232
Chapter 5
CHAPTER 5
HIGHERORDER LINEAR DIFFERENTIAL EQUATIONS
SECTION 5.1
INTRODUCTION:
SECONDORDER LINEAR EQUATIONS
In this section the central ideas of the theory of linear differential equations are introduced and
illustrated concretely in the context of
secondorder
equations.
These key concepts include
superposition of solutions (Theorem 1), existence and uniqueness of solutions (Theorem 2),
linear independence, the Wronskian (Theorem 3), and general solutions (Theorem 4).
This
discussion of secondorder equations serves as preparation for the treatment of
n
th order linear
equations in Section 5.2.
Although the concepts in this section may seem somewhat abstract to
students, the problems set is quite tangible and largely computational.
In each of Problems 1–16 the verification that
y
1
and
y
2
satisfy the given differential equation
is a routine matter.
As in E
x
ample 2, we then impose the given initial conditions on the general
solution
y
=
c
1
y
1
+
c
2
y
2
.
This yields two linear equations that determine the values of the
constants
c
1
and
c
2
.
1.
Imposition of the initial conditions
(0)
0,
(0)
5
y
y
′
=
=
on the general solution
1
2
( )
x
x
y x
c e
c e
−
=
+
yields the two equations
1
2
1
2
0,
0
c
c
c
c
+
=
−
=
with solution
1
2
5/ 2,
5/ 2.
c
c
=
= −
Hence the desired particular solution is
y
(
x
)
=
5(
e
x

e

x
)/2.
2.
Imposition of the initial conditions
(0)
1,
(0)
15
y
y
′
= −
=
on the general solution
3
3
1
2
( )
x
x
y x
c e
c e
−
=
+
yields the two equations
1
2
1
2
1, 3
3
15
c
c
c
c
+
= −
−
=
with solution
1
2
2,
3.
c
c
=
=
Hence the desired particular solution is
y
(
x
)
=
2
e
3
x

3
e

3
x
.
3.
Imposition of the initial conditions
(0)
3,
(0)
8
y
y
′
=
=
on the general solution
1
2
( )
cos2
sin2
y x
c
x
c
x
=
+
yields the two equations
1
2
3, 2
8
c
c
=
=
with solution
1
2
3,
4.
c
c
=
=
Hence the desired particular solution is
y
(
x
)
=
3 cos 2
x
+ 4 sin 2
x
.
4.
Imposition of the initial conditions
(0)
10,
(0)
10
y
y
′
=
= −
on the general solution
1
2
( )
cos5
sin5
y x
c
x
c
x
=
+
yields the two equations
1
2
10, 5
10
c
c
=
= −
with solution
1
2
3,
4.
c
c
=
=
Hence the desired particular solution is
y
(
x
)
=
10 cos 5
x

2 sin 5
x.
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Section 5.1
233
5.
Imposition of the initial conditions
(0)
1,
(0)
0
y
y
′
=
=
on the general solution
2
1
2
( )
x
x
y x
c e
c e
=
+
yields the two equations
1
2
1
2
1,
2
0
c
c
c
c
+
=
+
=
with solution
1
2
2,
1.
c
c
=
= −
Hence the desired particular solution is
y
(
x
)
=
2
e
x

e
2
x
.
6.
Imposition of the initial conditions
(0)
7,
(0)
1
y
y
′
=
= −
on the general solution
2
3
1
2
( )
x
x
y x
c e
c e
−
=
+
yields the two equations
1
2
1
2
7, 2
3
1
c
c
c
c
+
=
−
= −
with solution
1
2
4,
3.
c
c
=
=
Hence the desired particular solution is
y
(
x
)
=
4
e
2
x
+ 3
e

3
x
.
7.
Imposition of the initial conditions
(0)
2,
(0)
8
y
y
′
= −
=
on the general solution
1
2
( )
x
y x
c
c e
−
=
+
yields the two equations
1
2
2
2,
8
c
c
c
+
= −
−
=
with solution
1
2
6,
8.
c
c
=
= −
Hence the desired particular solution is
y
(
x
)
=
6
−
8
e

x
.
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 Spring '08
 Zhu
 Differential Equations, Linear Equations, Equations, initial conditions, yp, general solution, ytrial

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