2350 Solution 7
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Nov 24, 2024
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MATH2350
Applied Linear Algebra and Differential Equations
Problem Set 7
L1, L2 (Fall 2023)
Page
1
of 13
Solution to Problem Set 7
1.
(a)
By definition, the Laplace transform of
?(?) = cosh 3?
is given by the improper integral
?(?) = ∫
?
−??
cosh 3? ??
+∞
0
= ∫
?
−??
?
3?
+ ?
−3?
2
??
+∞
0
=
1
2
∫
(?
(3−?)?
+ ?
−(3+?)?
)??
+∞
0
=
1
2
[
?
(3−?)?
3 − ?
−
?
−(3+?)?
3 + ?
]
0
+∞
=
1
2(? − 3)
+
1
2(? + 3)
for every
? > 3
.
(b)
By definition, the Laplace transform of
?(?) = ?
?
cos 2?
is given by the improper integral
?(?) = ∫
?
−??
?
?
cos 2? ??
+∞
0
= ∫
?
(1−?)?
cos 2? ??
+∞
0
= [?
(1−?)?
sin 2?
2
]
0
+∞
− ∫
sin 2?
2
(1 − ?)?
(1−?)?
??
+∞
0
=
? − 1
2
∫
?
(1−?)?
sin 2? ??
+∞
0
=
? − 1
2
([?
(1−?)?
− cos 2?
2
]
0
+∞
− ∫
− cos 2?
2
(1 − ?)?
(1−?)?
??
+∞
0
)
=
? − 1
2
(
1
2
+
1 − ?
2
∫
?
(1−?)?
cos 2? ??
+∞
0
) =
? − 1
4
−
(? − 1)
2
4
?(?)
for every
? > 1
.
Therefore
?(?) =
? − 1
4
1 +
(? − 1)
2
4
=
? − 1
(? − 1)
2
+ 4
for every
? > 1
.
Alternative solution
:
By definition, the Laplace transform of
?(?) = ?
?
cos 2?
is given by the improper
integral
?(?) = ∫
?
−??
?
?
cos 2? ??
+∞
0
= ∫
?
(1−?)?
?
2𝑖?
+ ?
−2𝑖?
2
??
+∞
0
=
1
2
∫
(?
(1−?+2𝑖)?
+ ?
(1−?−2𝑖)?
)??
+∞
0
=
1
2
[
?
(1−?+2𝑖)?
1 − ? + 2𝑖
+
?
(1−?−2𝑖)?
1 − ? − 2𝑖
]
0
+∞
=
1
2
lim
𝑅→+∞
(
?
(1−?+2𝑖)𝑅
1 − ? + 2𝑖
+
?
(1−?−2𝑖)𝑅
1 − ? − 2𝑖
) −
2(1 − ?)
(1 − ?)
2
+ 4
=
? − 1
(? − 1)
2
+ 4
for every
? > 1
.
MATH2350
Applied Linear Algebra and Differential Equations
Problem Set 7
L1, L2 (Fall 2023)
Page
2
of 13
(c)
By definition, the Laplace transform of
?(?) = {
1
if
1 ≤ ? ≤ 2
0
otherwise
is given by the improper integral
?(?) = ∫
?
−??
?(?)??
+∞
0
= ∫ ?
−??
1??
2
1
= [
?
−??
−?
]
1
2
=
?
−?
− ?
−2?
?
for every
? ≠ 0
, and
?(0) = 1
.
2.
(a)
By definition, if
𝑎 > −1
, then the Laplace transform of
?(?) = ?
𝑎
is given by the improper integral
?(?) = ∫
?
−??
?
𝑎
??
+∞
0
.
For each
? > 0
, if we let
? = ??
, then
?? = ???
and the integral becomes
?(?) = ∫
?
−??
?
𝑎
??
+∞
0
=
1
?
∫
?
−?
(
?
?
)
𝑎
??
+∞
0
=
1
?
𝑎+1
∫
?
−?
?
𝑎
??
+∞
0
=
Γ(𝑎 + 1)
?
𝑎+1
.
(b)
The Laplace transform of
?
1/2
is given by
ℒ{?
1/2
} =
Γ(1/2 + 1)
?
1/2+1
=
1
?
3/2
∫
?
−?
?
1/2
??
+∞
0
.
Let
? = ?
