Differential equations MOC

Reduction of order (homogenous second-order differential equation)

Reduction of order is a technique for finding the general solution of a homogenous second order linear DE1 when a particular ๐‘ฆ1๐‘ฅ solution is known.

We begin by assuming that ๐‘ฆ(๐‘ฅ) =๐‘ข(๐‘ฅ)๐‘ฆ1(๐‘ฅ) for some function ๐‘ข(๐‘ฅ) to be determined. It follows from this that

๐‘ฆโ€ฒ=๐‘ขโ€ฒ๐‘ฆ1+๐‘ข๐‘ฆโ€ฒ1๐‘ฆโ€ณ=๐‘ขโ€ณ๐‘ฆ1+2๐‘ขโ€ฒ๐‘ฆโ€ฒ1+๐‘ข๐‘ฆโ€ณ1

Given that ๐‘ฆ1 is indeed a solution, this substitution will reduce the DE to a first order separable DE on the independent variable ๐‘ขโ€ฒ (see reasoning below), which can be then used to determine the general solution.

Explanation

We write the DE as ๐ฟ[๐‘ฆ] =0, where the linear operation ๐ฟ is defined by

๐ฟ=๐ท2+๐‘๐ท+๐‘ž๐Ÿ™

We are given that ๐ฟ[๐‘ฆ1] =0. Then

๐ฟ[๐‘ข๐‘ฆ1]=๐ท2[๐‘ข๐‘ฆ1]+๐‘๐ท[๐‘ข๐‘ฆ1]+๐‘ž๐Ÿ™[๐‘ข๐‘ฆ1]=๐‘ฆ1๐ท2[๐‘ข]+2๐ท[๐‘ข]๐ท[๐‘ฆ1]+๐‘ข๐ท2[๐‘ฆ1]+๐‘๐‘ฆ1๐ท[๐‘ข]+๐‘๐‘ข๐ท[๐‘ฆ1]+๐‘ž๐‘ข๐‘ฆ1=๐‘ข(๐ท2[๐‘ฆ1]+๐‘๐ท[๐‘ฆ1]+๐‘ž๐Ÿ™[๐‘ฆ1])+๐‘ฆ1๐ท2[๐‘ข]+(๐‘๐‘ฆ1+2๐ท[๐‘ฆ1])๐ท[๐‘ข]=๐‘ข๐ฟ[๐‘ฆ1]+๐พ[๐ท[๐‘ข]]

where ๐พ is the second order linear operator

๐พ=๐‘ฆ1๐ท+(๐‘๐‘ฆ1+2๐‘ฆโ€ฒ1)๐Ÿ™

Method

In general, I have found it most effective to only substitute the particular solution ๐‘ฆ1 after the gathering different ๐‘ข terms. Consider the ODE2

๐‘ฅ2๐‘ฆโ€ณ+๐‘ฅ๐‘ฆโ€ฒ+(๐‘ฅ2โˆ’14)๐‘ฆ=0

with particular solution ๐‘ฆ1 =๐‘ฅโˆ’1/2sinโก๐‘ฅ, so

๐‘ฆโ€ฒ1=โˆ’12๐‘ฅโˆ’3/2sinโก๐‘ฅ+๐‘ฅโˆ’1/2cosโก๐‘ฅ๐‘ฆโ€ณ1=34๐‘ฅโˆ’5/2sinโก๐‘ฅโˆ’๐‘ฅโˆ’3/2cosโก๐‘ฅโˆ’๐‘ฅโˆ’1/2sinโก๐‘ฅ

We assume ๐‘ฆ =๐‘ข๐‘ฆ1 is a solution to the ODE, thus

0=๐‘ฅ2(๐‘ขโ€ณ๐‘ฆ1+2๐‘ขโ€ฒ๐‘ฆโ€ฒ1+๐‘ข๐‘ฆโ€ณ1)+๐‘ฅ(๐‘ขโ€ฒ๐‘ฆ1+๐‘ข๐‘ฆโ€ฒ1)+(๐‘ฅ2โˆ’14)๐‘ข๐‘ฆ1=๐‘ขโ€ณ(๐‘ฅ2๐‘ฆ1)+๐‘ขโ€ฒ(2๐‘ฅ2๐‘ฆโ€ฒ1+๐‘ฅ๐‘ฆ1)+๐‘ข(๐‘ฅ2๐‘ฆโ€ณ1+๐‘ฅ๐‘ฆโ€ฒ1+๐‘ฅ2๐‘ฆ1โˆ’14๐‘ฆ1)=๐‘ขโ€ณ(๐‘ฅ3/2sinโก๐‘ฅ)+๐‘ขโ€ฒ(โˆ’๐‘ฅ1/2sinโก๐‘ฅ+2๐‘ฅ3/2cosโก๐‘ฅ+๐‘ฅ1/2sinโก๐‘ฅ)+๐‘ข(34๐‘ฅโˆ’1/2sinโก๐‘ฅโˆ’๐‘ฅ1/2cosโก๐‘ฅโˆ’๐‘ฅ3/2sinโก๐‘ฅโˆ’12๐‘ฅโˆ’1/2sinโก๐‘ฅ+๐‘ฅ1/2cosโก๐‘ฅ+๐‘ฅ3/2sinโก๐‘ฅโˆ’14๐‘ฅโˆ’1/2sinโก๐‘ฅ)=๐‘ฅ3/2sinโก(๐‘ฅ)๐‘‘๐‘ขโ€ฒ๐‘‘๐‘ฅ+2๐‘ฅ3/2cosโก(๐‘ฅ)๐‘ขโ€ฒ

which is a separable ODE of order 2.

Practice problems


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Footnotes

  1. Perhaps generalisable to higher orders when (๐‘› โˆ’1)/๐‘› solutions are given? โ†ฉ

  2. 2017. Elementary differential equations and boundary value problems, p. 133 (ยง3.4 problem 22) โ†ฉ