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- à 5.3èReal Roots- Fourth Order, Lïear, Constant Coefficient
- èè Differential Equation
- äèFïd ê general solution
-
- â y»»»» - 36y»» = 0
- The characteristic equation
- mÅ - 36mì = 0
- Facërs ïëè mì(m - 6)(m + 6) = 0
- The solutions areèm = 0, 0, -6, 6
- The general solution is
- C¬ + C½x +èC¼eúæ╣ + C«eæ╣
-
- éS è The LINEAR, HOMOGENEOUS, CONSTANT COEFFICIENT, FOURTH ORDER
- DIFFERENTIAL EQUATION can be written ï ê form
- Ay»»»» + By»»» + Cy»» + Dy» + Ey = 0
- where A, B, C, D å E are constants.
- è As with ê correspondïg SECOND ORDER differential
- equations, an assumption is made that ê form ç ê solutions
- is
- y = e¡╣
- Differentiatïg å substitutïg yields
- (AmÅ + BmÄ + Cmì + Dm + E)e¡╣ = 0
- As e¡╣ is never zero, it can be cancelled yieldïg ê
- CHARACTERISTIC EQUATION
- AmÅ + BmÄ + Cmì + Dm + E = 0
-
- èèQUARTIC EQUATIONS with real coefficients fall ïë one ç
- three categories :
- a) ALL roots are REAL (possibly repeated)
- b) TWO reals (possibly repeated) å a COMPLEX
- CONJUGATE PAIR
- c) TWO pairs ç COMPLEX CONJUGATES (possibly
- repeated)
-
- This section will cover ê case ç all real roots
- å Section 5.4 will cover ê cases where êre are complex
- roots.
-
- èèIf all FOURS ROOTS are REAL, êre are FOUR subcases
- a) 4 distïct roots, say l, m, n, g
- The general solution is
- C¬e╚╣ + C½e¡╣ + C¼eⁿ╣ + C½e╩╣
-
- b) 2 repeated roots, say m, m å two oêr
- disticnt roots, say n, l.èThe general solution is
- C¬e¡x + C½xe¡╣ + C¼eⁿ╣ + C«e╚╣
-
- c) 3 repeated roots, say m, m, m å one oêr
- root, say n
- The general solution is
- C¬e¡x + C½xe¡╣ + C¼xìe¡╣ + C«eⁿ╣
-
- d) 4 repeated roots, say m, m, m, m.èThe general
- solution is
- C¬e¡x + C½xe¡╣ + C¼xìe¡╣ + C«xÄe¡╣
-
- As with ê second order, non-homogeneous differential
- equation, solvïg a fourth order, NON-HOMOGENEOUS differential
- equation is done ï two parts.
-
- 1) Solve ê HOMOGENEOUS differential equation for a
- GENERAL SOLUTION with FOUR ARBITRARY CONSTANTS
-
- 2) Fïd ANY PARTICULAR SOLUTION ç ê NON-HOMOGENEOUS
- differential equation.èAs disucssed ï CHAPTER 5, êre are
- two maï techniques for fïdïg a particular solution.
-
- A) METHOD OF UNDETERMINED COEFFICIENTS
- This technique is used when ê non-homogeneous
- term is
- 1)è A polynomial
- 2)è A real exponential
- 3)è A sïe or cosïe times a real exponential
- 4)è A lïear combïation ç ê above.
- This technique is explaïed ï à 4.3 å can be
- for ANY ORDER differential equation.
-
- B) METHOD OF VARIATION OF PARAMETERS
- This technique is valid for an ARBITRARY NON-HOMOGEN-
- EOUS TERM.èIt does require ê ability ë evaluate
- N ïtegrals for an Nth order differential equaën.
- As ê order ç ê differential equation ïcreses,
- ê ïtegrals become messier ï general.èThe second
- order version is discussed ï à 4.4.
