Math 211 Math 211 Lecture #34 Inhomgeneous Equations Forced - PowerPoint PPT Presentation

1 Math 211 Math 211 Lecture #34 Inhomgeneous Equations Forced Harmonic Motion November 15, 2002

2 Inhomogeneous Equations Inhomogeneous Equations 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

3 Method of Undetermined Coefficients Method of Undetermined Coefficients y ′′ + py ′ + qy = f ( t ) The mantra for finding a particular solution is as follows: • 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

4 Exponential Forcing Term Exponential Forcing Term y ′′ + py ′ + qy = Ce bt • 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

5 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

6 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 . • Then x p ( t ) = Re( z ( t )) and y p ( t ) = Im( z ( t )) . Return Trigonometric forcing

7 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

8 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. � Particular solution: y ( t ) = − 5 − 2 t. • General solution y ( t ) = − 5 − 2 t + C 1 e t + C 2 e 2 t . Return

9 Exceptional Cases Exceptional Cases • Example: y ′′ − 3 y ′ + 2 y = 3 e t . � Try y ( t ) = ae t � The method does not work because e t is a solution to the associated homogeneous equation. • Try y ( t ) = ate t � Particular solution: y p ( t ) = − 3 te t . • General solution: y ( t ) = − 3 te t + C 1 e t + C 2 e 2 t . • If the suggested particular solution does not work, multiply it by t and try again. Return

10 Combination Forcing Term Combination Forcing Term Example y ′′ + 5 y ′ + 6 y = 2 e 2 t − 5 cos t • Solve 1 + 6 y 1 = 2 e 2 t y ′′ 1 + 5 y ′ y ′′ 2 + 5 y ′ 2 + 6 y 2 = − 5 cos t • Set y ( t ) = y 1 ( t ) + y 2 ( t ) . Theorem Previous UDC

11 Forced Harmonic Motion Forced Harmonic Motion Assume an oscillatory forcing term: y ′′ + 2 cy ′ + ω 2 0 y = A cos ωt • A is the forcing amplitude • ω is the forcing frequency • ω 0 is the natural frequency. • c is the damping constant. Return

12 Forced Undamped Motion Forced Undamped Motion y ′′ + ω 2 0 y = A cos ωt • Homogeneous equation y ′′ + ω 2 0 y = 0 � General solution y ( t ) = C 1 cos ω 0 t + C 2 sin ω 0 t. � If ω = ω 0 we have an exceptional case. Return

13 • ω � = ω 0 y ′′ + ω 2 0 y = A cos ωt � Look for a particular solution of the form x p ( t ) = a cos ωt + b sin ωt. � We find A x p ( t ) = 0 − ω 2 cos ωt. ω 2 Return

14 • ω � = ω 0 � General solution A x ( t ) = C 1 cos ω 0 t + C 2 sin ω 0 t + 0 − ω 2 cos ωt. ω 2 � Initial conditions x (0) = x ′ (0) = 0 ⇒ A x ( t ) = 0 − ω 2 [cos ωt − cos ω 0 t ] . ω 2 0 − ω 2 = 17 . � Example: ω 0 = 9 , ω = 8 , A = ω 2 x ( t ) = cos 9 t − cos 8 t. Return Homogeneous Inhomogeneous

15 • ω � = ω 0 ω = ω 0 + ω δ = ω 0 − ω � Set . and 2 2 ⇒ ω = ω − δ and ω 0 = ω + δ, and A x ( t ) = 0 − ω 2 [cos ωt − cos ω 0 t ] ω 2 = A sin δt sin ωt. 2 ωδ Return ICsol

16 • ω � = ω 0 � Example: ω = 8 . 5 δ = 0 . 5 . and � Envelope: Slow oscillation with frequency δ , � � A sin δt � � � . ± � � 2 ωδ � � Fast oscillation with frequency ω and varying amplitude. � Beats. Return

17 • ω = ω 0 y ′′ + ω 2 0 y = A cos ω 0 t. � We have an exceptional case. Try x p ( t ) = t [ a cos ωt + b sin ωt ] . � We find A x p ( t ) = t sin ω 0 t. 2 ω 0 � General solution x ( t ) = C 1 cos ω 0 t + C 2 sin ω 0 t + A t sin ω 0 t. 2 ω 0 Return

18 • ω = ω 0 � Initial conditions x (0) = x ′ (0) = 0 ⇒ A x ( t ) = t sin ω 0 t. 2 ω 0 ◮ Example: ω 0 = 5 , and A = 2 ω 0 = 10 . x ( t ) = t sin 5 t. � Oscillation with increasing amplitude. � First example of resonance. ◮ Forcing at the natural frequency can cause oscillations that grow out of control. Return