Math 211 Math 211 Lecture #33 Harmonic Motion Inhomogeneous - PowerPoint PPT Presentation

9 Underdamped Case Underdamped Case • c < ω 0 • Two complex roots λ and λ , where λ = − c + iω and 0 − c 2 . � ω 2 ω = • General solution x ( t ) = e − ct [ C 1 cos ωt + C 2 sin ωt ]

9 Underdamped Case Underdamped Case • c < ω 0 • Two complex roots λ and λ , where λ = − c + iω and 0 − c 2 . � ω 2 ω = • General solution x ( t ) = e − ct [ C 1 cos ωt + C 2 sin ωt ] . = Ae − ct cos( ωt − φ )

10 Overdamped Case Overdamped Case Return

10 Overdamped Case Overdamped Case • c > ω 0 , Return

10 Overdamped Case Overdamped Case • c > ω 0 , so two real roots Return

10 Overdamped Case Overdamped Case • c > ω 0 , so two real roots � c 2 − ω 2 λ 1 = − c − 0 � c 2 − ω 2 λ 2 = − c + 0 . Return

10 Overdamped Case Overdamped Case • c > ω 0 , so two real roots � c 2 − ω 2 λ 1 = − c − 0 � c 2 − ω 2 λ 2 = − c + 0 . • λ 1 < λ 2 < 0 . Return

10 Overdamped Case Overdamped Case • c > ω 0 , so two real roots � c 2 − ω 2 λ 1 = − c − 0 � c 2 − ω 2 λ 2 = − c + 0 . • λ 1 < λ 2 < 0 . • General solution x ( t ) = C 1 e λ 1 t + C 2 e λ 2 t . Return

11 Critically Damped Case Critically Damped Case Return

11 Critically Damped Case Critically Damped Case • c = ω 0 Return

11 Critically Damped Case Critically Damped Case • c = ω 0 • One negative real root λ = − c with multiplicity 2. Return

11 Critically Damped Case Critically Damped Case • c = ω 0 • One negative real root λ = − c with multiplicity 2. • General solution x ( t ) = e − ct [ C 1 + C 2 t ] . Return

12 Inhomogeneous Equations Inhomogeneous Equations y ′′ + py ′ + qy = f ( t ) Return

12 Inhomogeneous Equations Inhomogeneous Equations y ′′ + py ′ + qy = f ( t ) • The corresponding homogeneous equation is y ′′ + py ′ + qy = 0 Return

12 Inhomogeneous Equations Inhomogeneous Equations y ′′ + py ′ + qy = f ( t ) • The corresponding homogeneous equation is y ′′ + py ′ + qy = 0 � We know how to find a fundamental set of solutions y 1 and y 2 . Return

12 Inhomogeneous Equations Inhomogeneous Equations y ′′ + py ′ + qy = f ( t ) • The corresponding homogeneous equation is y ′′ + py ′ + qy = 0 � We know how to find a fundamental set of solutions y 1 and y 2 . � The general solution of the homogeneous equation is y h ( t ) = C 1 y 1 ( t ) + C 2 y 2 ( t ) . Return

13 Theorem: Assume Return Homogeneous equation

13 Theorem: Assume • y p ( t ) is a particular solution to the inhomogeneous equation y ′′ + py ′ + qy = f ( t ); Return Homogeneous equation

13 Theorem: Assume • y p ( t ) is a particular solution to the inhomogeneous equation y ′′ + py ′ + qy = f ( t ); • y 1 ( t ) & y 2 ( t ) is a fundamental set of solutions to the homogeneous equation y ′′ + py ′ + qy = 0 . Return Homogeneous equation

13 Theorem: Assume • y p ( t ) is a particular solution to the inhomogeneous equation y ′′ + py ′ + qy = f ( t ); • y 1 ( t ) & y 2 ( t ) is a fundamental set of solutions to the homogeneous equation y ′′ + py ′ + qy = 0 . Then the general solution to the inhomogeneous equation is y ( t ) = y p ( t ) + C 1 y 1 ( t ) + C 2 y 2 ( t ) . Return Homogeneous equation

14 Method of Undetermined Coefficients Method of Undetermined Coefficients y ′′ + py ′ + qy = f ( t ) Return

14 Method of Undetermined Coefficients Method of Undetermined Coefficients y ′′ + py ′ + qy = f ( t ) • If the forcing term f ( t ) has a form which is replicated under differentiation, then look for a particular solution of the same general form as the forcing term. Return

