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- à 3.2 Distïct, Real Roots ç ê Characteristic Equation
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- äèèFïd ê general solution ç ê homogeneious,
- èèèèèèèèdifferential equation.
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- â è The differential equation
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- y»» - 6y» + 5y = 0
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- è has ê general solution
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- C¬e╣ + C½eÉ╣
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- éS The lïear, second order, constant coefficient, homogenous
- differential equation
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- ay»» + by» + cy = 0
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- has solutions ç ê formèe¡╣èwhere m is a solution ç ê
- CHARACTERISTIC EQUATION
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- amì + bm + c = 0
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- When this quadratic equation has two distïct, real roots, ê
- GENERAL SOLUTION is ç ê form
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- y = C¬e¡╣ + C½eⁿ╣
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- where m å n are ê distïct real solutions
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- 1 y»»è+è4y»è+è3yè=è0
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- A) C¬eÅ╣ + C½eÄ╣ B) C¬e╣ + C½eÄ╣
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- C) C¬eú╣ + C½eúÄ╣ D) C¬eúÅ╣ + C½eúÄ╣
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- ü Forè
- y»» + 4y» + 3y = 0,
- ê characteristic equation is
- mì + 4m + 3 = 0
- This facërs ïë
- (m + 1)(m + 3) = 0
- The solutions are
- m = -1, -3
- The general solution is
- C¬eú╣ + C½eúÄ╣
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- Ç C
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- 2 y»» - 2y» - 8y = 0
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- A) C¬eì╣ + C½eúÅ╣ B) C¬eúì╣ + C½eúÅ╣
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- C) C¬eì╣ + C½eÅ╣ D) C¬eúì╣ + C½eÅ╣
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- ü Forè
- y»» - 2y» - 8y = 0,
- ê characteristic equation is
- mì - 2m - 8 = 0
- This facërs ïë
- (m + 2)(m - 4) = 0
- The solutions are
- m = -2, 4
- The general solution is
- C¬eúì╣ + C½eÅ╣
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- Ç D
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- 3 y»»è-è4yè=è0
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- A) C¬e╣ + C½eÅ╣ B) C¬eú╣ + C½eúÅ╣
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- C) C¬eúì╣ + C½eì╣ D) C¬ + C½eÅ╣
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- ü Forè
- y»»è-è4y = 0,
- ê characteristic equation is
- mì - 4 = 0
- This facërs ïë
- (m + 2)(m - 2) = 0
- The solutions are
- m = -2, 2
- The general solution is
- C¬eúì╣ + C½e║╣
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- Ç C
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- 4 y»»è-è4y»è=è0
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- A) C¬e╣ + C½eÅ╣ B) C¬ + C½eúÅ╣
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- C) C¬eúì╣ + C½eì╣ D) C¬ + C½eÅ╣
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- ü Forè
- y»»è-è4y» = 0,
- ê characteristic equation is
- mì - 4m = 0
- This facërs ïë
- m(m - 4) = 0
- The solutions are
- m = 0, 4
- The general solution is, as eò = 1,
- C¬ + C½eÅ╣èè
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- Ç D
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- 5 2y»» - y» - 6y = 0
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- A) C¬eÄ╣»ì + C½eì╣ B) C¬eúÄ╣»ì + C½eì╣
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- B) C¬eÄ╣»ì + C½eúì╣ D) C¬eúÄ╣»ì + C½eúì╣
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- ü Forè
- 2y»»è-èy»è-è6yè=è0,
- ê characteristic equation is
- 2mì - m - 6 = 0
- This facërs ïë
- (2m + 3)(m - 2) = 0
- The solutions are
- m = -3/2, 2
- The general solution is
- C¬eúÄ╣»ì + C½eì╣èè
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- Ç B
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- 6 y»» - 4y» + 2y = 0
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- A)è C¬eÑìóáìª╣ + C½eÑìúáìª╣ B)è C¬eÑúìóáìª╣ + C½eÑúìúáìª╣
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- C)è C¬eúÅ╣ + C½eì╣ D)è C¬eÅ╣ + C½eúì╣
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- ü Forè
- y»» - 4y» + 2y = 0,
- ê characteristic equation is
- mì - 4m + 2 = 0
- This does NOT facër å ê quadratic formula must be used
- 4 ± √[(-4)ì - 4(1)(2)]
- èèm = ────────────────────────
- è 2(1)
- 4 ± √[16 - 8]
- èèè= ───────────────
- 2
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- èèè= [4 ± √8] / 2
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- èèè= 2 ± √2èare ê solutions
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- The general solution is
- C¬eÑìóáìª╣ + C½eÑìúáìª╣
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- Ç A
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- äèèSolve ê followïg ïitial value problem.
