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LS.2
Homogeneous
Linear
Systems
with
Constant
Coeﬃcients
1.
Using
matrices
to
solve
linear
systems.
The
naive
way
to
solve
a
linear
system
of
ODE’s
with
constant
coeﬃcients
is
by
elimi
nating
variables,
so
as
to
change
it
into
a
single
higherorder
equation.
For
instance,
if
x
=
x
+
3
y
(1)
�
y
=
x
−
y
we
can
eliminate
x
by
solving
the
second
equation
for
x
,
getting
x
=
y
+
y
�
,
then
replacing
x
everywhere
by
y
+
y
�
in
the
first
equation.
This
gives
y
��
−
4
y
=
0
;
the
characteristic
equation
is
(
r
−
2)(
r
+
2)
=
0,
so
the
general
solution
for
y
is
2
t
−
2
t
y
=
c
1
e
+
c
2
e
.
From
this
we
get
x
from
the
equation
x
=
y
+
y
�
originally
used
to
eliminate
x
;
the
whole
solution
to
the
system
is
then
−
2
t
x
= 3
c
1
e
2
t
−
c
2
e
(2)
2
t
−
2
t
y
=
c
1
e
+
c
2
e
.
We
now
want
to
introduce
linear
algebra
and
matrices
into
the
study
of
systems
like
the
one
above.
Our
first
task
is
to
see
how
the
above
equations
look
when
written
using
matrices
and
matrix
multiplication.
When
we
do
this,
the
system
(1)
and
its
general
solution
(2)
take
the
forms
x
1
3
x
=
(4)
y
1
−
1
y
,
−
2
t
3
c
1
e
2
t
−
c
2
e
−
1
−
2
t
(5)
x
=
c
1
e
2
t
+
c
2
e
−
2
t
=
c
1
3
e
2
t
+
c
2
e
.
y
1
1
Study
the
above
until
it
is
clear
to
you
how
the
matrices
and
column
vectors
are
being
used
to
write
the
system
(1)
and
its
solution
(2).
Note
that
when
we
multiply
the
column
vectors
by
scalars
or
scalar
functions,
it
does
not
matter
whether
we
write
them
behind
or
in
front
of
the
column
vector;
the
way
it
is
written
above
on
the
right
of
(5)
is
the
one
usually
used,
since
it
is
easiest
to
read
and
interpret.
We
are
now
going
to
show
a
new
method
of
solving
the
system
(1),
which
makes
use
of
the
matrix
form
(4)
for
writing
it.
We
begin
by
noting
from
(5)
that
two
particular
solutions
to
the
system
(4)
are
3
2
t
(6)
e
and
−
1
e
−
2
t
.
1
1
6
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7
LS.2
HOMOGENEOUS
LINEAR
SYSTEMS
Based
on
this,
our
new
method
is
to
look
for
solutions
to
(4)
of
the
form
x
(7)
=
a
1
e
�t
,
y
a
2
where
a
1
,
a
2
and
�
are
unknown
constants.
We
substitute
(7)
into
the
system
(4)
to
deter
mine
what
these
unknown
constants
should
be.
This
gives
a
1
a
1
�t
(8)
�
e
�t
=
1
3
e
.
a
2
1
−
1
a
2
We
can
cancel
the
factor
e
�t
from
both
sides,
getting
3
a
1
(9)
�
a
1
=
1
.
a
2
1
−
1
a
2
We
have
to
solve
the
matrix
equation
(9)
for
the
three
constants.
It
is
not
very
clear
how
to
do
this.
When
faced
with
equations
in
unfamiliar
notation,
a
reasonable
strategy
is
to
rewrite
them
in
more
familiar
notation.
If
we
try
this,
(9)
becomes
the
pair
of
equations
�a
1
=
a
1
+
3
a
2
(10)
�a
2
=
a
1
−
a
2
.
Technically
speaking,
these
are
a
pair
of
nonlinear
equations
in
three
variables.
The
trick
in
solving
them
is
to
look
at
them
as
a
pair
of
linear
equations
in
the
unknowns
a
i
,
with
�
viewed
as
a
parameter.
If
we
think
of
them
this
way,
it
immediately
suggests
writing
them
in
standard
form
(1
−
�
)
a
1
+
3
a
2
= 0
(11)
a
1
+
(
−
1
−
�
)
a
2
= 0
.
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 Fall '09
 vogan
 Linear Algebra, Linear Systems, Matrices, characteristic equation

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