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 ๐ฆ1๐ฅ solution is known.
We begin by assuming that ๐ฆ(๐ฅ) =๐ข(๐ฅ)๐ฆ1(๐ฅ) for some function ๐ข(๐ฅ) to be determined. It follows from this that
๐ฆโฒ=๐ขโฒ๐ฆ1+๐ข๐ฆโฒ1๐ฆโณ=๐ขโณ๐ฆ1+2๐ขโฒ๐ฆโฒ1+๐ข๐ฆโณ1
Given that ๐ฆ1 is indeed a solution, this substitution will reduce the DE to a first order separable DE on the independent variable ๐ขโฒ (see reasoning below),
which can be then used to determine the general solution.
Explanation
We write the DE as ๐ฟ[๐ฆ] =0, where the linear operation ๐ฟ is defined by
๐ฟ=๐ท2+๐๐ท+๐๐
We are given that ๐ฟ[๐ฆ1] =0.
Then
๐ฟ[๐ข๐ฆ1]=๐ท2[๐ข๐ฆ1]+๐๐ท[๐ข๐ฆ1]+๐๐[๐ข๐ฆ1]=๐ฆ1๐ท2[๐ข]+2๐ท[๐ข]๐ท[๐ฆ1]+๐ข๐ท2[๐ฆ1]+๐๐ฆ1๐ท[๐ข]+๐๐ข๐ท[๐ฆ1]+๐๐ข๐ฆ1=๐ข(๐ท2[๐ฆ1]+๐๐ท[๐ฆ1]+๐๐[๐ฆ1])+๐ฆ1๐ท2[๐ข]+(๐๐ฆ1+2๐ท[๐ฆ1])๐ท[๐ข]=๐ข๐ฟ[๐ฆ1]+๐พ[๐ท[๐ข]]
where ๐พ is the second order linear operator
๐พ=๐ฆ1๐ท+(๐๐ฆ1+2๐ฆโฒ1)๐
Method
In general, I have found it most effective to only substitute the particular solution ๐ฆ1 after the gathering different ๐ข terms.
Consider the ODE2
๐ฅ2๐ฆโณ+๐ฅ๐ฆโฒ+(๐ฅ2โ14)๐ฆ=0
with particular solution ๐ฆ1 =๐ฅโ1/2sinโก๐ฅ, so
๐ฆโฒ1=โ12๐ฅโ3/2sinโก๐ฅ+๐ฅโ1/2cosโก๐ฅ๐ฆโณ1=34๐ฅโ5/2sinโก๐ฅโ๐ฅโ3/2cosโก๐ฅโ๐ฅโ1/2sinโก๐ฅ
We assume ๐ฆ =๐ข๐ฆ1 is a solution to the ODE, thus
0=๐ฅ2(๐ขโณ๐ฆ1+2๐ขโฒ๐ฆโฒ1+๐ข๐ฆโณ1)+๐ฅ(๐ขโฒ๐ฆ1+๐ข๐ฆโฒ1)+(๐ฅ2โ14)๐ข๐ฆ1=๐ขโณ(๐ฅ2๐ฆ1)+๐ขโฒ(2๐ฅ2๐ฆโฒ1+๐ฅ๐ฆ1)+๐ข(๐ฅ2๐ฆโณ1+๐ฅ๐ฆโฒ1+๐ฅ2๐ฆ1โ14๐ฆ1)=๐ขโณ(๐ฅ3/2sinโก๐ฅ)+๐ขโฒ(โ๐ฅ1/2sinโก๐ฅ+2๐ฅ3/2cosโก๐ฅ+๐ฅ1/2sinโก๐ฅ)+๐ข(34๐ฅโ1/2sinโก๐ฅโ๐ฅ1/2cosโก๐ฅโ๐ฅ3/2sinโก๐ฅโ12๐ฅโ1/2sinโก๐ฅ+๐ฅ1/2cosโก๐ฅ+๐ฅ3/2sinโก๐ฅโ14๐ฅโ1/2sinโก๐ฅ)=๐ฅ3/2sinโก(๐ฅ)๐๐ขโฒ๐๐ฅ+2๐ฅ3/2cosโก(๐ฅ)๐ขโฒ
which is a separable ODE of order 2.
Practice problems
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