33
PART II: LINEAR EQUATIONS WITH CONSTANT COEFFICIENTS
15.
Basic concepts
15.1.
Linear equations.
The standard form of a second order linear equation is
L
[
y
]
≡
y
±±
+
p
(
t
)
y
±
+
q
(
t
)
y
=
g
(
t
)
.
The map
y
±→
L
[
y
]isa
diﬀerential operation
.I
f
g
(
t
)
≡
0, then the equation
L
[
y
]=0
is called
homogenous
. The operation
L
has
constant coeﬃcients
if
p
(
t
) = const and
q
(
t
) = const.
The standard form of
n
th order equations is
L
[
y
]=
g
(
t
) with
L
[
y
y
(
n
)
+
p
1
(
t
)
y
(
n

1)
+
···
+
p
0
(
t
)
y.
15.2.
Initial value problem.
To solve the IVP(
t
0
,y
0
±
0
) for a second order equa
tion [not necessarily linear] is to Fnd a solution
y
(
t
) such that
y
(
t
0
)=
y
0
±
(
t
0
y
±
0
.
In the case of
n
th order equations, the initial conditions have the form
y
(
t
0
y
0
±
(
t
0
y
±
0
, ..., y
(
n

1)
(
t
0
y
(
n

1)
0
.
Theorem.
Consider a linear equation
L
[
y
g
. Suppose that
g
and the coeﬃcients
of
L
are continuous functions in some interval
I
. Then
(i) all solutions extend to the whole interval,
(ii) every IVP with
t
0
∈
I
has a unique solution deFned on
I
.
15.3.
Linear properties.
Assume that the coeﬃcients of a diﬀerential operation
L
are in
C
(
I
). Then we have a linear map (”operator”)
L
:
C
(
n
)
(
I
)
→
C
(
I
)
.
Notation:
ker
L
=
{
y
:
L
[
y
}
.
In other words, the kernel of
L
is the set of all solutions of the homogeneous
equation.
Lemma.
ker
L
is a linear space of dimension
n
.
Proof:
(n=2) ±ix some
t
0
, and consider the map
R
2
→
ker
L
:(
y
0
±
0
)
±→
y
(
t
)
,
where
y
(
t
) is the solution of the corresponding IVP. Applying the main theorem we
see that this map is linear, 1to1, and ”onto”.
±
Recall that
n
elements (”vectors”)
y
1
,...,y
n
of a linear space are (linearly) inde
pendent if
C
1
y
1
+
+
C
n
y
n
=0
⇒
C
1
=
=
C
n
.
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If the linear space has dimension
n
, then
n
independent vectors form a basis: every
vector
y
has a unique representation as a linear combination of
y
j
’s. Any
n
+1
vectors of an
n
dimensional space are dependent.
Let
y
1
,...,y
n
be solutions of an
n
th order equation
L
[
y
] = 0. Then the collection
y
1
n
is called a
fundamental system of solutions
if the functions
y
j
are inde
pendent as elements of ker
L
. It follows that every solution of
L
[
y
] = 0 has a unique
representation as a linear combination
y
(
t
)=
C
1
y
1
(
t
)+
···
+
C
n
y
n
(
t
)
.
Example.
The functions 1
,t,.
..,t
n

1
form a fundamental system of solutions of
the equation
y
(
n
)
= 0. The
general solution
(
≡
the set of all solutions) is the set of
polynomials of degree
<n
.
Lemma.
If
y
*
is a solution of
L
[
y
]=
g
, then the general solution of
L
[
y
g
is
y
∈
y
*
+ker
L.
[”particular integral” + ”complementary function”]
15.4.
Reduction of order.
The following substitution is quite useful.
Theorem.
If we know some solution
y
1
of a homogeneous equation
L
[
y
]=0
of
order
n
, then the substitution
y
=
y
1
v
, (where
v
=
v
(
t
)
is a new unknown function)
transforms the equation
L
[
y
g
into a linear equation of order
n

1
for
v
±
.
Proof:
Consider the case
n
=3,
L
[
y
]
≡
y
±±±
+
p
1
y
±±
+
p
2
y
±
+
p
3
y
=
g,
[the coeﬃcients are not necessarily constant]. We have
L
[
y
y
±±±
1
v
+3
y
±±
1
v
±
y
±
1
v
±±
+
y
1
v
±±±
+
p
1
y
1
v
+
p
1
y
±
1
v
±
+
p
1
y
1
v
±±
+
p
2
y
±
1
v
+
p
2
y
1
v
±
+
p
3
y
1
v
=
g.
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 Spring '10
 323
 Laplace, Cos, Complex number, general solution

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