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

Chapter 17
Concept Check Answers
Cut here and keep for reference
1. (a) Write the general form of a second-order homogeneous
linear differential equation with constant coefficients.
ay0 1 by9 1 cy − 0
where a, b, and c are constants and a ± 0.
(b) Write the auxiliary equation.
ar 2 1 br 1 c − 0
(c) How do you use the roots of the auxiliary equation to
solve the differential equation? Write the form of the
solution for each of the three cases that can occur.
If the auxiliary equation has two distinct real roots r 1 and
r 2, the general solution of the differential equation is
y − c 1e r 1x 1 c 2e r 2 x
If the roots are real and equal, the solution is
y − c 1e rx 1 c 2 xe rx
where r is the common root.
If the roots are complex, we can write r 1 − 1 i and
r 2 − 2 i, and the solution is
y − e x sc 1 cos x 1 c 2 sin xd
2. (a) What is an initial-value problem for a second-order differential
equation?
An initial-value problem consists of finding a solution y of
the differential equation that also satisfies given conditions
ysx 0d − y 0 and y9sx 0d − y 1, where y 0 and y 1 are constants.
(b) What is a boundary-value problem for such an equation?
A boundary-value problem consists of finding a solution y
of the differential equation that also satisfies given boundary
conditions ysx 0d − y 0 and ysx 1d − y 1.
3. (a) Write the general form of a second-order nonhomogeneous
linear differential equation with constant
coefficients.
ay0 1 by9 1 cy − Gsxd, where a, b, and c are constants
and G is a continuous function.
(b) What is the complementary equation? How does it help
solve the original differential equation?
The complementary equation is the related homogeneous
equation ay0 1 by9 1 cy − 0. If we find the general solution
y c of the complementary equation and y p is any particular
solution of the nonhomogeneous differential equation,
then the general solution of the original differential
equation is ysxd − y psxd 1 y csxd.
(c) Explain how the method of undetermined coefficients
works.
To determine a particular solution y p of
ay0 1 by9 1 cy − Gsxd, we make an initial guess that
y p is a general function of the same type as G. If Gsxd
is a polynomial, choose y p to be a general polynomial
of the same degree. If Gsxd is of the form Ce k x , choose
y psxd − Ae k x . If Gsxd is C cos kx or C sin kx, choose
y psxd − A cos kx 1 B sin kx. If Gsxd is a product of functions,
choose y p to be a product of functions of the same
type. Some examples are:
Gsxd
x 2
e 2x
sin 2x
xe 2x
y p sxd
Ax 2 1 Bx 1 C
Ae 2x
A cos 2x 1 B sin 2x
sAx 1 Bde 2x
We then substitute y p , y p9, and y p0 into the differential equation
and determine the coefficients.
If y p happens to be a solution of the complementary equation,
then multiply the initial trial solution by x (or x 2 if
necessary).
If Gsxd is a sum of functions, we find a particular solution
for each function and then y p is the sum of these.
The general solution of the differential equation is
ysxd − y psxd 1 y csxd
(d) Explain how the method of variation of parameters
works.
We write the solution of the complementary equation
ay0 1 by9 1 cy − 0 as y csxd − c 1y 1sxd 1 c 2y 2sxd,
where y 1 and y 2 are linearly independent solutions. We
then take y psxd − u 1sxd y 1sxd 1 u 2sxd y 2sxd as a particular
solution, where u 1sxd and u 2sxd are arbitrary functions.
After computing y p9, we impose the condition that
u 19y 1 1 u 29y 2 − 0 (1)
and then compute y p0. Substituting y p , y p9, and y p0 into the
original differential equation gives
asu 19y 19 1 u 29y 29d − G (2)
We then solve equations (1) and (2) for the unknown
functions u 19 and u 29. If we are able to integrate
these functions, then a particular solution is
y psxd − u 1sxd y 1sxd 1 u 2sxd y 2sxd and the general solution
is ysxd − y psxd 1 y csxd.
4. Discuss two applications of second-order linear differential
equations.
The motion of an object with mass m at the end of a spring
is an example of simple harmonic motion and is described by
the second-order linear differential equation
m d 2 x
dt 2 1 kx − 0
(continued)
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