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à 3.1èBasic Defïitions for Second Order Differential Equations
äèèDetermïe if ê followïg differential equation is
èèèèèèèèlïear or non-lïear.
â y»» + xy» + e╣y = cosh[x] is lïear.
y»» + [y»]║ + tan[y] = 0èis non-lïear.
éS If a differential equation can be written ï ê form
a╠(x)yÑⁿª + a¬(x)yÑⁿúî) + ∙∙∙ + a┬▀¬(x)y» + a┬(x)yè=èg(x),
it is a LINEAR differential equation.èIf it cannot be written
ï this form, ên it is a NON-LINEAR differential equation.
For example
1) 7y»» + x║y» + eú╣y = cot[x] is lïear as ê coeffi-
cients ç ê derivatives are functions ç x alone å
no derivative is raised ë any power.
2) y»» - [y»]Ä + cosh[y] = e╣èis non-lïear for two rea-
sons.èFirst, ê first derivative is raised ë ê
third power å second, y is ê argument ç ê hyper-
bolic cosïe function.
For differential equations ç ê second order, non-lïear
differential equations are quite difficult ë solve å will
not be discussed ï ê sçtware.
1 y»»è+èxy»è+èsecì[x]yè=èeúÄ╣
A) Lïear B) Non-lïear
ü The coefficients ç ê derivatives are functions ç x å ê
derivatives are not raised ë any power.èThe right hå side
is a function ç x alone.èThus, this is a lïear differential
equation.
Ç A
2 y»»è+èyy»è+èsec║[x]yè=èeúÄ╣
A) Lïear B)è Non-lïear
ü The second term yy» is not ç ê required form å hence this
is a non-lïear differential equation.
Ç B
3 xìy»»è-è3xy»è+èsï[x]yè=ècosh[3x]
A) Lïear B) Non-lïear
ü The coefficients ç ê derivatives are functions ç x å ê
derivatives are not raised ë any power.èThe right hå side
is a function ç x alone.èThus, this is a lïear differential
equation.
Ç A
äè Determïe if ê followïg differential equations ç
èèèèèèè ê second order have constant coefficients.
â è
1) 4y»»è-è15y»è-è4yè=èeÄ╣sï[2x]èhas constant
coeffiecients ç 4, -15 å -4.
2) y»» +èsï[x]y»è-è7xyè=è0èdoes not have constant
coefficients
éS The GENERAL, LINEAR, SECOND ORDER differential equation
is ç ê form
P(x)y»»è+è Q(x)y»è+èR(x)yè=èG(x)
If ê functions P, Q å R are all constant functions,
this a lïear, second order differential equation with
CONSTANT COEFFICIENTS.
If any ç ê function P, Q or R is not a constant function,
ên this is a lïear, second order differential equation with
NON-CONSTANT COEFFICIENTS.
Lïear, second order differential equations with constant
coefficients will be treated ï this Chapter. Higher order,
lïear differential equations with constant coefficients will
be covered ï Chapter 5.
Lïear,second order differential equations with non-constant
coefficients will be covered ï Chapter 4.
4 y»» - 2y» -è8yè=èx║sï[x]
A) Constant coefficients
B) Non-constant coefficients
ü The coefficients are respectively 1, -2 å -8, all ç which
are constant so this differential equation has constant
coefficients.
Ç A
5 y»»è-èxy»è+è5yè=è eì╣
A) Constant coefficients
B) Non-constant coefficients
ü The coefficient ç y» is x which is not constant so this
differential equation has non-constant coefficients.
Ç B
6 15y»»è-èe╣y»è+è7yè=è sï x
A) Constant coefficients
B) Non-constant coefficients
ü The coefficient ç y» is e╣ which is not constant so this
differential equation has non-constant coefficients.
