Mathematical Methods for Physics and Engineering : A Comprehensive Guide

HIGHER-ORDER ORDINARY DIFFERENTIAL EQUATIONS

For example, if we consider the second-order case with boundary conditions

y(a)=α,y(b)=βthen a suitable change of variable is

u=y−(mx+c),

wherey=mx+cis the straight line through the points (a, α)and(b, β), for which

m=(α−β)/(a−b)andc=(βa−αb)/(a−b). Alternatively, if the boundary

conditions for our second-order equation arey(0) =y′(0) =γthen we would

make the same change of variable, but this timey=mx+cwould be the straight

line through (0,γ) with slopeγ,i.e.m=c=γ.

Solution method.Require that the Green’s functionG(x, z)obeys the original ODE,

but with the RHS set to a delta functionδ(x−z). This is equivalent to assuming

thatG(x, z)is given by the complementary function of the original ODE, with the

constants replaced by functions ofz; these functions are different forx

z. Now require also thatG(x, z)obeys the given homogeneous boundary conditions

and impose the continuity conditions given in (15.64) and (15.65). The general

solution to the original ODE is then given by (15.60). For inhomogeneous boundary

conditions, make the change of dependent variableu=y−h(x),whereh(x)is a

polynomial obeying the given boundary conditions.

15.2.6 Canonical form for second-order equations

In this section we specialise fromnth-order linear ODEs with variable coefficients

to those of order 2. In particular we consider the equation

d^2 y

dx^2

+a 1 (x)

dy

dx

+a 0 (x)y=f(x), (15.70)

which has been rearranged so that the coefficient ofd^2 y/dx^2 is unity. By making

the substitutiony(x)=u(x)v(x) we obtain

v′′+

(

2 u′

u

+a 1

)

v′+

(

u′′+a 1 u′+a 0 u

u

)

v=

f

u

, (15.71)

where the prime denotes differentiation with respect tox. Since (15.71) would be

much simplified if there were no term inv′,letuschooseu(x) such that the first

factor in parentheses on the LHS of (15.71) is zero, i.e.

2 u′

u

+a 1 =0 ⇒ u(x)=exp

{

−^12

∫

a 1 (z)dz

}

. (15.72)

We then obtain an equation of the form

d^2 v

dx^2

+g(x)v=h(x), (15.73)