��
�
�
�
2.
Higherorder
Linear
ODE’s
2A.
Secondorder
Linear
ODE’s:
General
Properties
2A1.
On
the
right
below
is
an
abbreviated
form
of
the
ODE
on
the
left:
(*)
y
+
p
(
x
)
y
�
+
q
(
x
)
y
=
r
(
x
)
Ly
=
r
(
x
)
;
where
L
is
the
differential
operator
:
L
=
D
2
+
p
(
x
)
D
+
q
(
x
)
.
a)
If
u
1
and
u
2
are
any
two
twicedifferentiable
functions,
and
c
is
a
constant,
then
L
(
u
1
+
u
2
)
=
L
(
u
1
)
+
L
(
u
2
)
and
L
(
c
u
)
=
c
L
(
u
)
.
Operators
which
have
these
two
properties
are
called
linear
.
Verify
that
L
is
linear,
i.e.,
that
the
two
equations
are
satisfied.
b)
Show
that
if
y
p
is
a
solution
to
(*),
then
all
other
solutions
to
(*)
can
be
written
in
the
form
y
=
y
c
+
y
p
,
where
y
c
is
a
solution
to
the
associated
homogeneous
equation
Ly
=
0.
2A2.
a)
By
eliminating
the
constants,
find
a
secondorder
linear
homogeneous
ODE
whose
x
+
c
2
e
2
x
general
solution
is
y
=
c
1
e
.
b)
Verify
for
this
ODE
that
the
IVP
consisting
of
the
ODE
together
with
the
initial
conditions
y
(
x
0
)
=
y
0
,
y
(
x
0
)
=
y
�
y
0
, y
constants
0
0
is
always
solvable.
2A3.
a)
By
eliminating
the
constants,
find
a
secondorder
linear
homogeneous
ODE
whose
y
=
c
1
x
+
c
2
x
2
general
solution
is
.
b)
Show
that
there
is
no
solution
to
the
ODE
you
found
in
part
(a)
which
satisfies
the
initial
conditions
y
(0)
=
1
,
y
�
(0)
=
1
.
c)
Why
doesn’t
part
(b)
contradict
the
existence
theorem
for
solutions
to
secondorder
linear
homogeneous
ODE’s?
(Book:
Theorem
2,
p.
110.)
2A4.
Consider
the
ODE
y
��
+
p
(
x
)
y
+
q
(
x
)
y
=
0.
a)
Show
that
if
p
and
q
are
continuous
for
all
x
,
a
solution
whose
graph
is
tangent
to
the
x
axis
at
some
point
must
be
identically
zero,
i.e.,
zero
for
all
x
.
b)
Find
an
equation
of
the
above
form
having
x
2
as
a
solution,
by
calculating
its
derivatives
and
finding
a
linear
equation
connecting
them.
Why
isn’t
part
(a)
contradicted,
since
the
function
x
2
has
a
graph
tangent
to
the
x
axis
at
0?
1
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
�
�
�
�
��
2
18.03
EXERCISES
2A5.
Show
that
the
following
pairs
of
functions
are
linearly
independent,
by
calculating
their
Wronskian.
m
1
x
m
2
x
mx
a)
e
,
e
,
m
1
=
m
2
b)
e
,
xe
mx
(can
m
=
0?)
⇒
2A6.
Consider
y
1
=
x
2
and
y
2
=
x

x

.
(Sketch
the
graph
of
y
2
.)
a)
Show
that
W
(
y
1
,
y
2
)
�
0
(i.e.,
is
identically
zero).
b)
Show
that
y
1
and
y
2
are
not
linearly
dependent
on
any
interval
(
a,
b
)
containing
0.
Why
doesn’t
this
contradict
theorem
3b,
p.
116
in
your
book?
2A7.
Let
y
1
and
y
2
be
two
solutions
of
y
��
+
p
(
x
)
y
+
q
(
x
)
y
=
0.
dW
a)
Prove
that
=
−
p
(
x
)
W
,
where
W
=
W
(
y
1
,
y
2
),
the
Wronskian.
dx
b)
Prove
that
if
p
(
x
)
=
0,
then
W
(
y
1
,
y
2
)
is
always
a
constant.
c)
Verify
(b)
by
direct
calculation
for
y
��
+
k
2
y
= 0
, k
=
0,
whose
general
solution
is
⇒
y
1
=
c
1
sin
kx
+
c
2
cos
kx
.
2B.
Reduction
of
Order
2B1.
Find
a
second
solution
y
2
to
y
��
−
2
y
+
y
=
0,
given
that
one
solution
is
y
1
=
e
x
,
by
three
methods:
a)
putting
y
2
=
ue
x
and
determining
u
(
x
)
by
substituting
into
the
ODE;
b)
determining
W
(
y
1
,
y
2
)
using
Exercise
2A7a,
and
from
this
getting
y
2
;
c)
by
using
the
general
formula
y
2
=
y
1
1
2
e
−
p
dx
dx
.
This is the end of the preview.
Sign up
to
access the rest of the document.
 Fall '09
 vogan
 Ode, Complex number, IVP, general solution

Click to edit the document details