1/2
, then
? = ?
2
and
?? = 2???
, so
ℒ{?
1/2
} =
1
?
3/2
∫
?
−𝑥
2
?(2?)??
+∞
0
=
1
?
3/2
([−??
−𝑥
2
]
0
+∞
+ ∫
?
−𝑥
2
??
+∞
0
)
=
1
?
3/2
(0 +
√𝜋
2
) =
√𝜋
2?
3/2
for every
? > 0
.
The Laplace transform of
?
−1/2
is given by
ℒ{?
−1/2
} =
Γ(−1/2 + 1)
?
−1/2+1
=
1
?
1/2
∫
?
−?
?
−1/2
??
+∞
0
.
Let
? = ?
1/2
, then
? = ?
2
and
?? = 2???
, so
ℒ{?
−1/2
} =
1
?
1/2
∫
?
−𝑥
2
1
?
(2?)??
+∞
0
=
2
?
1/2
∫
?
−𝑥
2
??
+∞
0
=
2
?
1/2
⋅
√𝜋
2
=
√𝜋
√?
for every
? > 0
.
Alternatively
:
Since
?
−1/2
= 2
?
??
?
1/2
, the Laplace transform of
?
−1/2
is given by
ℒ{?
−1/2
} = 2ℒ {
?
??
?
1/2
} = 2(?ℒ{?
1/2
} − 0
1/2
) = 2 (? ⋅
√𝜋
2?
3/2
− 0) =
√𝜋
√?
for every
? > 0
.
MATH2350
Applied Linear Algebra and Differential Equations
Problem Set 7
L1, L2 (Fall 2023)
Page
3
of 13
3.
(a)
Let
?(?)
be the solution to the initial value problem, and denote its Laplace transform by
ℒ{?(?)} = ?(?)
.
Taking Laplace transforms on both sides of the given ODE, we have
ℒ{?
′′
} + 3ℒ{?
′
} + 2ℒ{?} = 4ℒ{?}
, i.e.
[?
2
?(?) − ??(0) − ?
′
(0)] + 3[??(?) − ?(0)] + 2?(?) =
4
?
2
.
Since
?(0) = 0
and
?
′
(0) = 8
, we have
(?
2
+ 3? + 2)?(?) − 8 =
4
?
2
.
So
?(?) =
8?
2
+ 4
?
2
(?
2
+ 3? + 2)
=
8?
2
+ 4
?
2
(? + 1)(? + 2)
.
To decompose
?(?)
into partial fractions, we suppose that
?(?) =
8?
2
+ 4
?
2
(? + 1)(? + 2)
=
?
?
+
?
?
2
+
?
? + 1
+
?
? + 2
and obtain an identity
??(? + 1)(? + 2) + ?(? + 1)(? + 2) + ??
2
(? + 2) + ??
2
(? + 1) = 8?
2
+ 4.
Putting
? = 0
,
? = −1
and
? = −2
into the identity, we get
? = 2
,
? = 12
and
? = −9
respectively.
Comparing the coefficient of
?
3
, we get
? + ? + ? = 0
, so
? = −3
.
Therefore,
?(?) = −
3
?
+
2
?
2
+
12
? + 1
−
9
? + 2
.
Now taking inverse Laplace transforms, we obtain the solution
?(?) = ℒ
−1
{?(?)} = −3 + 2? + 12?
−?
− 9?
−2?
.
(b)
Let
?(?)
be the solution to the initial value problem, and denote
?(?) = ℒ{?(?)}
.
Taking Laplace
transforms on both sides of the given ODE, we get
ℒ{?
(4)
} − 4ℒ{?} = ℒ{0}
, i.e.
[?
4
?(?) − ?
3
?(0) − ?
2
?
′
(0) − ??
′′
(0) − ?
′′′
(0)] − 4?(?) = 0.
Since
?(0) = 1
,
?
′
(0) = 0
,
?
′′
(0) = −2
and
?
′′′
(0) = 0
, we have
(?
4
− 4)? − ?
3
+ 2? = 0
.
So
?(?) =
?
3
− 2?
?
4
− 4
=
?
?
2
+ 2
.
Now taking inverse Laplace transforms, we obtain the solution
?(?) = ℒ
−1
{?(?)} = ℒ
−1
{
?
?
2
+ 2
} = cos √2
?.