-
- 1èè y»»»» + 6y»»» + 11y»» + 6y» = 0
-
- A)è C¬ + C½e╣ + C¼eì╣ + C«eÄ╣
- B) C¬ + C½e╣ + C¼eì╣ + C«eúÄ╣
- C)è C¬ + C½e╣ + C¼eúì╣ + C«eúÄ╣
- D)è C¬ + C½eú╣ + C¼eúì╣ + C«eúÄ╣
- ü èèFor ê differential equation
- y»»»» + 6y»»» + 11y»» + 6y» = 0
- ê CHARACTERISTIC EQUATION is
- mÅ + 6mÄ + 11mì + 6m = 0
- This facërs ïë
- m(m + 1)(m + 2)(m + 3) = 0
- This has ê solutions
- m = 0, -1, -2, -3
- Thus ê general solution is
- C¬ + C½eú╣ + C¼eúì╣ + C«eúÄ╣
-
- ÇèD
-
- 2 y»»»» - 15y»» + 10y» + 24y = 0
-
- A)è C¬e╣ + C½ì╣ + C¼eÄ╣ + C«eÅ╣
- B) C¬eú╣ + C½úì╣ + C¼eÄ╣ + C«eÅ╣
- C) C¬e╣ + C½ì╣ + C¼eúÄ╣ + C«eúÅ╣
- D) C¬eú╣ + C½ì╣ + C¼eÄ╣ + C«eúÅ╣
- ü èèFor ê differential equation
- y»»»» - 15y»» + 10y» + 24y = 0
- ê CHARACTERISTIC EQUATION is
- mÅ - 15mì + 10m + 24 = 0
- This facërs ïë
- (m + 1)(m - 2)(m - 3)(m + 4) = 0
- This has ê solutions
- m = -1, 2, 3, -4
- Thus ê general solution is
- C¬eú╣ + C½eì╣ + C¼eÄ╣ + C«eúÅ╣
-
- ÇèD
-
- è3 y»»»» - 13y»» + 36y = 0
-
- A)è C¬eì╣ + C½xeì╣ + C¼eÄ╣ + C«xeÄ╣
- B) C¬eúì╣ + C½eì╣ + C¼eúÄ╣ + C«eÄ╣
- C) C¬e╣ + C½xe╣ + C¼eÅ╣ + C«eö╣
- D) C¬e╣ + C½xe╣ + C¼eæ╣ + C«xeæ╣
- ü èèFor ê differential equation
- y»»»» - 13y»» + 36y = 0
- ê CHARACTERISTIC EQUATION is
- mÅ - 13mìè+ 36 = 0
- This facërs ïë
- (mì - 4)(mì - 9) = 0
- These each facër ïë
- (m - 2)(m + 2)(m - 3)(m + 3) = 0
- This has ê solutions
- m =è-2, 2, -3, 3
- Thus ê general solution is
- C¬eúì╣ + C½eì╣ + C¼eúÄ╣ + C«eÄ╣
-
- ÇèB
-
- 4 y»»»» + 8y»»» + 23y»» + 32y» + 12y = 0
-
- A)è C¬eú╣ + C½eúì╣ + C¼eúÄ╣ + C«eúÅ╣ è
- B) C¬eú╣ + C½xeú╣ + C¼eúì╣ + C«eúÄ╣
- C) C¬eú╣ + C½eúì╣ + C¼xeúì╣ + C«eúÄ╣
- D) C¬eú╣ + C½eúì╣ + C¼eúÄ╣ + C«xeúÄ╣
- ü èèFor ê differential equation
- y»»»» + 8y»»» + 23y»» + 32y» + 12y = 0
- ê CHARACTERISTIC EQUATION is
- mÅ + 8mÄ + 23mì + 32m + 12è= 0
- This facërs ïë
- (mì + 4m + 4)(mì + 4m + 3) = 0
- å furêr ë
- (m + 2)(m + 2)(m + 1)(m + 3)
- This has ê solutions
- m = -1, -2, -2, -3
- Thus ê general solution is
- C¬eú╣ + C½eúì╣ + C¼xeúì╣ + C«eúÄ╣
-
- ÇèC
-
- S 5 y»»»» - y»» = 0
-
- A) C¬ + C½x + C¼xì + C«e╣
- B) C¬ + C½x + C¼eú╣ + C«e╣
- C) C¬ + C½eú╣ + C¼xeú╣ + C«e╣
- D) C¬ + C½eú╣ + C¼e╣ + C«xe╣
- ü èèFor ê differential equation
- y»»»» - y»» = 0
- ê CHARACTERISTIC EQUATION is
