James Stewart-Calculus_ Early Transcendentals-Cengage Learning (2015)

Section 17.2 Nonhomogeneous Linear Equations 1161
is called the complementary equation and plays an important role in the solution of the
original nonhomogeneous equation (1).
3 Theorem The general solution of the nonhomogeneous differential equation
(1) can be written as
ysxd − y p sxd 1 y c sxd
where y p is a particular solution of Equation 1 and y c is the general solution of
the complementary Equation 2.
Proof We verify that if y is any solution of Equation 1, then y 2 y p is a solution of
the complementary Equation 2. Indeed
asy 2 y p d0 1 bsy 2 y p d9 1 csy 2 y p d − ay0 2 ay p 0 1 by9 2 by p 9 1 cy 2 cy p
− say0 1 by9 1 cyd 2 say p 0 1 by p 9 1 cy p d
− Gsxd 2 Gsxd − 0
This shows that every solution is of the form ysxd − y p sxd 1 y c sxd. It is easy to
check that every function of this form is a solution.
■
We know from Section 17.1 how to solve the complementary equation. (Recall that
the solution is y c − c 1 y 1 1 c 2 y 2 , where y 1 and y 2 are linearly independent solutions of
Equa tion 2.) Therefore Theorem 3 says that we know the general solution of the nonhomogeneous
equation as soon as we know a particular solution y p . There are two methods
for finding a particular solution: The method of undetermined coefficients is straightforward
but works only for a restricted class of functions G. The method of variation of
parameters works for every function G but is usually more difficult to apply in practice.
The Method of Undetermined Coefficients
We first illustrate the method of undetermined coefficients for the equation
ay0 1 by9 1 cy − Gsxd
where Gsx) is a polynomial. It is reasonable to guess that there is a particular solution
y p that is a polynomial of the same degree as G because if y is a polynomial, then
ay0 1 by9 1 cy is also a polynomial. We therefore substitute y p sxd − a polynomial (of
the same degree as G) into the differential equation and determine the coefficients.
Example 1 Solve the equation y0 1 y9 2 2y − x 2 .
SOLUtion The auxiliary equation of y0 1 y9 2 2y − 0 is
r 2 1 r 2 2 − sr 2 1dsr 1 2d − 0
with roots r − 1, 22. So the solution of the complementary equation is
y c − c 1 e x 1 c 2 e 22x
Since Gsxd − x 2 is a polynomial of degree 2, we seek a particular solution of the form
y p sxd − Ax 2 1 Bx 1 C
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