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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
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- If s = -2è40 = -20Dèi.e.èD = -2
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- If s = 0è 14è= -4B + 2C - 2Dè=è-4B + 2 + 4
- èèèèèè8è= -4Bèi.e.èB = -2
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- If s = 1è 16 = -3A - 3B + 6C - 2D = -3A + 6 + 6 + 4
- èèèèè 0 = -3Aè i.e.èA =è0
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- èèèèèè1èèèè 1èèèèè1è
- Y =è- 2 ──────è+è─────è- 2 ─────
- èèèèèsì+1èèè s-2èèèès+2
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- The reverse transform can be done, usïg ê table, ë yield
- ê specific solution
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- y =è- 2sï[t] + eì▐ - 2eúì▐
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- ÇèC
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