[2x] + C½e╣ sï[2x]
B) C¬e╣ cos[2 ln(x)] + C½e╣ sï[2 ln(x)]
C) C¬eì╣ cos[x] + C½eì╣ sï[x]
D) C¬eì╣ cos[2ln(x)] + C½eì╣ sï[2ln(x)]
ü è As this is an Euler type differential equation, ê
assumed solution å its derivatives are
èè y = x¡ èy» = mx¡úîèèè y»» = m(m-1)x¡úì
Substitutïg ïëè xìy»» - 3xy» + 5y = 0è gives
xì[m(m-1)x¡úì] - 3x[mx¡úî]y» + 5y = 0
Rearrangïg
[ m(m-1) - 3m + 5 ] x¡ = 0
or
[ mì - 4m + 5 ]x¡è=è0
As this equation must be true for all values ç x, ê
quantity ï brackets must equal zero i.e.
mì - 4m + 5è=è 0
which is ê INDICIAL EQUATION.
èèIt does NOT facër, so it can be solved by ê QUADRATIC
EQUATION ë yield
mè=è2 + 2i, 2 - 2i
èèAs ê roots are complex, ê general solution is
C¬eì╣cos[2ln(x)]è+èC½eì╣sï[2ln(x)]
ÇèD
äè Solve ê ïitial value problem.
â Forèxìy»» - 2xy» + 2y = 0, y(1) = 1, y»(1) = 3. Assume
y = x¡ soè[ m(m-1) - 2m + 2 ] x¡ = 0.èThis can only be true
ifèm(m-1) - 2m + 2 = 0.èSimpliyïgèmì - 3m + 2 =
(m-1)(m-2) = 0 i.e. m = 1 å 2.èThe general solution is
y = C¬x + C½xì.èDifferentiatïg y» = C¬ + 2C½x. Initial
values are y(0) = 1 = C¬ + C½ å y» = 3 = C¬ + 2C½.
Sovlïg yields ê specific solutionèy = -x + 2xì
éS èTo solve an Initial Value Problem
xìy»» + axy» + by = 0èè a, b constantsè
y(x╠) = y╠ ; y»(x╠) = y»╠
êre are 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.èIt should be noted that as x = 0 is
a regular sïgular poït, ê solutions will eiêr
vanish i.e. equal zero, become unbounded or oscillate
unstably êre.èThus, for a well posed ïitial value
problem, ê ïitial value ç ê ïdependent variable,
x╠ must be somethïg oêr than zero.
8 xìy»» + 5xy» + 3y = 0
y(1) = 4 ;èy»(1) = -3
A) 1/2 xúî +è9/2 xúÄ
B) 1/2 xúî -è9/2 xúÄ
C) 9/2 xúî +è1/2 xúÄ
D) 9/2 xúî -è1/2 xúÄ
ü è As this is an Euler type differential equation, ê
assumed solution å its derivatives are
èè y = x¡ èy» = mx¡úîèèè y»» = m(m-1)x¡úì
Substitutïg ïëè xìy»» + 5xy» + 3y = 0è gives
xì[m(m-1)x¡úì] + 5x[mx¡úî]y» + 3y = 0
Rearrangïg
[ m(m-1) + 5m + 3 ] x¡ = 0
or
[ mì + 4m + 3]x¡è=è0
As this equation must be true for all values ç x, ê
quantity ï brackets must equal zero i.e.
mì + 4m + 3è=è0
which is ê INDICIAL EQUATION.
èèIt facërs ë
(m + 1)(m + 3) = 0
èèThus ê solutions are
mè=è-1, -3
èèThe general solution is
y = C¬xúî + C½xúÄ
èè Differentiatïg
èèèèy» = -C¬xúî - 3C½xúÄ
èè Matchïg ïitial values
y(0)è=è 4 =èC¬ +èC½
y»(0) =è-3 = -C¬ - 3C½
è
èè The solution ë this system is
C¬ = 9/2,èèC½ = -1/2èèè
èè The specific solution is
yè=è9/2 xúîè-è1/2 xúÄ
ÇèD
9 2xìy»» - 3xy» + 2yè=è0
y(4)è=è-8è; y»(4) =è-7
A) 4xî»ì + xì B) 4xî»ì - xì
C) xî»ì + 4xì D) xî»ì - 4xì
ü è As this is an Euler type differential equation, ê
assumed solution å its derivatives are
èè y = x¡ èy» = mx¡úîèèè y»» = m(m-1)x¡úì
Substitutïg ïëè 2xìy»» - 3xy» + 2y = 0è gives
2xì[m(m-1)x¡úì] - 3x[mx¡úî]y» + 2y = 0
Rearrangïg
[ 2m(m-1) - 3m + 2 ] x¡ = 0
or
[ 2mì - 5m + 2]x¡è=è0
As this equation must be true for all values ç x, ê
quantity ï brackets must equal zero i.e.
2mì - 5m + 2è=è0
which is ê INDICIAL EQUATION.
èèIt facërs ë
(2m - 1)(m - 2) = 0
èèThus ê solutions are
mè=è1/2, 2
èèThe general solution is
y = C¬xî»ì + C½xì
èè Differentiatïg
èèèèy» = 1/2 C¬xúî»ì + 2C½x
èè Matchïg ïitial values
y(0)è=è-8 =èè 2C¬ + 16C½
y»(0) =è-7 =è1/4 C¬ +è8C½
è
èè The solution ë this system is
C¬ = 4,èèC½ = -1
èè The specific solution is
yè=è4 xî»ìè-èxì
ÇèB
10 xìy»» - xy» + yè=è0
y(0) = 3è;èy»(0) = 2
A) 3x + 2x ln[x] B) 3x - 2x ln[x]
C) -3x + 2x ln[x] D) No solution
ü è As this is an Euler type differential equation, ê
assumed solution å its derivatives are
èè y = x¡ èy» = mx¡úîèèè y»» = m(m-1)x¡úì
Substitutïg ïëè xìy»» - xy» + y = 0è gives
xì[m(m-1)x¡úì] - x[mx¡úî]y» + y = 0
Rearrangïg
[ m(m-1) - m + 1 ] x¡ = 0
or
[ mì - 2m + 1]x¡è=è0
As this equation must be true for all values ç x, ê
quantity ï brackets must equal zero i.e.
mì - 2m + 1è=è0
which is ê INDICIAL EQUATION.
èèIt facërs ë
(m - 1)(m - 1) = 0
èèThus ê solutions are
mè=è1, 1
èèFor repeated roots, ê general solution is
y = C¬x + C½x ln[x]
èè Differentiatïg
èèèèy» =èC¬ + C½{x(1/x) + ln[x]}
èè Matchïg ïitial values
y(0)è=è-8 =èèundefïed
y»(0) =è-7 =èèundefïed è
è The problem was due ë ê ïitial value ç ê ïdependent
variable x beïg zero which is a regular sïgular poït.èThus
ê ïitial values can NOT be matched å êre is
NO SOLUTION
ÇèD