15 Exponential Forcing Term Exponential Forcing Term y ′′ + py ′ + qy = Ce at Return

15 Exponential Forcing Term Exponential Forcing Term y ′′ + py ′ + qy = Ce at • Example: y ′′ + 3 y ′ + 2 y = 4 e − 3 t Return

15 Exponential Forcing Term Exponential Forcing Term y ′′ + py ′ + qy = Ce at • Example: y ′′ + 3 y ′ + 2 y = 4 e − 3 t • Try y p ( t ) = ae − 3 t ; a to be determined. Return

15 Exponential Forcing Term Exponential Forcing Term y ′′ + py ′ + qy = Ce at • Example: y ′′ + 3 y ′ + 2 y = 4 e − 3 t • Try y p ( t ) = ae − 3 t ; a to be determined. � Particular solution: y p ( t ) = 2 e − 3 t . Return

15 Exponential Forcing Term Exponential Forcing Term y ′′ + py ′ + qy = Ce at • Example: y ′′ + 3 y ′ + 2 y = 4 e − 3 t • Try y p ( t ) = ae − 3 t ; a to be determined. � Particular solution: y p ( t ) = 2 e − 3 t . • Homogeneous equation: y ′′ + 3 y ′ + 2 y = 0 . Return

15 Exponential Forcing Term Exponential Forcing Term y ′′ + py ′ + qy = Ce at • Example: y ′′ + 3 y ′ + 2 y = 4 e − 3 t • Try y p ( t ) = ae − 3 t ; a to be determined. � Particular solution: y p ( t ) = 2 e − 3 t . • Homogeneous equation: y ′′ + 3 y ′ + 2 y = 0 . � Fundamental set of solutions: e − 2 t & e − t . Return

15 Exponential Forcing Term Exponential Forcing Term y ′′ + py ′ + qy = Ce at • Example: y ′′ + 3 y ′ + 2 y = 4 e − 3 t • Try y p ( t ) = ae − 3 t ; a to be determined. � Particular solution: y p ( t ) = 2 e − 3 t . • Homogeneous equation: y ′′ + 3 y ′ + 2 y = 0 . � Fundamental set of solutions: e − 2 t & e − t . • General solution to the inhomogeneous equation: y ( t ) = 2 e − 3 t + C 1 e − t + C 2 e − 2 t . Return

16 Trigonometric Forcing Term Trigonometric Forcing Term y ′′ + py ′ + qy = A cos ωt + B sin ωt Return

16 Trigonometric Forcing Term Trigonometric Forcing Term y ′′ + py ′ + qy = A cos ωt + B sin ωt • Example: y ′′ + 4 y ′ + 5 y = 4 cos 2 t − 3 sin 2 t Return

16 Trigonometric Forcing Term Trigonometric Forcing Term y ′′ + py ′ + qy = A cos ωt + B sin ωt • Example: y ′′ + 4 y ′ + 5 y = 4 cos 2 t − 3 sin 2 t • Try y p ( t ) = a cos 2 t + b sin 2 t Return

16 Trigonometric Forcing Term Trigonometric Forcing Term y ′′ + py ′ + qy = A cos ωt + B sin ωt • Example: y ′′ + 4 y ′ + 5 y = 4 cos 2 t − 3 sin 2 t • Try y p ( t ) = a cos 2 t + b sin 2 t � Particular solution: y p ( t ) = [28 cos 2 t + 29 sin 2 t ] / 65 . Return

16 Trigonometric Forcing Term Trigonometric Forcing Term y ′′ + py ′ + qy = A cos ωt + B sin ωt • Example: y ′′ + 4 y ′ + 5 y = 4 cos 2 t − 3 sin 2 t • Try y p ( t ) = a cos 2 t + b sin 2 t � Particular solution: y p ( t ) = [28 cos 2 t + 29 sin 2 t ] / 65 . • Homogeneous equation: y ′′ + 4 y ′ + 5 y = 0 Return

16 Trigonometric Forcing Term Trigonometric Forcing Term y ′′ + py ′ + qy = A cos ωt + B sin ωt • Example: y ′′ + 4 y ′ + 5 y = 4 cos 2 t − 3 sin 2 t • Try y p ( t ) = a cos 2 t + b sin 2 t � Particular solution: y p ( t ) = [28 cos 2 t + 29 sin 2 t ] / 65 . • Homogeneous equation: y ′′ + 4 y ′ + 5 y = 0 � Fund. set of sol’ns: e − 2 t cos t & e − 2 t sin t . Return

