Hint:
For the first part write as
y
′
−
ay
= 0 and multiply both sides by
e
−
at
.
Then explain
why you get
d
dt
(
e
−
at
y
(
t
)
)
= 0
, y
(
t
0
) = 0
.
Now you finish the argument. To show uniqueness in the second part, suppose
y
′
= (
a
+
ib
)
y, y
(0) = 0
and verify this requires
y
(
t
) = 0. To do this, note
y
′
= (
a
−
ib
)
y,
y
(0) = 0
and that
d
dt

y
(
t
)

2
=
y
′
(
t
)
y
(
t
) +
y
′
(
t
)
y
(
t
)
=
(
a
+
ib
)
y
(
t
)
y
(
t
) + (
a
−
ib
)
y
(
t
)
y
(
t
)
=
2
a

y
(
t
)

2
,

y

2
(
t
0
) = 0
Thus from the first part

y
(
t
)

2
= 0
e
−
2
at
= 0
.
Finally observe by a simple computation that
(16.8) is solved by (16.9). For the last part, write the equation as
y
′
−
ay
=
f
and multiply both sides by
e
−
at
and then integrate from
t
0
to
t
using the initial condition.
24.
↑
Now consider
A
an
n
×
n
matrix. By Schur’s theorem there exists unitary
Q
such that
Q
−
1
AQ
=
T
where
T
is upper triangular. Now consider the first order initial value problem
x
′
=
A
x
,
x
(
t
0
) =
x
0
.
Show there exists a unique solution to this first order system.
Hint:
Let
y
=
Q
−
1
x
and so
the system becomes
y
′
=
T
y
,
y
(
t
0
) =
Q
−
1
x
0
(16.10)
Now letting
y
= (
y
1
,
· · ·
, y
n
)
T
,
the bottom equation becomes
y
′
n
=
t
nn
y
n
, y
n
(
t
0
) =
(
Q
−
1
x
0
)
n
.
Then use the solution you get in this to get the solution to the initial value problem which
occurs one level up, namely
y
′
n
−
1
=
t
(
n
−
1)(
n
−
1)
y
n
−
1
+
t
(
n
−
1)
n
y
n
, y
n
−
1
(
t
0
) =
(
Q
−
1
x
0
)
n
−
1
Continue doing this to obtain a unique solution to (16.10).
Saylor URL:
The Saylor Foundation
16.4.
EXERCISES
349
25.
↑
Now suppose Φ (
t
) is an
n
×
n
matrix of the form
Φ (
t
) =
(
x
1
(
t
)
· · ·
x
n
(
t
)
)
(16.11)
where
x
′
k
(
t
) =
A
x
k
(
t
)
.
Explain why
Φ
′
(
t
) =
A
Φ (
t
)
if and only if Φ (
t
) is given in the form of (16.11). Also explain why if
c
∈
F
n
,
y
(
t
)
≡
Φ (
t
)
c
solves the equation
y
′
(
t
) =
A
y
(
t
)
.
26.
↑
In the above problem, consider the question whether all solutions to
x
′
=
A
x
(16.12)
are obtained in the form Φ (
t
)
c
for some choice of
c
∈
F
n
.
In other words, is the general
solution to this equation Φ (
t
)
c
for
c
∈
F
n
? Prove the following theorem using linear algebra.
Theorem 16.4.1
Suppose
Φ (
t
)
is an
n
×
n
matrix which satisfies
Φ
′
(
t
) =
A
Φ (
t
)
.
Then the general solution to (16.12) is
Φ (
t
)
c
if and only if
Φ (
t
)
−
1
exists for some
t.
Fur
thermore, if
Φ
′
(
t
) =
A
Φ (
t
)
,
then either
Φ (
t
)
−
1
exists for all
t
or
Φ (
t
)
−
1
never exists for any
t.
(det (Φ (
t
)) is called the Wronskian and this theorem is sometimes called the Wronskian alter
native.)
Hint:
Suppose first the general solution is of the form Φ (
t
)
c
where
c
is an arbitrary constant
vector in
F
n
. You need to verify Φ (
t
)
−
1
exists for some
t
. In fact, show Φ (
t
)
−
1
exists for every
t
. Suppose then that Φ (
t
0
)
−
1
does not exist. Explain why there exists
c
∈
F
n
such that there
is no solution
x
to
c
= Φ (
t
0
)
x
By the existence part of Problem 24 there exists a solution to
x
′
=
A
x
,
x
(
t
0
) =
c
but this cannot be in the form Φ (
t
)
c
. Thus for every
t
, Φ (
t
)
−
1
exists. Next suppose for some
t
0
,
Φ (
t
0
)
−
1
exists. Let
z
′
=
A
z
and choose
c
such that
z
(
t
0
) = Φ (
t
0
)
c
Then both
z
(
t
)
,
Φ (
t
)
c
solve
x
′
=
A
x
,
x
(
t
0
) =
z
(
t
0
)
Apply uniqueness to conclude
z
= Φ (
t
)
c
. Finally, consider that Φ (
t
)
c
for