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à 8.2è LaPlace Transform Solution ç Second Order Equations
èèè
äè Solve ê ïitial value problem via LaPlace transforms
â èèForèy» + 3y = 2 ;èy(0) = -3 .èTakïg ê LaPlace
transform ç this differential equation yields [Y = ÿ{y}]
èsY + 3 + 3Y = 2/s.èRearrangïg (s+3)Y = (2-3s)/sèor
Y = (2-3s)/[s(s+3]èUsïg partial fraction decomposition
yieldsèY = 2/3 1/sè-è11/3 1/(s+3).èUsïg ê table ë do
ê ïverse transform givesèy = 2/3 - 11/3 eúÄ▐ 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 å ê next 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» - 3y = 0èèy(0) = 4
A) y = 4eÄ▐ B) y = -4eÄ▐
C) y = 4eúÄ▐ D) y = -4eúÄ▐
ü è Takïg ê LaPlace transform ç ê differential equation
y» - 3y = 0
yields via ê lïearity property, ê derivative property
å callïg Y = ÿ{ y }
èsY - y(0) - 3Y = 0
Substitutïg ê ïitial value å rearrangïg
è (s-3)Y - 4 = 0
Or
èèè 4èèèèè 1
Y =è─────è=è4 ─────
èèès-3èèèè s-3
This is already ï ê desired form, so ê reverse transform
can be done ë yield ê specific solution
y =è4eÄ▐
ÇèA
2 y» - 3yè=èeÄ▐è;èy(0) = 4 è
A) y = (t+4)eÄ▐ B) y = (t-4)eÄ▐
C) y = (t+4)eúÄ▐ D) y = (t-4)eúÄ▐
ü è Takïg ê LaPlace transform ç ê differential equation
y» - 3y = eÄ▐
yields via ê lïearity property, ê derivative property,
å callïg Y = ÿ{ y }
èsY - y(0) - 3Y = ÿ{ eÄ▐ }è=è1/ s-3
Substitutïg ê ïitial value å rearrangïg
èèèèèèèèè1
è (s-3)Y - 4 = ─────
èèèèèèèè s-3
èèèèèèèè1èèè
è (s-3)Yè=è───── + 4
èèèèèèè s-3èèè
èèèèè1èèèè 4èèèè
Yè=è────────è+ ─────
èèè (s-3)ìèèès-3
Usïg ê reverse transform yields ê specific solution
y =èteÄ▐ + 4eÄ▐ = (t+4)eÄ▐
ÇèA
3 y» - 3yè= sï[t]è y(0) = 4
A) 41/10 eÄ▐ + 1/10 cos[t] + 9/10 sï[t]
B) 41/10 eÄ▐ + 1/10 cos[t] - 9/10 sï[t]
C) 41/10 eÄ▐ - 1/10 cos[t] + 9/10 sï[t]
D) 41/10 eÄ▐ - 1/10 cos[t] - 9/10 sï[t]
ü è Takïg ê LaPlace transform ç ê differential equation
y» - 3y = sï[t]
yields via ê lïearity property, ê derivative property,
å callïg Y = ÿ{ y }
èèèèèèèèèèèèèèèèèèè1
èsY - y(0) - 3Y = ÿ{ sï[t] }è=è──────
èèèèèèèèèèèèèèèèèèsì+1
Substitutïg ê ïitial value å rearrangïg
èèèèèèèèè 1
è (s-3)Y - 4 = ──────
èèèèèèèè sì+3
èèèèèèèè 1èèèèèè4sì+5
è (s-3)Yè=è────── + 4è=è───────
èèèèèèè sì+3èèèèèèsì+3
èèèèè4sì+5èèèè
Yè=è────────────
èèè (s-3)(sì+1)
è Usïg partial fraction decomposition requires
èèèè 4sì+5èèèè Aèèè Bs+C
èèè───────────è=è───è+è──────
èèè(s-3)(sì+1)èè s-3èèèsì+1
where A, B, C are undetermïed constants.
