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à 8.3è LaPlace Transform Solution ç Higher Order Equations
èè
äè Solve ê ïitial value problem via LaPlace transforms
â èèForèy»»»» - y = 0 ;èy(0) = -3; y»(0) = 0; y»»(0) = -1;
y»»»(0) = 0.èTakïg ê LaPlace transform ç this
differential equation yields [Y = ÿ{y}] sÅY - sÄ + sè- Y = 0.è
Rearrangïg (sÅ - 1)Y = sÄ - sèor Y = (sÄ-s)/[sÅ-1]èUsïg
partial fraction decomposition yieldsèY = s/[sì+1].èUsïg
ê table ë do ê ïverse transform givesèè y = cos[t]
as ê specific solution ç ê ïitial value problem.
éSè The LAPLACE TRANSFORM can be used ë directly solve an
Initial Value Problem which has a lïear, constant coeffi-
cient differential equation.èThis is due ë ê transform
property ç ê derivative function ç order n.
ÿ{fÑⁿª(t)} = sⁿÿ{f(t)} - sⁿúîf(0) - ∙∙∙
èèèèèèèèèèèè- sfÑⁿú²ª(0) - fÑⁿúîª(0)
As is seen, ê n-1 ïitial conditions are embedded ï ê
transform.èThis is different from ê usual technique for
solvïg ïitial value problems ç first fïdïg a GENERAL
SOLUTION ç ê differential equation å ên substitutïg
ê ïitial ïdependent variable ïë ê general solution
å its derivatives å ên solvïg for ê n arbitrary
constants.
èèPart ç ê ease ç ê LaPlace transform technique is ë
use a table ç tranforms.èThe followïg table should be
copied for use ï ê problems ç this section.
èèèèèèèèèè1
1.èèèÿ{ 1 }è=è───
èèèèèèèèèès
èèèèèèèèèè n!
2.èèèÿ{ tⁿ } =è──────
èèèèèèèèèèsⁿóî
èèèèèèèèèè 1
3.èèèÿ{ e╜▐ } = ─────
èèèèèèèèèès-a
èèèèèèèèèèèès
4.èèèÿ{cos[at]} = ───────
èèèèèèèèèèèsì+aì
èèèèèèèèèèèèa
5. ÿ{sï[at]} = ───────
èèèèèèèèèèèsì+aì
èèèèèèèèèèèè s
6. ÿ{cosh[at]} = ───────
èèèèèèèèèèè sì-aì
èèèèèèèèèèèè a
7. ÿ{sïh[at]} = ───────
èèèèèèèèèèè sì-aì
èèèèèèèèèèèèèès-a
8. ÿ{e╜▐cos[bt]} = ───────────
èèèèèèèèèèèè (s-a)ì+bì
èèèèèèèèèèèèèè b
9. ÿ{e╜▐sï[bt]} = ───────────
èèèèèèèèèèèè (s-a)ì+bì
10. ÿ{fÑⁿª(t)} = sⁿÿ{f(t)} - sⁿúîf(0) - ∙∙∙
èèèèèèèèèèèè- sfÑⁿú²ª(0) - fÑⁿúîª(0)
11.è ÿ{ C¬f¬(t) + C½f½(t) } = C¬ÿ{ f¬(t) } + C½ÿ{ f½(t) }
èè The basic technique is ê usual transform process
1) Transform ê problem ë a different but related
variable
2) Solve ê transformed problem ï terms ç ê
related variable.
3) Transform back ë ê origïal variable ë get ê
solution ë ê origïal problem.
èèThese steps will be illustrated ï solvïg ê ïitial
value problem
y»» - yè=è0
y(0)è= 3
y»(0) = 2
1)èèTake ê LaPlace transform ç ê entire differential
equation å use ê DERIVATIVE property å ê LINEARITY
property
ÿ{ y»» - y }è=èÿ{ 0 }
By lïearity
ÿ{ y»» } - ÿ{ y }è=è0
By ê derivative property
sìÿ{ y }è- sy(0)è-èy»(0)è-èÿ{ y }è=è0
Substitutïg for ê ïitial values å settïg Y = ÿ{y}
sìYè-è3sè-è2è- Yè=è0
2) Solve for Y(s) å use PARTIAL FRACTION DECOMPOSITION
ë write Y as a sum ç fractions whose denomïaërs are lïear
terms or irreducible (over ê reals) quadratic terms.
