Differential Equations Part 2 NOTES

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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 𝑑, the dependent variable is π‘₯, and π‘Ž2 , π‘Ž1 , π‘Ž0 are coefficients that can be either
constants or functions of 𝑑. Differential operator, 𝐷 is defined to be the derivative, =
𝑑
𝑑𝑑
.
2
(π‘Ž2 𝐷 + π‘Ž1 𝐷 + π‘Ž0 )π‘₯ = 0
B) Auxiliary equation is:
π‘Ž2 𝐷 2 + π‘Ž1 𝐷 + π‘Ž0 = 0
C) The roots (r1 and r2) for this auxiliary equation are given by the solution to the quadratic equation.
π‘Ÿ1 =
π‘Ÿ2 =
−π‘Ž1 +√π‘Ž12 −4π‘Ž2 π‘Ž0
2π‘Ž2
−π‘Ž1 −√π‘Ž12 −4π‘Ž2 π‘Ž0
2π‘Ž2
D) The solution is then found by using the roots.
If π‘Ÿ1 ≠ π‘Ÿ2 ,
then π‘₯ = 𝐢1 𝑒 π‘Ÿ1 𝑑 + 𝐢2 𝑒 π‘Ÿ2 𝑑
If π‘Ÿ1 = π‘Ÿ2 = π‘Ÿ, then π‘₯ = 𝐢1 𝑒 π‘Ÿπ‘‘ + 𝐢2 𝑑𝑒 π‘Ÿπ‘‘
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.
(π‘Ž2 𝐷 2 + π‘Ž1 𝐷 + π‘Ž0 )π‘₯ = 𝑓(𝑑)
B) The general solution will be the sum of the complementary and particular solutions.
π‘₯ = π‘₯π‘π‘œπ‘šπ‘π‘™π‘’π‘šπ‘’π‘›π‘‘π‘Žπ‘Ÿπ‘¦ + π‘₯π‘π‘Žπ‘Ÿπ‘‘π‘–π‘π‘’π‘™π‘Žπ‘Ÿ
C) Complementary solution is found by following the procedure in part 1. Particular solution is found by
successive integration.
π‘’π‘π‘Žπ‘Ÿπ‘‘π‘–π‘π‘’π‘™π‘Žπ‘Ÿ = 𝑒 −𝐼 ∫ 𝑓(𝑑) 𝑒 𝐼 𝑑𝑑 where 𝑒 𝐼 = 𝑒 ∫ −π‘Ÿ1 𝑑𝑑
π‘₯π‘π‘Žπ‘Ÿπ‘‘π‘–π‘π‘’π‘™π‘Žπ‘Ÿ = 𝑒 −𝐼 ∫ π‘’π‘π‘Žπ‘Ÿπ‘‘π‘–π‘π‘’π‘™π‘Žπ‘Ÿ 𝑒 𝐼 𝑑𝑑 where 𝑒 𝐼 = 𝑒 ∫ −π‘Ÿ2 𝑑𝑑
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