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- â èFor ê ïitial value problem
- y»» + 5y» + 6y = 0è y(0) = 3 ;èy»(0) = -2
- The general solution is
- y = C¬eúÄ╣ + C½eúì╣
- Substitutïg x = 0 ïë ê solution å its derivative yields
- C¬ = -4 ; C½ = 7
- Thus ê solution ë ê ïitial value problem is
- y = -4eúÄ╣ + 7eúì╣
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- éS èTo solve an Initial Value Problem
- ay»» + by» + cy = 0è
- y(x╠) = y╠ ; y»(x╠) = y»╠
- has two stages.
- 1) Fïd a general solution ç ê differential equation.
- As this is a second order, differential equation,
- ê general solution will have TWO ARBITRARY CONSTANTS
- 2) Substitute ê INITIAL VALUE ç ê ïdependent
- variable ïë ê general solution å its deriviative
- å set êm equal ë ê TWO INITIAL CONDITIONS.èThis
- produces two lïear equations ï two unknowns (ê
- arbitrary constants).èSolvïg this system yields ê
- value ç ê constants å ê solution ç ê ïitial
- value problem.
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- 7 y»» - 4y» + 3y = 0èè
- y(0) = 2è;èy»(0) = 0
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- A) 2e╣ B) 2eÄ╣
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- C) 3eú╣ - eúÄ╣ D) 3e╣ - eÄ╣
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- üèè For ê ïitial value problem
- y»» - 4y» + 3y = 0è
- y(0) = 2 ;èy»(0) = 0
- The characteristic equation is
- mì - 4m + 3 = 0
- This facërs ë
- (m - 1)(m - 3) = 0
- The solutions are
- m = 1, 3
- The general solution is
- y = C¬e╣ + C½eÄ╣
- Substitutïg x = 0 ïë ê solution å its derivative yields
- y(0)è=èC¬ +èC½ = 2
- y»(0) =èC¬ + 3C½ = 0
- Solvïg this system yields
- C¬ = 3 ; C½ = -1
- Thus ê solution ë ê ïitial value problem is
- y = 3e╣ - eÄ╣
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- Ç D
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- 8 6y»» - 7y» - 3y = 0è
- y(0) = -1è;èy»(0) = 4
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- A)è -3eú╣»Ä + 2eÄ╣»ì B)è 2eú╣»Ä - 3eÄ╣»ì
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- C)è -3e╣»Ä = 2eúÄ╣»ì D)è 2e╣»Ä - 3eúÄ╣»ì
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- üèè For ê ïitial value problem
- 6y»» - 7y» - 3y = 0è
- y(0) = -1 ;èy»(0) = 4
- The characteristic equation is
- 6mì - 7m - 3 = 0
- This facërs ë
- (3m + 1)(2m - 3) = 0
- The solutions are
- m = -1/3, 3/2
- The general solution is
- y = C¬eú╣»Ä + C½eÄ╣»ì
- Substitutïg x = 0 ïë ê solution å its derivative yields
- y(0)è=è C¬è +èC½è = -1
- y»(0) =è-C¬/3 + 3C½/2 =è4
- Solvïg this system yields
- C¬ = -3 ; C½ = 2
- Thus ê solution ë ê ïitial value problem is
- y = -3eú╣»Ä + 2eÄ╣»ì
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- Ç A
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- 9 y»» - y = 0èè
- y(2) = 3 ;èy»(2) = -2
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- A)è5/2 eúìeú╣ + 1/2 e║e╣ B)è1/2 eúìeú╣ - 5/2 eìe╣
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- C)è5/2 eìeú╣ + 1/2 eú║e╣ D)è1/2 eìeú╣ - 5/2 eúìe╣
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- üèè For ê ïitial value problem
- y»» - y = 0è
- y(2) = 3 ;èy»(2) = -2
- The characteristic equation is
- mì - 1 = 0
- This facërs ë
- (m + 1)(m - 1) = 0
- The solutions are
- m = -1, 1
- The general solution is
- y = C¬eú╣ + C½e╣
- Substitutïg x = 2 ïë ê solution å its derivative
- yields messier equations than previously but ê technique
- will be ê same
- y(2)è=è C¬eúì + C½eìè=è3
- y»(2) =è-C¬eúì + C½eìè= -2
- Solvïg this system yields
- C¬ = 5/2 eìè;èC½ = 1/2 eúì
- Thus ê solution ë ê ïitial value problem is
- y = 5/2 eìeú╣ + 1/2 eúìe╣
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- Ç C
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