Rearrangïg
(sì - 1)Yè=è3s + 2
Solvïg for Y
èèè 3s+2èèèèè3s+2
Yè=è──────è=è────────────
èèè sì-1èèè (s-1)(s+1)
The partial fraction decomposition is
èèè 3s + 2èèèèèAèèèè B
èè────────────è=è─────è+è─────
èè (s-1)(s+1)èèè s-1èèè s+1
where A å B are constants ë be determïed.
èèMultiplyïg both sides byè(s-1)(s+1) yields
èèè3s + 2è=èA(s+1) + B(s-1)
èèThere are several methods for solvïg for A å B.
Probably ê easiest, particularly when lïear facërs are
ïvolved is ë substitute strategic values ç s.èFor this
case substitute values ç s that make ê multiplyïg facërs
zeroèi.e.ès = -1, 1
s = 1è 5 = 2Aè i.e.èA = 5/2
s = -1è-1 = -2B i.e.èB = 1/2
Thusèèèèè5è 1èèè 1è 1
èèèèYè=è─ ─────è+è─ ─────è
èèèèèèè2ès-1èèè2ès+1
3) Use ê table ë take ê ïverse transform i.e. go
from ê transformed solution Y(s) back ë ê orgïal
solution y(t).èLook ï ê table ë fïd ê transform
given with its specific value ç constant(s) å write it
ï terms ç ê origïal function ç t.èThe lïearity
property holds ï both directions.
Usïg ê transform
èèèèèèèèèè 1
èèèèÿ{ e╜▐ } = ─────
èèèèèèèèèès-a
with a = 1èfor 1/s-1èå a = -1 for 1/s+1,
ê specific solution becomes
y = 5/2 e▐ + 1/2 eú▐
1 y»»» + 2y»» = 0;èy(0) = -6; y»(0) = 1; y»»(0) = 8
A) y = 8 + 5t + 2eúì▐
B) y = 8 + 5t - 2eúì▐
C) y = 8 - 5t + 2eúì▐
D) y = -8 + 5t + 2eúì▐
ü è Takïg ê LaPlace transform ç ê differential equation
y»»» + 2y»» = 0
yields via ê lïearity property, ê derivative property
å callïg Y = ÿ{ y }
èsÄY - sìy(0) - sy»(0) - y»»(0) + 2[sìY -sy(0) - y»(0)] = 0
Substitutïg ê ïitial values å rearrangïg
èsÄY + 6sì - s - 8 + 2[sìY + 6s - 1] = 0
èè (sÄ+2sì)Yè= -6sì - 11s + 10
Or
èèè-6sì-11s+10
Y =è─────────────
èèèèsì(s+2)
In order ë put this ï a form where ê denomïaërs are eiêr
lïear or irreducible quadratic facërs, ê method ç
PARTIAL FRACTION DECOMPOSITION is used.
èè -6sì-11s+10èèè Aèèè BèèèèC
èè─────────────è=è───è+è───è+è─────
èèè sì(s+2)èèèè sèèè sìèèès+2
where A, B, C are undetermïed constants.è
Multiplyïg both sides by ê least common denomïaër
sì(s+2)
-6sì-11s+10 = As(s+2) + B(s+2) + Csì
If s = 0è 10 = 2Bèèi.e.èB = 5
If s = -2è 8 = 4Cèèi.e.èC = 2
If s = 1è -7 = 3A + 3B + C = 2A + 15 + 2
èèèèè -24 = 3Aè i.e.èA = -8
èèèè 1èèèè1èèèè 1
Y =è-8 ───è+ 5 ───è+ 2 ─────
èèèè sèèèèsìèèè s+2
The reverse transform can be done, usïg ê table, ë yield
ê specific solution
y =è-8 + 5t + 2eúì▐
ÇèD
2 y»»» - 3y»» + 2y»è=è6eÄ▐
y(0) = 5;èy»(0) = 6;èy»»(0) = 16
A) y = 3 - e▐ + 2eì▐ + eÄ▐
B) y = 3 + e▐ + 2eì▐ - eÄ▐
C) y = 3 + e▐ - 2eì▐ + eÄ▐
D) y = 3 + e▐ + 2eì▐ + eÄ▐
ü è Takïg ê LaPlace transform ç ê differential equation
y»»» - 3y»» + 2y» =è6eÄ▐
yields via ê lïearity property, ê derivative property
å callïg Y = ÿ{ y }
èsÄY - sìy(0) - sy»(0) - y»»(0) - 3[sìY -sy(0) - y»(0)]
+ 2[ sY -y(0)]è=èÿ{ eÄ▐ }è=è1/ s-3
Substitutïg ê ïitial values å rearrangïg
èsÄY - 5sì - 6s - 16 - 3[sìY - 5s - 6]
èèèèèèèèèèèèèè1
èèèèè+ 2[sY - 5]è=è─────
èèèèèèèèèèèèè s-3
èèèèèèèèèèèèèèèèèèèè1
èè (sÄ-3sì+2s)Yè= 5sì - 9s + 8è+è─────
èèèèèèèèèèèèèèèèèèè s-3
èèèèèèèèèèè 5sÄ - 24sì + 35s - 18
èèèèèèèèYè=è───────────────────────
èèèèèèèèèèèèès(s-1)(s-2)(s-3)
In order ë put this ï a form where ê denomïaërs are eiêr
lïear or irreducible quadratic facërs, ê method ç
PARTIAL FRACTION DECOMPOSITION is used.