Ç B
äèDetermïe if ê followïg lïear, second order
èèèèèèèdifferential equations are homogenous or non-homogeneous
â 1) xìy»» - 3xy» + 4yè=è0 is homogeneous
2) y»»è-è3y»è-è10yè=èsï[x] is non-homogeneous
éSè The GENERAL, LINEAR, SECOND ORDER differential equation
is ç ê form
P(x)y»»è+è Q(x)y»è+èR(x)yè=èG(x)
If G(x) = 0 for all x, this is a HOMOGENEOUS differential
equation.è
IfèG(x) ƒ 0 for some x, this a NON-HOMOGENEOUS differential
equation.
7 xìy»»è+è(x - 1)y»è=èy sï[x]
A) Homogeneous
B) Non-homogeneous
ü This differential equation can be rearranged ë
xìy»»è+è(x - 1)y»è-èsï[x] yè=è0
å hence is homogeneous.
Ç A
8 y»»è-èxìè-è2yè=è0
A) Homogeneous
B) Non-homogeneous
ü This can be rearragned ë
y»»è-è2yè=èxì
å hence is non-homogeneous
Ç B
äèèDetermïe which ç ê followïg are solutions ç ê
èèèèèèèègiven ïitial value problems.
â è For ê ïitial value problem
y»» - 3y» - 4y = 0è
y(0) = 3 ;èy'(0) = 5
The solution is
y = -2/5 eúÅ╣è+è17/5 e╣
éS
For a differential equation ç ê second order, an Initial
Value Problem consists ç 3 parts
1) A second order differential equation
2) Two INITIAL CONDITIONS which specifyè
a)èa value ç ê solution y at a specific value
èèç ê ïdependent variable, say y(x╠) = y╠
b)èa value ç ê derivative ç ê solution, agaï
èèat a specific value ç ê ïdependent variable,
èèsayèy»(x╠) = y╠»
If ê ïitial conditions are well posed, êre will be a
UNIQUE solution ë this ïitial value problem.èTwo ïitial
conditions are needed as ê two differentiations needed ë
ë produce ê second order differential equation will take
any part ç ê solution which is ç ê form
C¬x + C½
å differentiate it ë zero.èThus two ïtegrations (å hence
two constants ç ïtegration) are necessary ë fïd ê
solutions.èThus two pieces ç ïitial ïformation are needed
ë specify ê values for êse two constants ç ïtegration.
9 y»» =è12x ;èy(0) = -2 ;èy»(0) = 3
A) y = 2xÄ + 3x + 2 B) y = 2xÄ + 3x - 2
C) y = 2xÄ - 3x + 2 D) y = 2xÄ - 3x - 2
ü For yè = 2xÄ + 3x - 2è ênèy(0) = -2
y»è= 6xì + 3èèè ênèy»(0) = 3
y»» = 12x
Thus all three conditions are met å this is ê solution.
Ç B
10 y»» = sï[x] ;èy(╥) = -╥ ; y»(╥) = -3
A) y = -sï[x] + 4x + 3╥ B)è y = -sï[x] + 4x - 3╥
C) y = -sï[x] - 4x + 3╥ D)è y = -sï[x] - 4x - 3╥
ü For y = -sï[x] - 4x + 3╥ ên y(╥) = -╥
y» = -cos[x] - 4 ên y»(╥) = 1 - 4 = -3
y»» = sï[x]
Thus all three conditions are met å this is ê solution
ë this ïitial value problem.
Ç C
11 y»» -èy»è-è2yè=è0 ;èy(0) = 3 ;èy»(0) = 0
A) y = eú╣ + 2eì╣ B) y = 2eú╣ + eì╣
C) y = eú╣ - 2eì╣ D) y = -eú╣ + 2eì╣
ü Forè yè = 2eú╣ + eì╣èên y(0) = 2 + 1 = 3
y»è= -2eú╣ + 2eì╣èên y»(0) = -2 + 2 = 0
y»» = 2eú╣ + 4eì╣
y»» - y» - 2y = 2eú╣ + 4eì╣ + 2eú╣ - 2eì╣ - 4eú╣ - 2eì╣
è
èèè= (2 + 2 - 4)eú╣ + (4 - 2 -2)eì╣
èèè= 0
Thus all three conditions are met å this is ê
solution ë ê ïitial value problem.
Ç B
äèèFor ê followïg differential equations, give ê
èèèèèèèèsolutions ç ê characteristic equation.