(c)
Let
?(?)
be the solution to the initial value problem, and denote
?(?) = ℒ{?(?)}
.
Taking Laplace
transforms on both sides of the given ODE, we get
ℒ{?
′′
} + ?
2
ℒ{?} = ℒ{cos 2?}
, i.e.
[?
2
?(?) − ??(0) − ?
′
(0)] + ?
2
?(?) =
?
?
2
+ 4
.
Since
?(0) = 1
and
?
′
(0) = 0
, we have
(?
2
+ ?
2
)?(?) − ? =
?
?
2
+4
.
So
?(?) =
1
?
2
+ ?
2
(?
+
?
?
2
+ 4
) =
?
?
2
+ ?
2
+
1
?
2
− 4
(
?
?
2
+ 4
−
?
?
2
+ ?
2
)
.
Now taking inverse Laplace transforms, we obtain the solution
?(?) = ℒ
−1
{?(?)} = cos ??
+
1
?
2
− 4
(cos 2? − cos ??)
=
1
?
2
− 4
cos 2? +
?
2
− 5
?
2
− 4
cos ??.
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MATH2350
Applied Linear Algebra and Differential Equations
Problem Set 7
L1, L2 (Fall 2023)
Page
4
of 13
4.
By definition, the Laplace transform of the function
?
𝑎?
?(?)
is given by the improper integral
ℒ{?
𝑎?
?(?)} = ∫
?
−??
?
𝑎?
?(?)??
+∞
0
= ∫
?
−(?−𝑎)?
?(?)??
+∞
0
= ?(? − 𝑎)
for every
? > 𝑎
.
(a)
The inverse Laplace transform of
?(?)
is given by
ℒ
−1
{?(?)} = ℒ
−1
{
12
(? − 1)
3
} = 2ℒ
−1
{
3!
(? − 1)
3
} = 2?
3
?
?
.
(b)
The inverse Laplace transform of
?(?)
is given by
ℒ
−1
{?(?)} = ℒ
−1
{
1 − 2?
?
2
+ 4? + 5
} = ℒ
−1
{
−2(? + 2) + 5
(? + 2)
2
+ 1
}
= −2ℒ
−1
{
? + 2
(? + 2)
2
+ 1
} + 5ℒ
−1
{
1
(? + 2)
2
+ 1
}
= −2?
−2?
cos ? + 5?
−2?
sin ?.
(c)
First note that
?(?) =
(2? − 4)?
−?
?
2
− 4? + 3
= ?
−?
2? − 4
(? − 1)(? − 3)
= ?
−?
(
1
? − 1
+
1
? − 3
).
Now let
?(?)
be the inverse Laplace transform of the last factor, i.e.
?(?) = ℒ
−1
{
1
? − 1
+
1
? − 3
} = ?
?
+ ?
3?
.
Then the inverse Laplace transform of
?(?)
is given by
ℒ
−1
{?(?)} = ?
1
(?)?(? − 1) = {
0
if
0 ≤ ? < 1
?
?−1
+ ?
3(?−1)
if
? ≥ 1
.
5.
Let
?: [0, +∞) → ℝ
be the function
?(?) = ∫ ?(?)??
?
0
.
Then
?
is differentiable on
(0, +∞)
with
?
′
(?) = ?(?)
according to the Fundamental Theorem of Calculus.
So,
ℒ{?(?)} = ℒ{?
′
(?)} = ?ℒ{?(?)} − ?(0) = ?ℒ{?(?)}
for
? > 0
.
Therefore,
ℒ {∫ ?(?)??
?
0
} = ℒ{?(?)} =
1
?
ℒ{?(?)}
for
? > 0
.
∎
MATH2350
Applied Linear Algebra and Differential Equations
Problem Set 7
L1, L2 (Fall 2023)
Page
5
of 13
6.
(a)
By definition of Laplace transform, we have
?(?) = ∫
?
−??
?(?)??
+∞
0
.
Differentiating both sides with respect to
?
, we have
?
′
(?) =
?
??
∫
?
−??
?(?)??
+∞
0
= ∫
?
??
(?
−??
?(?))??
+∞
0
= ∫
−??
−??
?(?)??
+∞
0
= ℒ{−??(?)}.
(b)
The derivative of
?
is given by
?
′
(?) =
?
