Mathematical Methods for Physics and Engineering : A Comprehensive Guide

HIGHER-ORDER ORDINARY DIFFERENTIAL EQUATIONS

It is worth noting that, even if a higher-order ODE is not exact in its given form,

it may sometimes be made exact by multiplying through by some suitable function,

anintegrating factor, cf. subsection 14.2.3. Unfortunately, no straightforward

method for finding an integrating factor exists and one often has to rely on

inspection or experience.

Solve

x(1−x^2 )

d^2 y

dx^2

− 3 x^2

dy

dx

−xy=x. (15.48)

It is easily shown that (15.48) is not exact, but we also see immediately that by multiplying
it through by 1/xwe recover (15.44), which is exact and is solved above.

Another important point is that an ODE need not be linear to be exact,

although no simple rule such as (15.43) exists if it is not linear. Nevertheless, it is

often worth exploring the possibility that a non-linear equation is exact, since it

could then be reduced in order by one and may lead to a soluble equation. This

is discussed further in subsection 15.3.3.

Solution method.For a linear ODE of the form (15.41) check whether it is exact

using equation (15.43). If it is not then attempt to find an integrating factor which

when multiplying the equation makes it exact. Once the equation is exact write the

LHS as a derivative as in (15.42) and, by expanding this derivative and comparing

with the LHS of the ODE, determine the functionsbm(x)in (15.42). Integrate the

resulting equation to yield another ODE, of one order lower. This may be solved or

simplified further if the new ODE is itself exact or can be made so.

15.2.3 Partially known complementary function

Suppose we wish to solve thenth-order linear ODE

an(x)

dny

dxn

+···+a 1 (x)

dy

dx

+a 0 (x)y=f(x), (15.49)

andwehappentoknowthatu(x) is a solution of (15.49) when the RHS is

set to zero, i.e.u(x) is one part of the complementary function. By making the

substitutiony(x)=u(x)v(x), we can transform (15.49) into an equation of order

n−1indv/dx. This simpler equation may prove soluble.

In particular, if the original equation is of second order then we obtain

a first-order equation indv/dx, which may be soluble using the methods of

section 14.2. In this way both the remaining term in the complementary function

and the particular integral are found. This method therefore provides a useful

way of calculating particular integrals for second-order equations with variable

(or constant) coefficients.