- mÅ - mì = 0
- This facërs ïë
- mì(mì - 1) = 0
- This furêr facërs ë
- mì(m - 1)(m + 1) = 0
- This has ê solutions
- m =è0, 0,-1, 1
- Thus ê general solution is
- C¬ + C½x + C¼eú╣ + C«e╣
-
- ÇèB
-
- 6 y»»»» - 6y»»» + 12y»» - 8y» = 0
-
- A) C¬ + C½x + C¼xì + C«eì╣
- B) C¬ + C½x + C¼eì╣ + C«xeì╣
- C) C¬ + C½eì╣ + C¼xeì╣ + C«xìeì╣
- D) C¬eì╣ + C½xeì╣ + C¼xìeì╣ + C«xÄeì╣
- ü èèFor ê differential equation
- y»»»» - 6y»»» + 12y»» - 8y» = 0
- ê CHARACTERISTIC EQUATION is
- mÅ - 6mÄ + 12mì - 8m = 0
- This facërs ïë
- m(m - 2)(m - 2)(m - 2) = 0
- or
- m(m - 2)Ä = 0
- This has ê solutions
- m =è0, 2, 2, 2
- Thus ê general solution is
- C¬ + C½eì╣ + C¼xeì╣ + C«xìeì╣
-
- ÇèC
-
- 7 y»»»» + 4y»»» + 6y»» + 4y» + y = 0
-
- A) C¬ + C½x + C¼xì + C«eú╣
- B) C¬ + C½x + C¼eú╣ + C«xeú╣
- C) C¬ + C½eú╣ + C¼xeú╣ + C«xìeú╣
- D) C¬eú╣ + C½xeú╣ + C¼xìeú╣ + C«xÄeú╣
- ü èèFor ê differential equation
- y»»»» + 4y»»» + 6y»» + 4y» + y = 0
- ê CHARACTERISTIC EQUATION is
- mÅ + 4mÄ + 6mì + 4m + 1 = 0
- This facërs ïë
- (mì + 2m + 1)(mì + 2m + 1) = 0
- or
- (m + 1)Å = 0
- This has ê solutions
- m =è-1, -1, -1, -1
- Thus ê general solution is
- C¬eú╣ + C½xeú╣ + C¼xìeú╣ + C«xÄeú╣
-
- ÇèD
-
- äè Solve ê ïitial value problem
-
- â è For ê Initial Value Problem,
- y»»»» - 4y»» + 5y = 0
- y(0) = 0, y»(0) = -11, y»»(0) = -3èy»»»(0) = -53
- The general solution isè C¬eú╣ + C½e╣ + C¼eúì╣ + C«eì╣
- Differentiatïg å substitutïg 0 for x produces a system ç
- four equations ï ê four constants.èSolvïg this system
- gives ê solutionèè y = -eú╣ + 2e╣ + 3eì╣ - 4eúì╣
-
- éSèèAs ê GENERAL SOLUTON ç a FOURTH ORDER differential
- equation has FOUR ARBITRARY CONSTANTS, for an Initial Value
- Problem ë completely specify which member ç this four
- parameter family ç curves requires four INITAL VALUES.
- è The ståard ïitial values problem for a fourth order,
- lïear, constant coefficient differential equation is
- Ay»»»» + By»»» + Cy»» + Dy» + Ey = g(x)
- èèèy(x╠) =èèy╠
- èè y»(x╠) =è y»╠
- èèy»»(x╠) =èy»»╠
- è y»»»(x╙) = y»»»╙
- èèAs with ê second order, ïital value problem, solvïg
- this problem is a 2 step process
-
- 1)èèSolve ê differential equation ë produce a general
- solution with four arbitrary constants.