??
arctan
3
? + 2
=
1
1 + (
3
? + 2
)
2
⋅
−3
(? + 2)
2
= −
3
(? + 2)
2
+ 9
.
Now let
?(?) = ℒ
−1
{?(?)}
.
According to the result from (a), we have
−??(?) = ℒ
−1
{?
′
(?)} = −?
−2?
sin 3? .
Therefore
?(?) = ?
−1
?
−2?
sin 3?.
(c)
According to the result from (a), we have
ℒ{? sin 𝑎?} = −
?
??
ℒ{sin 𝑎?} = −
?
??
𝑎
?
2
+ 𝑎
2
= − (−
𝑎
(?
2
+ 𝑎
2
)
2
⋅ 2?) =
2𝑎?
(?
2
+ 𝑎
2
)
2
.
(d)
According to the result from (a), we have
ℒ{?
3
sin ?} = (−1)
3
?
3
??
3
ℒ{sin ?} = −
?
3
??
3
1
?
2
+ 1
=
24?
3
− 24?
(?
2
+ 1)
4
and
ℒ{??
3?
cos 2?} = −
?
??
ℒ{?
3?
cos 2?} = −
?
??
? − 3
(? − 3)
2
+ 2
2
=
?
2
− 6? + 5
((? − 3)
2
+ 4)
2
.
7.
The given function
?
can be written as
?(?) = 1 + ?
𝜋
2
(?)(sin ? − 1) + ?
3𝜋
(?)((? − 3𝜋)?
−?
− sin ?)
= 1 + ?
𝜋
2
(?) (sin (? −
𝜋
2
+
𝜋
2
) − 1) + ?
3𝜋
(?) ((? − 3𝜋)?
−(?−3𝜋+3𝜋)
− sin(? − 3𝜋 + 3𝜋))
= 1 + ?
𝜋
2
(?) (cos (? −
𝜋
2
) − 1) + ?
3𝜋
(?)(?
−3𝜋
(? − 3𝜋)?
−(?−3𝜋)
+ sin(? − 3𝜋)),
so its Laplace transform is
?(?) = ℒ{1} + ℒ {?
𝜋
2
(?) (cos (? −
𝜋
2
) − 1)} + ℒ{?
3𝜋
(?)(?
−3𝜋
(? − 3𝜋)?
−(?−3𝜋)
+ sin(? − 3𝜋))}
= ℒ{1} + ?
−
𝜋
2
?
ℒ{cos ? − 1} + ?
−3𝜋?
ℒ{?
−3𝜋
??
−?
+ sin ?}
=
1
?
+ ?
−
𝜋
2
?
(
?
?
2
+ 1
−
1
?
) + ?
−3𝜋?
(
?
−3𝜋
(?
+
1)
2
+
1
?
2
+ 1
)
for
? >
0
.
MATH2350
Applied Linear Algebra and Differential Equations
Problem Set 7
L1, L2 (Fall 2023)
Page
6
of 13
8.
(a)
Let
?(?)
be the solution to the initial value problem, and denote its Laplace transform by
ℒ{?(?)} = ?(?)
.
Taking Laplace transforms on both sides of the given ODE, we have
ℒ{?
(4)
} + 2ℒ{?
′′
} + ℒ{?} = 50ℒ{?
2?
}
, i.e.
[?
4
?(?) − ?
3
?(0) − ?
2
?
′
(0) − ??
′′
(0) − ?
′′′
(0)] + 2[?
2
?(?) − ??(0) − ?
′
(0)] + ?(?) =
50
? − 2
.
Since
?(0) = ?
′
(0) = ?
′′
(0) = ?
′′′
(0) = 0
, we have
(?
4
+ 2?
2
+ 1)?(?) =
50
?−2
.
So
?(?) =
50
(? − 2)(?
2
+ 1)
2
.
To decompose
?(?)
into partial fractions, we suppose that
?(?) =
1
(? − 2)(?
2
+ 1)
2
=
?
? − 2
+
?? + ?
?
2
+ 1
+
?? + ?
(?
2
+ 1)
2
and obtain an identity
?(?
2
+ 1)
2
+ (?? + ?)(? − 2)(?
2
+ 1) + (?? + ?)(? − 2) = 50.
Putting
? = 2
, we get
? = 2
.