èè 5sÄ-24sì+35s-18èèè AèèèèBèèèè Cèèè D
èè─────────────────è=è───è+è─────è+è───── + ─────
èè s(s-1)(s-2)(s-3)èèèsèèè s-1èèè s-2èè s-3
where A, B, C, D are undetermïed constants.è
Multiplyïg both sides by ê least common denomïaër
s(s-1(s-2)(s-3)
5sÄ-24sì+35s-18 = A(s-1)(s-2)(s-3) + Bs(s-2)(s-3)
èèèèèèèèè + Cs(s-1)(s-3) + Ds(s-1)(s-2)
If s = 0è-18 = -6Aèèi.e.èA = 3
If s = 1è -2 = 2Bèè i.e.èB = -1
If s = 2è -4 = -2Cèèi.e.èC = 2
If s = 3èè6 = 6Dèè i.e.èD = 1
èèèè 1èèèè1èèèèè1èèèè 1
Y =è 3 ───è-è─────è+ 2 ─────è+è─────
èèèè sèèè s-1èèèès-2èèè s-3
The reverse transform can be done, usïg ê table, ë yield
ê specific solution
y =è3 - e▐ + 2eì▐ + eÄ▐
ÇèA
3 y»»» -2y»» + y» - 2yè=è=
y(0) = -6è y»(0) = -5è y»»(0) = -14
A) 2cos[t] + 3sï[t] + 4eì▐
B) -2cos[t] + 3sï[t] - 4eì▐
C) 2cos[t] - 3sï[t] + 4eì▐
D) 2cos[t] + 3sï[t] - 4eì▐
ü è Takïg ê LaPlace transform ç ê differential equation
y»»» - 2y»» + y» - 2yè=è0
yields via ê lïearity property, ê derivative property
å callïg Y = ÿ{ y }
èsÄY - sìy(0) - sy»(0) - y»»(0) - 2[sìY -sy(0) - y»(0)]
+ [ sY -y(0)] -2Yè=è0
Substitutïg ê ïitial values å rearrangïg
èsÄY + 6sì + 5s + 14 - 2[sìY + 6s + 5]
èèèèèèèèèèèè
èèèèè+ [sY + 6] - 2Yè=è0
èèèèèèèèèèèè èèèèèèèèèèèèèèèèèè
èè (sÄ-2sì+s-2)Yè= -6sì + 7s - 10
èèèèèèèèèèèèèèèèèè
èèèèèèèèèèè - 6sì + 7s - 10
èèèèèèèèYè=è─────────────────
èèèèèèèèèèèè(sì+1)(s-2)
In order ë put this ï a form where ê denomïaërs are eiêr
lïear or irreducible quadratic facërs, ê method ç
PARTIAL FRACTION DECOMPOSITION is used.
