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 𝑦1π‘₯ solution is known.

We begin by assuming that 𝑦(π‘₯) =𝑒(π‘₯)𝑦1(π‘₯) for some function 𝑒(π‘₯) to be determined. It follows from this that

𝑦′=𝑒′𝑦1+𝑒𝑦′1𝑦″=𝑒″𝑦1+2𝑒′𝑦′1+𝑒𝑦″1

Given that 𝑦1 is indeed a solution, this substitution will reduce the DE to a first order separable DE on the independent variable 𝑒′ (see reasoning below), which can be then used to determine the general solution.

Explanation

We write the DE as 𝐿[𝑦] =0, where the linear operation 𝐿 is defined by

𝐿=𝐷2+𝑝𝐷+π‘žπŸ™

We are given that 𝐿[𝑦1] =0. Then

𝐿[𝑒𝑦1]=𝐷2[𝑒𝑦1]+𝑝𝐷[𝑒𝑦1]+π‘žπŸ™[𝑒𝑦1]=𝑦1𝐷2[𝑒]+2𝐷[𝑒]𝐷[𝑦1]+𝑒𝐷2[𝑦1]+𝑝𝑦1𝐷[𝑒]+𝑝𝑒𝐷[𝑦1]+π‘žπ‘’π‘¦1=𝑒(𝐷2[𝑦1]+𝑝𝐷[𝑦1]+π‘žπŸ™[𝑦1])+𝑦1𝐷2[𝑒]+(𝑝𝑦1+2𝐷[𝑦1])𝐷[𝑒]=𝑒𝐿[𝑦1]+𝐾[𝐷[𝑒]]

where 𝐾 is the second order linear operator

𝐾=𝑦1𝐷+(𝑝𝑦1+2𝑦′1)πŸ™

Method

In general, I have found it most effective to only substitute the particular solution 𝑦1 after the gathering different 𝑒 terms. Consider the ODE2

π‘₯2𝑦″+π‘₯𝑦′+(π‘₯2βˆ’14)𝑦=0

with particular solution 𝑦1 =π‘₯βˆ’1/2sin⁑π‘₯, so

𝑦′1=βˆ’12π‘₯βˆ’3/2sin⁑π‘₯+π‘₯βˆ’1/2cos⁑π‘₯𝑦″1=34π‘₯βˆ’5/2sin⁑π‘₯βˆ’π‘₯βˆ’3/2cos⁑π‘₯βˆ’π‘₯βˆ’1/2sin⁑π‘₯

We assume 𝑦 =𝑒𝑦1 is a solution to the ODE, thus

0=π‘₯2(𝑒″𝑦1+2𝑒′𝑦′1+𝑒𝑦″1)+π‘₯(𝑒′𝑦1+𝑒𝑦′1)+(π‘₯2βˆ’14)𝑒𝑦1=𝑒″(π‘₯2𝑦1)+𝑒′(2π‘₯2𝑦′1+π‘₯𝑦1)+𝑒(π‘₯2𝑦″1+π‘₯𝑦′1+π‘₯2𝑦1βˆ’14𝑦1)=𝑒″(π‘₯3/2sin⁑π‘₯)+𝑒′(βˆ’π‘₯1/2sin⁑π‘₯+2π‘₯3/2cos⁑π‘₯+π‘₯1/2sin⁑π‘₯)+𝑒(34π‘₯βˆ’1/2sin⁑π‘₯βˆ’π‘₯1/2cos⁑π‘₯βˆ’π‘₯3/2sin⁑π‘₯βˆ’12π‘₯βˆ’1/2sin⁑π‘₯+π‘₯1/2cos⁑π‘₯+π‘₯3/2sin⁑π‘₯βˆ’14π‘₯βˆ’1/2sin⁑π‘₯)=π‘₯3/2sin⁑(π‘₯)𝑑𝑒′𝑑π‘₯+2π‘₯3/2cos⁑(π‘₯)𝑒′

which is a separable ODE of order 2.

Practice problems


tidy | en | SemBr | review

Footnotes

  1. Perhaps generalisable to higher orders when (𝑛 βˆ’1)/𝑛 solutions are given? ↩

  2. 2017. Elementary differential equations and boundary value problems, p. 133 (Β§3.4 problem 22) ↩