Putting
? = 𝑖
, we get
(−? − 2?) + (? − 2?)𝑖 = 50
, so comparing the real and imaginary parts we get
? = −10
and
? = −20
.
Expanding the identity we have
(? + ?)?
4
+ (? − 2?)?
3
+ (2? + ? − 2? + ?)?
2
+ (? − 2? + ? − 2?)? + (? − 2? − 2?) = 50,
so comparing the coefficients of
?
4
and
?
3
on both sides, we get
? = −2
and
? = −4
.
Therefore,
?(?) =
2
? − 2
−
2?
?
2
+ 1
−
4
?
2
+ 1
−
10?
(?
2
+ 1)
2
−
20
(?
2
+ 1)
2
=
2
? − 2
−
2?
?
2
+ 1
−
4
?
2
+ 1
− 5 ⋅
2?
(?
2
+ 1)
2
+ 10 (
?
2
− 1
(?
2
+ 1)
2
−
?
2
+ 1
(?
2
+ 1)
2
)
=
2
? − 2
−
2?
?
2
+ 1
−
14
?
2
+ 1
− 5 ⋅
2?
(?
2
+ 1)
2
+ 10
?
2
− 1
(?
2
+ 1)
2
.
Now taking inverse Laplace transforms, we obtain the solution
?(?) = ℒ
−1
{?(?)}
= 2ℒ
−1
{
1
? − 2
} − 2ℒ
−1
{
?
?
2
+ 1
} − 14ℒ
−1
{
1
?
2
+ 1
} − 5ℒ
−1
{
2?
(?
2
+ 1)
2
} + 10ℒ
−1
{
?
2
− 1
(?
2
+ 1)
2
}
= 2?
2?
− 2 cos ? − 14 sin ? − 5? sin ? + 10? cos ?.
(b)
The non
-
homogeneous term
?(?)
on the right
-
hand side of the given ODE can be rewritten as
?(?) = ? + ?
2𝜋
(?)(2𝜋 cos ? − ?) = ? + ?
2𝜋
(?)[2𝜋 cos(? − 2𝜋) − (? − 2𝜋) − 2𝜋].
Let
?(?)
be the solution to the initial value problem, and denote its Laplace transform by
ℒ{?(?)} = ?(?)
.
Taking Laplace transforms on both sides of the given ODE, we have
ℒ{?
′′
} + ℒ{?} = ℒ{?} + ℒ{?
2𝜋
(?)[2𝜋 cos(? − 2𝜋) − (? − 2𝜋) − 2𝜋]},
i.e.
[?
2
?(?) − ??(0) − ?
′
(0)] + ?(?) =
1
?
2
+ ?
−2𝜋?
(
2𝜋?
?
2
+ 1
−
1
?
2
−
2𝜋
?
) .
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MATH2350
Applied Linear Algebra and Differential Equations
Problem Set 7
L1, L2 (Fall 2023)
Page
7
of 13
Since
?(0) = 0
and
?
′
(0) = 0
, we have
?(?) =
1
?
2
(?
2
+ 1)
+ ?
−2𝜋?
[
2𝜋?
(?
2
+ 1)
2
−
1
?
2
(?
2
+ 1)
−
2𝜋
?(?
2
+ 1)
]
= (
1
?
2
−
1
?
2
+ 1
) + ?
−2𝜋?
[𝜋 ⋅
2?
(?
2
+ 1)
2
− (
1
?
2
−
1
?
2
+ 1
) − 2𝜋 (
1
?
−
?
?
2
+ 1
)] .
Now let
ℎ(?)
be the inverse Laplace transform of the last factor, i.e.
?(?) = ℒ
−1
{𝜋 ⋅
2?
(?
2
+ 1)
2
− (
1
?
2
−
1
?
2
+ 1
) − 2𝜋 (
1
?
−
?
?
2
+ 1
)} = 𝜋? sin ? − (? − sin ?) − 2𝜋(1 − cos ?).
Then the solution to the initial value problem is given by
?(?) = ℒ
−1
{?(?)} = (? − sin ?) + ?
2𝜋
(?)?(? − 2𝜋)
= (? − sin ?) + ?
2𝜋
(?)[𝜋(? − 2𝜋) sin ? − (? − 2𝜋 − sin ?) − 2𝜋(1 − cos ?)]
= (? − sin ?) + ?