èèè- 6sì + 7s - 10èèè As+BèèèèC
èè ─────────────────è=è──────è+è─────
èèèè(sì+1)(s-2)èèèè sì+1èèè s-2
where A, B, C are undetermïed constants.è
Multiplyïg both sides by ê least common denomïaër
(sì+1)(s-2)
èèè- 6sì+7s-10è= As(s-2) + B(s-2) + C(sì+1)
If s = 2è -20 = 5Cèèi.e.èC = -4
If s = 0è -10 = -2B + Cè=è-2B - 4
èèèèèèèèè -6è= -2Bè i.e.èB = 3
If s = 1è -9è= -A - B + 2Cè=è-A - 3 - 8
èèèèèèèèèè2è= -Aèèi.e.èA = -2
èèèèè sèèèèè 1èèèèè1è
Y =è-2 ──────è+ 3 ──────è- 4 ─────
èèèè sì+1èèèèsì+1èèèès-2è
The reverse transform can be done, usïg ê table, ë yield
ê specific solution
y =è-2cos[t] + 3sï[t] - 4eì▐
ÇèB
4 y»»»» - y = 0è
y(0) = 2;èy»(0) = 1; y»»(0) = 2;èy»»»(0) = 3
A) y = sï[t] + eì▐
B) y = sï[t] - eì▐
C) y = -sï[t] + eì▐
D) y = -sï[t] - eì▐
ü è Takïg ê LaPlace transform ç ê differential equation
y»»»» - yè=è0
yields via ê lïearity property, ê derivative property
å callïg Y = ÿ{ y }
èsÅY - sÄy(0) - sìy»(0) - sy»»(0) - y(0) - Yè=è0
Substitutïg ê ïitial values å rearrangïg
èsÅY - 2sÄ - sì - 2s - 3 - Yè=è0
èèèèèèèèèèèè
èèèèèèèèèèèèèèèèèè
èè (sÅ-1)Yè= 2sÄ + sì + 2s + 3
èèèèèèèèèèèèèèèèèè
èèèèèèèèèèèè 2sÄ+sì+2s+3
èèèèèèèèYè=è─────────────────
èèèèèèèèèèè (sì+1)(s-1)(s+1)
In order ë put this ï a form where ê denomïaërs are eiêr
lïear or irreducible quadratic facërs, ê method ç
PARTIAL FRACTION DECOMPOSITION is used.
èèèè2sÄ+sì+2s+3èèèèèAs+BèèèèCèèèè D
èè ──────────────────è=è──────è+è─────è+è─────
èèè(sì+1)(s-1)(s+1)èèè sì+1èèè s-1èèè s+1
where A, B, C, D are undetermïed constants.è
Multiplyïg both sides by ê least common denomïaër
(sì+1)(s-1)(s+1)
2sÄ+sì+2s+3è= As(s-1)(s+1) + B(s-1)(s+1)
+ C(sì+1)(s+1)è+èD(sì+1)(s-1)
If s = 1è 8 = 4Cèèi.e.èC = 2
If s = -1è 0 = -2Dèi.e.èD = 0
If s = 0è 3è= -B + C - 2Dè=è-A + 2
èèèèèèèèè 1 = -Bèèi.e.èB = -1
If s = 2è 27 = 6A + 3B + 15C + 5D = 6A - 3 + 30
èèèèèè0è6Aèèi.e.èA = 0
èèèèè1èèèèè1èè
Y =è- ──────è+ 2 ────
èèèèsì+1èèèès-1èèè
The reverse transform can be done, usïg ê table, ë yield
ê specific solution
y =è-sï[t] + 2e▐
ÇèC
5 y»»»» - y»»è=è0
y(0) = 2; y»(0) = 7; y»»(0) = -1; y»»»(0) = 9
A) 3 + 2t + 4e▐ + 5eú▐
B) 3 - 2t + 4e▐ - 5eú▐
C) -3 + 2t - 4e▐ + 5eú▐
D) -3 - 2t - 4e▐ - 5eú▐
ü è Takïg ê LaPlace transform ç ê differential equation
y»»»» - y»»è=è0
yields via ê lïearity property, ê derivative property
å callïg Y = ÿ{ y }
èsÅY - sÄy(0) - sìy»(0) - sy»»(0) - y(0)
- [ sìY - sy(0) - y»(0) ]è =è0
Substitutïg ê ïitial values å rearrangïg
èsÅY - 2sÄ - 7sì + s - 9 - sìY + 2s + 7è=è0
èèèèèèèèèèèè
èèèèèèèèèèèèèèèèèè
èè (sÅ-sì)Yè= 2sÄ + 7sì - 3s + 2
èèèèèèèèèèèèèèèèèè
èèèèèèèèèèèè2sÄ+7sì-3s+2
èèèèèèèèYè=è───────────────
èèèèèèèèèèèèsì(s-1)(s+1)
In order ë put this ï a form where ê denomïaërs are eiêr
lïear or irreducible quadratic facërs, ê method ç
PARTIAL FRACTION DECOMPOSITION is used.
