Chapter
3.1,
Problem
2P
Stepbystep
solution
Step1072
Consider
the
functions
¢*
¢
*
¢**
and
the
ODE
y™—2)"—p'+2y=0
...
1)
The
objective
is
to
show
that
the
functions
¢*,¢*,
¢
are
solutions
and
form
a
basis
on
any
interval
by
using
Wronskians.
First,
show
that
the
given
functions
are
solutions
of
equation
(1).
In
the
equation
(1),
the
ODE
is
a
third
order
homogeneous
linear
ODE
with
constant
coefficients.
Therefore,
the
corresponding
characteristic
equation
of
(1),
is
given
as
follows:
A=227242=0
By
inspection
2
=1
is
a
root
of
the
characteristic
equation.
Since
*
—
2(1)'
~1+2=0
Hence,
(A1)
is
a
factor,
so
divide
;*
3%
7
+2
by
2]
tofind
the
other
roots.
A22
AP
20222
(2'+2%)
A=2
(2*+2)
22+2
(24+2)
0
Thus,
2'22"2+2=(A1)(A*22)=0
Theroots
of
;2
_,_2

are
A=1,2.
By
inspection.
Therefore,
the
three
roots
of
33
2,2
_j4+2=0
are
A=1,~land2.
The
roots
are
real
and
distinct
hence,
the
solution
of
(1)
is
l
y=ce'
+ee
+c
e
Therefore,
the
functions
¢*
¢
¢**
are
solutions
of
ODE
(1).
s
Step
2072
The
ODE
has
continuous
coefficients,
so
apply
the
following
stated
theorem:
Theorem
:
Letthe
ODE
y'"
+
p,
(x)y""
"
+..+
p,(x)
'+
p,
(x)
y
=0
...
(2)
have
the
coefficients
p
(x)
....
p,,
(x)
on an
open
interval
I.
Then
»
solutions
of
(2)
on

are
linearly
dependent
on

if
and
only
if
their
Wronskian
is
zero
for
some
x
=
x,
inl.
Furthermore,
if
there
is
an
x,
in

at
which
Wronskian
is
not
zero,
then
y,
.....
,y,
are
linearly
independent
on
I,
so
that
they
form
a
basis
of
solutions
of
(2)
on
I.
Now,
to
apply
this
theorem,
compute
the
Wronskian
of
the
functions
¢*,¢™*,¢**
as
follows:
Let
y,
=¢',y,=¢"and
y,
=¢™
Thus,
yl'
=e",
y"
=e",
y:'
=—e
",
yz'
=e
",
y;
=2¢**
and
y;
=
4™
N
V2
Vs
W
(32
05)
=91
25
2
wov:on
Substituting
the
values
to
get
the
following:
o
et
o
x
x
2\
_
x_
_x
52
Wle',e
e
)=le"—e
"
2e
o
et
4et
x
2x
I
e'
e
=e*[l
—e
™
2¢
Factorout
.
from
first
column.
1
e
4¢™
1
1é™
=e'e*l
1
2¢*
Factorout
.+
from
second
column.
11
4e™
I
11
=e
l
1
2
Factor
out
2+
from
third
column.
1
14
=e*
[l(—4
2)1(42)+1(1+
l)]
Expand
the
determinant.
=—be"
Exponential
function
never
zero.
#0
forall
x
Hence,
by
above
stated
theorem,
the
functions
¢*
¢
*,¢**
are
linearly
independent
and
so
they
form
a
basis
of
solutions
of
the
ODE
y"—2y"—y'+2y=0.