Mathematical Methods for Physics and Engineering : A Comprehensive Guide

HIGHER-ORDER ORDINARY DIFFERENTIAL EQUATIONS

If the original equation (15.1) hasf(x) = 0 (i.e. it is homogeneous) then of

course the complementary functionyc(x) in (15.3) is already the general solution.

If, however, the equation hasf(x)= 0 (i.e. it is inhomogeneous) thenyc(x)isonly

one part of the solution. The general solution of (15.1) is then given by

y(x)=yc(x)+yp(x), (15.7)

whereyp(x)istheparticular integral,whichcanbeanyfunction that satisfies (15.1)

directly, provided it is linearly independent ofyc(x). It should be emphasised for

practical purposes thatanysuch function, no matter how simple (or complicated),

is equally valid in forming the general solution (15.7).

It is important to realise that the above method for finding the general solution

to an ODE by superposing particular solutions assumes crucially that the ODE

is linear. For non-linear equations, discussed in section 15.3, this method cannot

be used, and indeed it is often impossible to find closed-form solutions to such

equations.

15.1 Linear equations with constant coefficients

If theamin (15.1) are constants rather than functions ofxthen we have

an

dny

dxn

+an− 1

dn−^1 y

dxn−^1

+···+a 1

dy

dx

+a 0 y=f(x). (15.8)

Equations of this sort are very common throughout the physical sciences and

engineering, and the method for their solution falls into two parts as discussed

in the previous section, i.e. finding the complementary functionyc(x) and finding

the particular integralyp(x). Iff(x) = 0 in (15.8) then we do not have to find

a particular integral, and the complementary function is by itself the general

solution.

15.1.1 Finding the complementary functionyc(x)

The complementary function must satisfy

an

dny

dxn

+an− 1

dn−^1 y

dxn−^1

+···+a 1

dy

dx

+a 0 y= 0 (15.9)

and containnarbitrary constants (see equation (15.3)). The standard method

for findingyc(x) is to try a solution of the formy=Aeλx, substituting this into

(15.9). After dividing the resulting equation through byAeλx, we are left with a

polynomial equation inλof ordern;thisistheauxiliary equationand reads

anλn+an− 1 λn−^1 +···+a 1 λ+a 0 =0. (15.10)