è Multiplyïg this equation by (s-3)(sì+1), ê least common
denomïaër yields
4sì+5è=èA(sì+1) + (Bs+C)(s-3)
If s = 3èè41 =è10Aèi.e.èA = 41/10
If s = 0èè 5 =èA - 3Cè=è41/10 - 3C
èè3C = 41/10 - 5 = -9/10èi.e. C = -3/10
If s = 1èè 9 = 2A - 2B - 2Cè= 82/10 - 2B + 6/10
èèèèèè2B = 88/10 - 9 = -2/10èi.e B = -1/10
Thusèèèèè41èè1èèèè1èè 1èèèè9èè 1
èèè Yè=è──── ─────è-è─── ──────è-è─── ──────
èèèèèèè10è s-3èèè 10èsì+1èèè 10èsì+1
Usïg ê reverse transform yields ê specific solution
èè 41èèèèè1èèèèèè 9
y =è── eÄ▐è-è── cos[t]è-è── sï[t]
èè 10èèèè 10èèèèèè10
ÇèD
4 y»» + 4yè=è0è y(0) = 3è y»(0) = -4
A) 3cos[2t] + 2sï[2t]
B) 3cos[2t] - 2sï[2t]
C) -3cos[2t] + 2sï[2t]
D) -3cos[2t] - 2sï[2t]
ü è Takïg ê LaPlace transform ç ê differential equation
y»» + 4yè=è0
yields via ê lïearity property, ê derivative property
å callïg Y = ÿ{ y }
èsìY - sy(0) - y»(0) + 4Y = 0
Substitutïg ê ïitial value å rearrangïg
èèèèèèèèè
è (sì+4)Y - 3s + 4 = 0
èèè
è (sì+4)Yè=è3s - 4èè
èèè 3s-4èèèèè sèèèèèè2s
Yè=è──────è=è 3 ──────è-è2 ──────
èèè sì+4èèèèèsì+4èèèè sì+4
Usïg ê reverse transform yields ê specific solution
y =è3cos[2t] - 2sï[2t]
ÇèB
5 y»» - y =èeì▐èè y(0) = 3è y»(0) = 2
A) 1/3 eì▐ + 2e▐ + 2/3eú▐
B) 1/3 eì▐ + 2e▐ - 2/3eú▐
C) 1/3 eì▐ - 2e▐ + 2/3eú▐
D) 1/3 eì▐ - 2e▐ - 2/3eú▐
ü è Takïg ê LaPlace transform ç ê differential equation
y»» - yè=èeì▐
yields via ê lïearity property, ê derivative property,
å callïg Y = ÿ{ y }
èsìY - sy(0) - y»(0) - Y = ÿ{ eì▐ }
Substitutïg ê ïitial value å rearrangïg
èèèèèèèèèèèè 1
è (sì-1)Y - 3s - 2 =è─────
èèèèèèèès-2
èèèèèèèèèèèèè 1èèè3sì-4s-3
è (sì-1)Yè=è3s + 2è+ ───── = ──────────èè
èèèèèèèèèèèèès-2èèèès-2
èèèèè3sì-4s-3èèè
Yè=è─────────────────
èèè (s-1)(s+1)(s-2)
è Usïg partial fraction decomposition requires
èèèè3sì-4s-3èèèèèèAèèèè Bèèèè C
èè─────────────────è=è─────è+è─────è+è─────
èè (s-1)(s+1)(s-2)èèè s+1èèè s-1èèè s-2
where A, B, C are undetermïed constants.
