Math 211 Math 211 Lecture #22 November 14, 2000 2 Second Order - PDF document

1 Math 211 Math 211 Lecture #22 November 14, 2000 2 Second Order Equations Second Order Equations y ′′ + py ′ + qy = 0 • Equivalent system: x ′ = A x , where � y � 0 � 1 � x = A = . and y ′ − q − p Definition: Two functions u ( t ) and v ( t ) are linearly independent if neither is a constant multiple of the other. Return 3 General Solution General Solution Theorem: Suppose that y 1 ( t ) & y 2 ( t ) are linearly independent solutions to the equation y ′′ + py ′ + qy = 0 . Then the general solution is y ( t ) = C 1 y 1 ( t ) + C 2 y 2 ( t ) . Definition: A set of two linearly independent solutions is called a fundamental set of solutions. Return 1 John C. Polking

4 Solutions to y ′′ + py ′ + qy = 0 . Solutions to y ′′ + py ′ + qy = 0 . • Look for exponential solutions y ( t ) = e λt . • Characteristic equation: λ 2 + pλ + q = 0 . • Characteristic polynomial: λ 2 + pλ + q . • Compare y ′′ + py ′ + qy = 0 ODE λ 2 + pλ + q = 0 ch. poly. System Return 5 Real Roots Real Roots • If λ is a root to the characteristic polynomial then y ( t ) = e λt is a solution. • If λ is a root to the characteristic polynomial of multiplicity 2, then y 1 ( t ) = e λt and y 2 ( t ) = te λt are linearly independent solutions. Solutions General solution Return 6 Complex Roots Complex Roots • If λ = α + iβ is a complex root of the characteristic equation, then so is λ = α − iβ . • A complex valued fundamental set of solutions is z ( t ) = e λt and z ( t ) = e λt . • A real valued fundamental set of solutions is x ( t ) = e αt cos βt y ( t ) = e αt sin βt. and Solutions General solution Return 2 John C. Polking

7 Examples Examples • y ′′ − 5 y ′ + 6 y = 0 . • y ′′ + 25 y = 0 . • y ′′ + 4 y ′ + 13 y = 0 . General solution Real roots Complex roots 8 The Vibrating Spring The Vibrating Spring Newton’s second law: ma = total force. • Forces acting: ⋄ Gravity mg . ⋄ Restoring force R ( x ) . ⋄ Damping force D ( v ) . ⋄ External force F ( t ) . Return 9 The Vibrating Spring (2) The Vibrating Spring (2) ma = mg + R ( x ) + D ( v ) + F ( t ) • Hooke’s law: R ( x ) = − kx. k > 0 is the spring constant. • Spring-mass equilibrium x 0 = mg/k. • Set y = x − x 0 . Equation becomes my ′′ = − ky + D ( y ′ ) + F ( t ) . VS1 Return 3 John C. Polking

10 The Vibrating Spring (3) The Vibrating Spring (3) • Damping force D ( y ′ ) = − µy ′ . • Equation becomes my ′′ = − ky − µy ′ + F ( t ) , or my ′′ + µy ′ + ky = F ( t ) , or y ′′ + µ my ′ + k my = 1 mF ( t ) . VS1 VS2 Return 11 RLC Circuit RLC Circuit L I + E C − I R LI ′′ + RI ′ + 1 C I = E ′ ( t ) , or I ′′ + R LC I = 1 1 L I ′ + LE ′ ( t ) . VS3 Return 12 Harmonic Motion (1) Harmonic Motion (1) • Spring: y ′′ + µ m y ′ + k m y = 1 m F ( t ) . • Circuit: I ′′ + R L I ′ + LC I = 1 1 L E ′ ( t ) . • Essentially the same equation. Use x ′′ + 2 cx ′ + ω 2 0 x = f ( t ) . • The equation for harmonic motion. VS3 RLC Return 4 John C. Polking

13 Harmonic Motion (2) Harmonic Motion (2) x ′′ + 2 cx ′ + ω 2 0 x = f ( t ) . • ω 0 is the natural frequency. � ⋄ Spring: ω 0 = k/m. � ⋄ Circuit: ω 0 = 1 /LC. • c is the damping constant. • f ( t ) is the forcing term. HM1 Return 14 Simple Harmonic Motion Simple Harmonic Motion No forcing, and no damping. x ′′ + ω 2 0 x = 0 • p ( λ ) = λ 2 + ω 2 0 , λ = ± iω 0 . Fundamental set of solutions x 1 ( t ) = cos ω 0 t & x 2 ( t ) = sin ω 0 t. Complex roots HM2 Return 15 Simple Harmonic Motion (2) Simple Harmonic Motion (2) General solution x ( t ) = C 1 cos ω 0 t + C 2 sin ω 0 t. • Every solution is periodic with period ω 0 . ⋄ ω 0 is the natural frequency. ⋄ The period is T = 2 π/ω 0 . HM2 SHM1 Return 5 John C. Polking