Mathematical Methods for Physics and Engineering : A Comprehensive Guide

HIGHER-ORDER ORDINARY DIFFERENTIAL EQUATIONS

15.2.4 Variation of parameters

The method of variation of parameters proves useful in finding particular integrals

for linear ODEs with variable (and constant) coefficients. However, it requires

knowledge of the entire complementary function, not just of one part of it as in

the previous subsection.

Suppose we wish to find a particular integral of the equation

an(x)

dny

dxn

+···+a 1 (x)

dy

dx

+a 0 (x)y=f(x), (15.53)

and the complementary function yc(x) (the general solution of (15.53) with

f(x) = 0) is

yc(x)=c 1 y 1 (x)+c 2 y 2 (x)+···+cnyn(x),

where the functionsym(x) are known. We now assume that a particular integral of

(15.53) can be expressed in a form similar to that of the complementary function,

but with the constantscmreplaced by functions ofx,i.e.weassumeaparticular

integral of the form

yp(x)=k 1 (x)y 1 (x)+k 2 (x)y 2 (x)+···+kn(x)yn(x). (15.54)

This will no longer satisfy the complementary equation (i.e. (15.53) with the RHS

set to zero) but might, with suitable choices of the functionski(x), be made equal

tof(x), thus producing not a complementary function but a particular integral.

Since we havenarbitrary functionsk 1 (x),k 2 (x),...,kn(x), but only one restric-

tion on them (namely the ODE), we may impose a furthern−1 constraints. We

can choose these constraints to be as convenient as possible, and the simplest

choice is given by

k′ 1 (x)y 1 (x)+k′ 2 (x)y 2 (x)+···+k′n(x)yn(x)=0

k′ 1 (x)y 1 ′(x)+k′ 2 (x)y 2 ′(x)+···+k′n(x)y′n(x)=0

..

. (15.55)

k 1 ′(x)y

(n−2)

1 (x)+k

′

2 (x)y

(n−2)

2 (x)+···+k

′

n(x)y

(n−2)

n (x)=0

k 1 ′(x)y( 1 n−1)(x)+k′ 2 (x)y( 2 n−1)(x)+···+k′n(x)y(nn−1)(x)=

f(x)

an(x)

,

where the primes denote differentiation with respect tox. The last of these

equations is not a freely chosen constraint; given the previousn−1 constraints

and the original ODE, it must be satisfied.

This choice of constraints is easily justified (although the algebra is quite

messy). Differentiating (15.54) with respect tox, we obtain

y′p=k 1 y 1 ′+k 2 y′ 2 +···+kny′n+[k′ 1 y 1 +k 2 ′y 2 +···+kn′yn],

where, for the moment, we drop the explicitx-dependence of these functions. Since