è Multiplyïg this equation by (s+1)(s-1)(s-2), ê least
common denomïaër yields
3sì-4s-3è=èA(s-1)(s-2) + B(s+1)(s-2) + C(s+1)(s-1)
Set s = -1è 4 = 6Aèi.e.èA = 2/3
Set s = 1è -4 = -2Bèi.e. B = 2
Set s= 2èè 1 = 3Cè i.e. C = 1/3
èèèè èè 2è 1èèèèè1èèè1è 1
Yè= ─ ─────è+ 2 ─────è+ ─ ─────
èè 3ès+1èèèès-1èè 3ès-2
Usïg ê reverse transform yields ê specific solution
y =è2/3 eú▐ + 2e▐ + 1/3 eì▐
ÇèA
6 y»» - 4y = 2t - 3èèy(0) = 3è y»(0) = -4
A) 3/4è+è1/2 tè+è1/4 eì▐è+ 2 eì▐
B) 3/4è+è1/2 tè+è1/4 eì▐è- 2 eì▐
C) 3/4è+è1/2 tè-è1/4 eì▐è+ 2 eì▐
D) 3/4è-è1/2 tè+è1/4 eì▐è+ 2 eúì▐
ü è Takïg ê LaPlace transform ç ê differential equation
y»» - 4yè=è2t - 3
yields via ê lïearity property, ê derivative property
å callïg Y = ÿ{ y }
èsìY - sy(0) - y»(0) - 4Y = ÿ{ 2t - 3 }
Substitutïg ê ïitial value å rearrangïg
èèèèèèèèèèèè2èèè 3
è (sì-4)Y - 3s + 4 =è───è-è───
èèèèèèèèsìèèès
èèèèèèèèèèèèè2èè 3èè 3sÄ - 4Äs -3s + 2
è (sì-4)Yè=è3s - 4è+ ─── - ─── = ───────────────────èè
èèèèèèèèèèèèèsìèèsèèèèèè sì
èèè 3sÄ-4sì-3s+2èèè
Yè=è──────────────
èèèèsì(s-2)(s+2)
è Usïg partial fraction decomposition requires
èè 3sÄ-4sì-3s+2èèè Aèèè Bèèè CèèèèD
èè──────────────è=è───è+è───è+ ────è+è─────
èè sì(s-2)(s+2)èèè sèèè sìèè s-2èèès+2
where A, B, C, D are undetermïed constants.
è Multiplyïg this equation by sì(s+2)(s-2), ê least
common denomïaër yields
è 3sÄ-4sì-3s+2 = As(s-2)(s+2) + B(s-2)(s+2) + Csì(s+2) + Dsì(s-2)
Set s = 0èè2è=è-4Bèi.e.èB = -1/2
Set s = -2è-32 = -16Dèi.e.èD = 2
Set s = 2è 4 = 16Cèi.e. C = 1/4
Set s= 1èè-2 = -3A - 3B + 3C - D = -3A + 3/2 + 3/4 - 2
èè-9/4 = -3Aèi.e.èA = 3/4
èèèè èè 3 1èè 1è1èè 1è 1èèèèè1è
Yè= ─ ─è-è─ ───è+ ─ ─────è+ 2 ─────
èè 4 sèè 2èsìèè4ès-2èèèès+2
Usïg ê reverse transform yields ê specific solution
y =è3/4 - 1/2ètè+è1/4 eì▐è+è2eúì▐
ÇèD
7 y»» - 4y» + 3y = 0èè y(0) = -2èèy»(0) = 5
A) 11/2 e▐ + 7/2 eÄ▐
B) 11/2 e▐ - 7/2 eÄ▐
C) -11/2 e▐ + 7/2 eÄ▐
D) -11/2 e▐ - 7/2 eÄ▐
ü è Takïg ê LaPlace transform ç ê differential equation
y»» - 4y» + 3yè=è0
yields via ê lïearity property, ê derivative property
å callïg Y = ÿ{ y }
èsìY - sy(0) - y»(0) - 4{ sY - y(0)} + 3Y = 0
Substitutïg ê ïitial value å rearrangïg
èèèèèèèèè
è (sì-4s+3)Y + 2s - 5 - 8 = 0
èèè
è (sì-4s+3)Yè=è-2s + 13èè
èèèè -2s+13èèèèè
Yè=è────────────è
èèè (s-1)(s-3)è
è Usïg partial fraction decomposition requires
èèè -2s+13èèèèèAèèèè Bè
èè────────────è=è─────è+è─────
èè (s-1)(s-3)èèè s-1èèè s-3
where A, B are undetermïed constants.
