For example, consider the function d = sin x. Its derivatives are
Here's an example of a function that does not have a finite family of derivatives: d = tan x. Its first four derivatives are
Notice that the nth derivative ( n ≥ 1) contains a term involving tan n‐1 x, so as higher and higher derivatives are taken, each one will contain a higher and higher power of tan x, so there is no way that all derivatives can be written in terms of a finite number of functions. The method of undetermined coefficients could not be applied if the nonhomogeneous term in (*) were d = tan x. So just what are the functions d( x) whose derivative families are finite? See Table 1.
Example 1: If d( x) = 5 x 2, then its family is { x 2, x, 1}. Note that any numerical coefficients (such as the 5 in this case) are ignored when determining a function's family.
Example 2: Since the function d( x) = x sin 2 x is the product of x and sin 2 x, the family of d( x) would consist of all products of the family members of the functions x and sin 2 x. That is,
Linear combinations of n functions . A linear combination of two functions y 1 and y 2 was defined to be any expression of the form
The central idea of the method of undetermined coefficients is this: Form the most general linear combination of the functions in the family of the nonhomogeneous term d( x), substitute this expression into the given nonhomogeneous differential equation, and solve for the coefficients of the linear combination.
Example 3: Find a particular solution of the differential equation
As noted in Example 1, the family of d = 5 x 2 is { x 2, x, 1}; therefore, the most general linear combination of the functions in the family is y = Ax 2 + Bx + C (where A, B, and C are the undetermined coefficients). Substituting this into the given differential equation gives
Now, combinbing like terms yields
In order for this last equation to be an identity, the coefficients of like powers of x on both sides of the equation must be equated. That is, A, B, and C must be chosen so that
The first equation immediately gives . Substituting this into the second equation gives , and finally, substituting both of these values into the last equation yields . Therefore, a particular solution of the given differential equation is