16 Trigonometric Forcing Term Trigonometric Forcing Term y ′′ + py ′ + qy = A cos ωt + B sin ωt • Example: y ′′ + 4 y ′ + 5 y = 4 cos 2 t − 3 sin 2 t • Try y p ( t ) = a cos 2 t + b sin 2 t � Particular solution: y p ( t ) = [28 cos 2 t + 29 sin 2 t ] / 65 . • Homogeneous equation: y ′′ + 4 y ′ + 5 y = 0 � Fund. set of sol’ns: e − 2 t cos t & e − 2 t sin t . • General solution to the inhomogeneous equation: Return

16 Trigonometric Forcing Term Trigonometric Forcing Term y ′′ + py ′ + qy = A cos ωt + B sin ωt • Example: y ′′ + 4 y ′ + 5 y = 4 cos 2 t − 3 sin 2 t • Try y p ( t ) = a cos 2 t + b sin 2 t � Particular solution: y p ( t ) = [28 cos 2 t + 29 sin 2 t ] / 65 . • Homogeneous equation: y ′′ + 4 y ′ + 5 y = 0 � Fund. set of sol’ns: e − 2 t cos t & e − 2 t sin t . • General solution to the inhomogeneous equation: y ( t ) = 28 cos 2 t + 29 sin 2 t + e − 2 t [ C 1 cos t + C 2 sin t ] . 65 Return

17 Complex Method Complex Method x ′′ + px ′ + qx = A cos ωt or y ′′ + py ′ + qy = A sin ωt. Return

17 Complex Method Complex Method x ′′ + px ′ + qx = A cos ωt or y ′′ + py ′ + qy = A sin ωt. • Solve z ′′ + pz ′ + qz = Ae iωt . Return

17 Complex Method Complex Method x ′′ + px ′ + qx = A cos ωt or y ′′ + py ′ + qy = A sin ωt. • Solve z ′′ + pz ′ + qz = Ae iωt . � Try z ( t ) = ae iωt . Return

17 Complex Method Complex Method x ′′ + px ′ + qx = A cos ωt or y ′′ + py ′ + qy = A sin ωt. • Solve z ′′ + pz ′ + qz = Ae iωt . � Try z ( t ) = ae iωt . • x p ( t ) = Re( z ( t )) and y p ( t ) = Im( z ( t )) . Return

18 Example Example x ′′ + 4 x ′ + 5 x = 4 cos 2 t Return

18 Example Example x ′′ + 4 x ′ + 5 x = 4 cos 2 t • Solve z ′′ + 4 z ′ + 5 z = 4 e 2 it . Return

18 Example Example x ′′ + 4 x ′ + 5 x = 4 cos 2 t • Solve z ′′ + 4 z ′ + 5 z = 4 e 2 it . � Try z ( t ) = ae 2 it . Return

18 Example Example x ′′ + 4 x ′ + 5 x = 4 cos 2 t • Solve z ′′ + 4 z ′ + 5 z = 4 e 2 it . � Try z ( t ) = ae 2 it . � Particular solution: z ( t ) = (4 − 32 i ) e 2 it / 65 . Return

18 Example Example x ′′ + 4 x ′ + 5 x = 4 cos 2 t • Solve z ′′ + 4 z ′ + 5 z = 4 e 2 it . � Try z ( t ) = ae 2 it . � Particular solution: z ( t ) = (4 − 32 i ) e 2 it / 65 . • Particular solution to the real equation: x p ( t ) = Re( z ( t )) Return

18 Example Example x ′′ + 4 x ′ + 5 x = 4 cos 2 t • Solve z ′′ + 4 z ′ + 5 z = 4 e 2 it . � Try z ( t ) = ae 2 it . � Particular solution: z ( t ) = (4 − 32 i ) e 2 it / 65 . • Particular solution to the real equation: x p ( t ) = Re( z ( t )) = [4 cos 2 t + 32 sin 2 t ] / 65 . Return

19 Polynomial Forcing Term Polynomial Forcing Term y ′′ + py ′ + qy = P ( t ) Return

19 Polynomial Forcing Term Polynomial Forcing Term y ′′ + py ′ + qy = P ( t ) • Example: y ′′ − 3 y ′ + 2 y = 1 − 4 t. Return

19 Polynomial Forcing Term Polynomial Forcing Term y ′′ + py ′ + qy = P ( t ) • Example: y ′′ − 3 y ′ + 2 y = 1 − 4 t. � Try y ( t ) = a + bt. Return