è Multiplyïg this equation by (s-1)(s-3), ê least
common denomïaër yields
èèè -2s + 13è=èA(s - 3)è+èB(s - 1)
For s = 1è 11 =è-2Aèi.e.èA = -11/2
For s = 3èè7 =è2Bè i.e.èB = 7/2
Thusèèèèè 11è 1èèè 7è 1
èèè Yè=è- ── ─────è+è─ ─────
èèèèèèèè2ès-1èèè2ès-3
Usïg ê reverse transform yields ê specific solution
y =è-11/2 e▐ + 7/2 eÄ▐
ÇèC
è8 y»» + 3y» + 2y = 7èè y(0) = 6è y»(0) = -3
A) 7/2è+è2 eú▐è+è1/2 eúì▐
B) 7/2è+è2 eú▐è-è1/2 eúì▐
C) 7/2è-è2 eú▐è+è1/2 eúì▐
D) -7/2è+è2 eú▐è+è1/2 eúì▐
ü è Takïg ê LaPlace transform ç ê differential equation
y»» + 3y» + 2yè=è7
yields via ê lïearity property, ê derivative property
å callïg Y = ÿ{ y }
èsìY - sy(0) - y»(0) + 3{ sY - y(0)} + 2Y = ÿ{ 7 }
Substitutïg ê ïitial value å rearrangïg
èèèèèèèèèèèèèèè 7
è (sì+3s+2)Y - 6s + 3 - 18 = ───
èèèèèèèèèèè s
èèèèèèèèèèèèèè 7èèè6sì + 15s + 7
è (sì+3s+2)Yè=è6s + 15 + ───è= ───────────────
èèèèèèèèèèèèèè sèèèèèès
èèèè6sì+15s+7èèèèè
Yè=è────────────è
èèè s(s+1)(s+2)è
è Usïg partial fraction decomposition requires
èèè6sì+15s+7èèè AèèèèBèèèè C
èè────────────è=è───è+è─────è+è─────
èè s(s+1)(s+2)èèèsèèè s+1èèè s+2
where A, B, C are undetermïed constants.
è Multiplyïg this equation by s(s+1)(s+2), ê least
common denomïaër yields
èè6sì+15s+7è=èA(s+1)(s+2) + Bs(s+2) + Cs(s+1)
For s = 0è 7 =è2Aèi.e.èA = 7/2
For s = -1è-2 = -Bèi.e.èB = 2
For s = -2è1 = 2Cèi.e.èC = 1/2
Thusèèèè7è1èèèè1èèè1è 1
èèè Yè= ─ ─── - 2 ───── +è─ ─────
èèèèèè2èsèèè s+1èè 2ès+2
Usïg ê reverse transform yields ê specific solution
y =è7/2 - 2eú▐ + 1/2 eúì▐
ÇèC
9 y»» + 2y» + 2yè=è0èèy(0) = 6èèy»(0) = -3
A) 6eú▐cos[t] + 3eú▐sï[t]
B) 6eú▐cos[t] - 3eú▐sï[t]
C) -6eú▐cos[t] + 3eú▐sï[t]
D) -6eú▐cos[t] - 3eú▐sï[t]
ü è Takïg ê LaPlace transform ç ê differential equation
y»» + 2y» + 2yè=è0
yields via ê lïearity property, ê derivative property
å callïg Y = ÿ{ y }
èsìY - sy(0) - y»(0) + 2{ sY - y(0)} + 2Y = 0
Substitutïg ê ïitial value å rearrangïg
èèèèèèèèèèèèèèè
è (sì+2s+2)Y - 6s + 3 - 6è=è0
èèèèèèèèèèè
è (sì+2s+2)Yè=è6s + 3
èèèèèèèèèèèè
èèèè6s+3èèèèè
Yè=è─────────è
èèè sì+2s+2è
Asèsì+2s+2 is an IRREDUCIBLE QUADRATIC, this is fïal form
if ê square is completed ï ê denomïaër
èèèèèè sèèèèèèè 1èèè
Yè=è6 ──────────è+è3 ──────────è
èèèè (s+1)ì+1èèèè (s+1)ì+1
Usïg ê reverse transform yields ê specific solution
y =è6eú▐cos[t] + 3eú▐sï[t]
ÇèA