-
- 2)èèCalculate ê first, second å third derivatives ç
- ê general solution.èThen substitue ê ïital value ç ê
- ïdependent variable, x╠ , ïë ê general solution å its first two
- derivatives.èThis will produce a system ç 4 equations ï
- ê four arbitrary constants.èSolvïg this system gives ê
- values ç ê four constants which gives ê specific
- solution ç ê ïitial value problem.
-
- 8è y»»»» - 4y»» = 0
- y(0) = 10èy»(0) = 9èy»»(0) = 4èy»»»(0) = 24
-
- A) 6 + 3x + eúì╣ + 2eì╣
- B) 6 + 3x + eúì╣ - 2eì╣
- C) 6 + 3x - eúì╣ + 2eì╣
- D) 6 - 3x + eúì╣ + 2eì╣
- ü èèFor ê differential equation
- y»»»» - 4y»» = 0
- ê CHARACTERISTIC EQUATION is
- mÅ - 4mì = 0
- This facërs ïë
- mì(m - 2)(m + 2) = 0
- This has ê solutions
- m =è0, 0, -2, 2
- Thus ê general solution is
- èy = C¬ + C½x + C¼eúì╣ + C«eì╣
- Differentiatïg
- y» = C½ - 2C¼eúì╣ + 2C«eì╣
- y»» =èèè4C¼eúì╣ + 4C«eì╣
- èèè y»»» =èè -8C¼eúì╣ + 8C«eì╣
- Substitutïg ê ïital value ç ê dependent variable 0
- èy(0) =è10 = C¬ +èC½ +èC¼ +èC«
- y»(0) =è 9 =èèè C½ - 2C¼ + 2C«
- y»»(0) =è 4 =èèèèèè4C¼ + 4C«
- èèè y»»»(0) =è24 =èèèèè- 8C¼ + 8C«
- Sovlïg this system ç equations yields
- C¬ = 6è C½ = 3èC¼ = -1èC« = 2
- Thus ê solution ç ê ïitial value problem is
- y = 6 + 3x - eúì╣ + 2eì╣
- ÇèC
-
- 9 y»»»» - 6y»»» + 11y»» - 6y» = 0
- y(0) = 2èy»(0) = -2èy»»(0) = -4èy»»»(0) = -14
- A) 4 + 3e╣ + 2eì╣ + eÄ╣
- B) 4 + 3e╣ + 2eì╣ - eÄ╣
- C) 4 + 3e╣ - 2eì╣ - eÄ╣
- D) 4 - 3e╣ + 2eì╣ - eÄ╣
- ü èèFor ê differential equation
- y»»»» - 6y»»» + 11y»» - 6y» = 0
- ê CHARACTERISTIC EQUATION is
- mÅ - 6mÄ + 11mì - 6m = 0
- This facërs ïë
- m(m - 1)(m - 2)(m - 3) = 0
- This has ê solutions
- m =è0, 1, 2, 3
- Thus ê general solution is
- èy = C¬ + C½e╣ + C¼eì╣ + C«eÄ╣
- Differentiatïg
- y» = C½e╣ + 2C¼eì╣ +è3C«eÄ╣
- y»» = C½e╣ + 4C¼eì╣ +è9C«eÄ╣
- èèè y»»» = C½e╣ + 8C¼eì╣ + 27C«eÄ╣
- Substitutïg ê ïital value ç ê dependent variable 0
- èy(0) =è 2 = C¬ +èC½ +èC¼ +èC«
- y»(0) =è-2 =èèè C½ + 2C¼ + 3C«
- y»»(0) =è-4 =èèè C½ + 4C¼ + 9C«
- èèè y»»»(0) = -14 =èèè C½ + 8C¼ + 27C«
- Sovlïg this system ç equations yields
- C¬ = 4è C½ = -3èC¼ = 2èC« = -1
- Thus ê solution ç ê ïitial value problem is
- y =è4 - 3e╣ + 2eì╣ - eÄ╣
- ÇèD
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