èèè2sÄ+7sì-3s+2èèè AèèèBèèèèCèèèè D
èè ──────────────è=è───è+ ───è+è─────è+è─────
èèèsì(s-1)(s+1)èèè sèèèsìèèès-1èèè s+1
where A, B, C, D are undetermïed constants.è
Multiplyïg both sides by ê least common denomïaër
è sì(s-1)(s+1)
2sÄ+7sì-3s+2è= As(s-1)(s+1) + B(s-1)(s+1)
+ Csì(s+1)è+èDsì(s-1)
If s = 0è 2 = -Bèèi.e.èB = -2
If s = -1è10 = -2Dèi.e.èD = -5
If s = 1è 8è= 2Cè i.e.èC =è4
If s = 2è 40 = 6A + 3B + 12C + 4D = 6A - 6 + 48 - 20
èèèèè 18 = 6Aè i.e.èA =è3
èèè 1èèèè1èèèè 1èèèèè1
Y = 3 ───è- 2 ───è+ 4 ─────è- 5 ─────
èèè sèèèèsìèèè s-1èèèès+1
The reverse transform can be done, usïg ê table, ë yield
ê specific solution
y =è3 - 2t + 4e▐ - 5eú▐
ÇèB
6 y»»»» - 3y»» - 4yè=è0
y(0) = -1;èy»(0) = 4; y»»(0) = -4;èy»»»(0) = 26
A) 2sï[t] + eì▐ + 2eúì▐
B) 2sï[t] - eì▐ - 2eúì▐
C) -2sï[t] + eì▐ - 2eúì▐
D) -2sï[t] - eì▐ + 2eúì▐
ü è Takïg ê LaPlace transform ç ê differential equation
y»»»» - 3y»» - 4yè=è0
yields via ê lïearity property, ê derivative property
å callïg Y = ÿ{ y }
èsÅY - sÄy(0) - sìy»(0) - sy»»(0) - y(0)
- 3[ sìY - sy(0) - y»(0) ] - 4Yè =è0
Substitutïg ê ïitial values å rearrangïg
èsÅY + sÄ - 4sì + 4s - 26 - 3sìY - 3s + 12 - 4Y =è0
èèèèèèèèèèèè
èèèèèèèèèèèèèèèèèè
èè (sÅ-3sì-4)Yè= -sÄ + 4sì - s + 14
èèèèèèèèèèèèèèèèèè
èèèèèèèèèèèè -sÄ+4sì-s+14
èèèèèèèèYè=è──────────────────
èèèèèèèèèèè (sì+1)(s-2)(s+2)
In order ë put this ï a form where ê denomïaërs are eiêr
lïear or irreducible quadratic facërs, ê method ç
PARTIAL FRACTION DECOMPOSITION is used.
èèèè-sÄ+4sì-s+14èèèèAs+BèèèèCèèèè D
èè ─────────────────è=è──────è+è─────è+è─────
èèè(sì+1)(s-1)(s+1)èèèsì+1èèè s-2èèè s+2è
where A, B, C, D are undetermïed constants.è
Multiplyïg both sides by ê least common denomïaër
è (sì+1)(s-2)(s+2)
-sÄ+4sì-s+14è= As(s-2)(s+2) + B(s-2)(s+2)
+ C(sì+1)(s+2)è+èD(sì+1)(s-2)
If s = 2è 20 = 20Cèèi.e.èC = 1
If s = -2è40 = -20Dèi.e.èD = -2
If s = 0è 14è= -4B + 2C - 2Dè=è-4B + 2 + 4
èèèèèè8è= -4Bèi.e.èB = -2
If s = 1è 16 = -3A - 3B + 6C - 2D = -3A + 6 + 6 + 4
èèèèè 0 = -3Aè i.e.èA =è0
èèèèèè1èèèè 1èèèèè1è
Y =è- 2 ──────è+è─────è- 2 ─────
èèèèèsì+1èèè s-2èèèès+2
The reverse transform can be done, usïg ê table, ë yield
ê specific solution
y =è- 2sï[t] + eì▐ - 2eúì▐
ÇèC