2.7.
2.7. Homogeneous Linear Diﬀerential Equations with Con
stant Coeﬃcients
1.
(a) T (p.137 corollary to Theorem 2.32)
(b) T (p.132 Theorem 2.28)
(c) F
(d) F (Any solution is a linear combination of
e
at
and
t
k
e
at
)
(e) T
(
∵
) If
x
and
y
are solutions of
p
(
D
) = 0,
then
p
(
D
)(
αx
+
βy
) =
αp
(
D
)
x
+
βp
(
D
)
y
= 0 + 0 = 0
,α,β
∈
F
∴
αx
+
is a solution of
p
(
D
) = 0
(f) F
(
∵
) It’s diﬀerent with the multiplicity of
c
i
(p.137 and 139, Theorem 2.33 and
2.34) (g) T (p.131)
2.
(a) F
Let
S
=
{
a
1+
t
2

a
∈
R
} ⇒
S
: 1dimensional subspace of
C
∞
But there is no homogeneous linear diﬀerential equation with constant coeﬃcients
(b) F
Let
{
t,t
2
}
is the solution of
y
00
+
ay
0
+
by
= 0
0 +
a
+
bt
= 0
⇒
a
=
b
= 0
140
PNUMATH
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document2.7.
then
y
00
+
ay
0
+
by
= 0 becomes
y
00
= 0
(
t
2
)
00
= 2 = 0
(cf)
y
000
= 0
⇒
D
3
= 0
⇒
t
= 0
e
0
t
,te
0
t
,t
2
e
0
t
i.e.
1
,t,t
2
∃
y
000
= 0
(c) T
Let
x
is a solution to the homogeneous linear diﬀerential equation with constant
coeﬃcients
P
(
D
)
y
= 0
Since
P
(
D
)
x
= 0
, P
(
D
)
x
0
=
P
(
D
)(
Dx
) =
P
(
D
)
Dx
=
DP
(
D
)
x
=
D
(0) = 0
∴
x
0
is also a solution to the equation
(d)T
Let
p
(
D
)
x
= 0 and
q
(
D
)
y
= 0
p
(
D
)
q
(
D
)(
x
+
y
) =
p
(
D
)
q
(
D
)
x
+
p
(
D
)
q
(
D
)
y
=
q
(
D
This is the end of the preview.
Sign up
to
access the rest of the document.
 Spring '09
 RUDYAK
 Derivative, 1 g, linear operator, Auxiliary Polynomial

Click to edit the document details