â For ê differential equation
y»» -è6y»'è+è8yè=è0
The characteristic equation is
mì - 6m + 8 = 0
This facërs ëè(m - 2)(m - 4) = 0 å so ê solutions
areèm = 2, 4.
éS A lïear, constant coefficient, homogeneous differential
equation
ay»» + by» + cy = 0
can be solved by assumïg a solution ç ê form
y = e¡╣
where m is a constant ë be determïed.èSubstitutïg this
possible solution ïë ê differential equation produces
am║e¡╣ + bme¡╣ + ce¡╣ = 0
or (am║ + bm + c)e¡╣ = 0
As ê range ç ê exponential function is always non-zero,
this equation can have a solution only if
am║ + bm + c = 0
This is known as ê CHARACTERTISTIC EQUATION .
As ê characteristic equation is a QUADRATIC EQUATION with
REAL coefficients, êre 3 possible types ç solution.èThese
can be categorized by ê DISCRIMINANT ç ê QUADRATIC
FORMULA
èè -b ± √(bì - 4ac)
m = ──────────────────
èèèèèè2a
The DISCRIMINANT is ê radicå under ê square root i.e.
ê discrimïant isèbì - 4ac.
1) If bì - 4ac > 0, êre are 2 REAL, DISTINCT ROOTS
2) If bì - 4ac = 0, êre are 2 REAL, REPEATED ROOTS
3) If bì - 4ac < 0, êre are 2 COMPLEX ROOTS which are
COMPLEX CONJUGATES
12 y»» + 5y» - 6y = 0
A) m = 2, 3 B) m = -2, -3
C) m = 1, -6 D) m = -1, 6
ü The characteristic equation is
mì + 5m - 6 = 0
This facërs ïë
(m - 1)(m + 6) = 0
So its solutions are
m = 1, -6
Ç C
13 2y»» - 11y» + 12y = 0
A) m = 3/2, 4 B) m = 3/2, -4
C) m = -3/2, 4 D) m = -3/2, -4
ü The characteristic equation is
2mì - 11m + 12 = 0
This facërs ïë
(2m - 3)(m - 4) = 0
So its solutions are
m = 3/2, 4
Ç A
14 y»» + 6y» + 9y = 0
A) m = 1, 9 B) m = -1, -9
C) m = 3, 3 D) m = -3, -3
ü The characteristic equation is
mì + 6m + 9 = 0
This facërs ïë
(m + 3)(m + 3) = (m + 3)ì = 0
So its solutions are repeated roots
m = -3, -3
Ç D
15 4y»» -12y» + 9y = 0
A) m = 2/3, 2/3 B) m = -2/3, -2/3
C) m = 3/2, 3/2 D)è m = -3/2, -3/2
ü The characteristic equation is
4mì -12m + 9 = 0
This facërs ïë
(2m - 3)(2m - 3) = (2m - 3)ì = 0
So its solutions are repeated roots
m = 3/2, 3/2
Ç C
16 y»» + 9y = 0
A) m = 2/3, 2/3 B) m = -2/3, -2/3
C) m = 3/2, 3/2 D)è m = ±3i
ü The characteristic equation is
mì + 9 = 0
This does NOT facër ïë real roots so ê quadratic formula
is used
èè -0 ± √[0ì - 4(1)(9)]
m = ──────────────────────
èèè2(1)
è=è±√(-36) / 2
è
è=è±6i / 2
è=è±3ièA pair ç complex conjugates
Ç D
17 y»» + 2y» + 2 = 0
A) m = 1, 2 B) m = -1, -2
C) m = 1 - i, 1 +èi D) m = -1 - i, -1 + i
ü The characteristic equation is
mì + 2m + 2 = 0
This does NOT facër ïë real roots so ê quadratic formula
is used
èè -2 ± √[2ì - 4(1)(2)]
m = ──────────────────────
èèè2(1)
è=è[-2 ± √(-4)] / 2
è
è=è[-2 ± 2i] / 2
è=è-1 ± ièA pair ç complex conjugates
Ç D