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60 CHAPTER 2 Linear Second-Order Equations

TABLE 2.1 Functions to Try for Y p (x) in the Method of Undetermined
Coefficients

f (x) Y p (x)
P(x) Q(x)
Ae cx Re cx
A cos(βx) C cos(βx) + D sin(βx)
A sin(βx) C cos(βx) + D sin(βx)
P(x)e cx Q(x)e cx
P(x)cos(βx) Q(x)cos(βx) + R(x)sin(βx)
P(x)sin(βx) Q(x)cos(βx) + R(x)sin(βx)
cx
cx
cx
P(x)e cos(βx) Q(x)e cos(βx) + R(x)e sin(βx)
cx
cx
cx
P(x)e sin(βx) Q(x)e cos(βx) + R(x)e sin(βx)
Table 2.1 provides a list of functions for a first try at Y p (x) for various functions f (x) that
might appear in the differential equation. In this list, P(x) is a given polynomial of degree n,
Q(x) and R(x) are polynomials of degree n with undetermined coefficients for which we must
solve, and c and β are constants.

2.3.3 The Principle of Superposition

Suppose we want to find a particular solution of

y + p(x)y + q(x)y = f 1 (x) + f 2 (x) + ··· + f N (x).

It is routine to check that, if Y j is a solution of

y + p(x)y + q(x)y = f j (x),
then Y 1 + Y 2 + ··· + Y N is a particular solution of the original differential equation.

EXAMPLE 2.17
Find a particular solution of

y + 4y = x + 2e −2x .
To find a particular solution, consider two problems:
Problem 1: y + 4y = x

Problem 2: y + 4y = 2e −2x
(x)=x/4 of Problem 1 and
Using undetermined coefficients, we find a particular solution Y p 1
(x) = e −2x /4 of Problem 2. A particular solution of the given differential
a particular solution Y p 2
equation is
1 1 −2x
Y p (x) = x + e .
4 4
Using this, the general solution is

1
−2x
y(x) = c 1 cos(2x) + c 2 sin(2x) + x + e .
4

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October 14, 2010 14:12 THM/NEIL Page-60 27410_02_ch02_p43-76