ezLecture: Differential Equations that Describe Oscillations NOTES Learning Goals: 1. To use an auxiliary equation to solve a homogeneous, second order linear differential equation. 2. To use successive integration to find a particular solution for an inhomogeneous, second order, linear differential equation. To help us we will be following the discussion by Boaz in Mathematical Methods in the Physical Sciences, Chapter 8, Sections 5 and 6. 1. Use an auxiliary equation to solve a homogeneous, second order linear differential equation. A) Start with an arbitrary, second order, homogeneous, linear differential equation. Let’s say that the independent variable is π‘, the dependent variable is π₯, and π2 , π1 , π0 are coefficients that can be either constants or functions of π‘. Differential operator, π· is defined to be the derivative, = π ππ‘ . 2 (π2 π· + π1 π· + π0 )π₯ = 0 B) Auxiliary equation is: π2 π· 2 + π1 π· + π0 = 0 C) The roots (r1 and r2) for this auxiliary equation are given by the solution to the quadratic equation. π1 = π2 = −π1 +√π12 −4π2 π0 2π2 −π1 −√π12 −4π2 π0 2π2 D) The solution is then found by using the roots. If π1 ≠ π2 , then π₯ = πΆ1 π π1 π‘ + πΆ2 π π2 π‘ If π1 = π2 = π, then π₯ = πΆ1 π ππ‘ + πΆ2 π‘π ππ‘ 2. Use successive integration to find a particular solution for an inhomogeneous, second order, linear differential equation. A) Start by defining an arbitrary, second order, inhomogeneous, linear differential equation. (π2 π· 2 + π1 π· + π0 )π₯ = π(π‘) B) The general solution will be the sum of the complementary and particular solutions. π₯ = π₯ππππππππππ‘πππ¦ + π₯ππππ‘πππ’πππ C) Complementary solution is found by following the procedure in part 1. Particular solution is found by successive integration. π’ππππ‘πππ’πππ = π −πΌ ∫ π(π‘) π πΌ ππ‘ where π πΌ = π ∫ −π1 ππ‘ π₯ππππ‘πππ’πππ = π −πΌ ∫ π’ππππ‘πππ’πππ π πΌ ππ‘ where π πΌ = π ∫ −π2 ππ‘