2𝜋
(?)[𝜋(? − 2𝜋) sin ? − (? − sin ?) + 2𝜋 cos ?]
= {
? − sin ?
if
0 ≤ ? < 2𝜋
𝜋(? − 2𝜋) sin ? + 2𝜋 cos ?
if
? ≥ 2𝜋
.
(c)
Let
?(?)
be the solution to the initial value problem, and denote its Laplace transform by
ℒ{?(?)} = ?(?)
.
Taking Laplace transforms on both sides of the given ODE, we have
ℒ{?
′′
} + 2ℒ{?
′
} + 3ℒ{?} = 4ℒ{sin ?} + ℒ{𝛿(? − 3𝜋)},
i.e.
[?
2
?(?) − ??(0) − ?
′
(0)] + 2[??(?) − ?(0)] + 3?(?) =
4
?
2
+ 1
+ ?
−3𝜋?
.
Since
?(0) = ?
′
(0) = 0
, we have
?(?) =
4
(?
2
+ 1)(?
2
+ 2? + 3)
+
?
−3𝜋?
?
2
+ 2? + 3
=
1
?
2
+ 1
−
?
?
2
+ 1
+
? + 1
(? + 1)
2
+ 2
+
?
−3𝜋?
(? + 1)
2
+ 2
.
Now taking inverse Laplace transforms, we obtain the solution
?(?) = ℒ
−1
{?(?)} = sin ? − cos ? + ?
−?
cos √2
? +
1
√2
?
3𝜋
(?)?
−(?−3𝜋)
sin (√2
(? − 3𝜋)).
(d)
Note that
𝛿(? − 2𝜋) cos ?
is just the same as
𝛿(? − 2𝜋)
.
Let
?(?)
be the solution to the initial value
problem, and denote its Laplace transform by
ℒ{?(?)} = ?(?)
.
Taking Laplace transforms on both sides of
the given ODE, we have
ℒ{?
′′
} + ℒ{?} = ℒ{𝛿(? − 2𝜋)}
, i.e.
[?
2
?(?) − ??(0) − ?
′
(0)] + ?(?) = ?
−2𝜋?
.
Since
?(0) = 0
and
?
′
(0) = 1
, we have
?(?) =
1
?
2
+ 1
+
?
−2𝜋?
?
2
+ 1
.
Now taking inverse Laplace transforms, we obtain the solution
?(?) = ℒ
−1
{?(?)} = sin ? + ?
2𝜋
(?) sin(? − 2𝜋) = {
sin ?
if
? < 2𝜋
2 sin ?
if
? ≥ 2𝜋
.
MATH2350
Applied Linear Algebra and Differential Equations
Problem Set 7
L1, L2 (Fall 2023)
Page
8
of 13
(e)
Let
?(?)
be the solution to the initial value problem, and denote its Laplace transform by
ℒ{?(?)} = ?(?)
.
Taking Laplace transforms on both sides of the given ODE, we have
ℒ{?
′′
} + ℒ{?} = ℒ {?
𝜋
2
(?)} + 3ℒ {𝛿 (? −
3𝜋
2
)} − ℒ{?
2𝜋
(?)},
i.e.
[?
2
?(?) − ??(0) − ?
′
(0)] + ?(?) =
?
−
𝜋
2
?
?
+ 3?
−
3𝜋
2
?
−
?
−2𝜋?
?
.
Since
?(0) = ?
′
(0) = 0
, we have
?(?) =
?
−
𝜋
2
?
?(?
2
+ 1)
+
3?
−
3𝜋
2
?
?
2
+ 1
−
?
−2𝜋?
?(?
2
+ 1)
= ?
−
𝜋
2
?
(
1
?
−
?
?
2
+ 1
) +
3?
−
3𝜋
2
?
?
2
+ 1
− ?
−2𝜋?
(
1
?
−
?
?
2
+ 1
).
Now taking inverse Laplace transforms, we obtain the solution
?(?) = ℒ
−1
{?(?)}
= ?
𝜋
2
(?) [1 − cos (? −
𝜋
2
)] + 3?
3𝜋
2
(?) sin (? −
3𝜋
2
) − ?
2𝜋
(?)[1 − cos(? −
2𝜋
)]
= ?
𝜋
2
(?)(1 − sin ?) + 3?
3𝜋
2
(?) cos ? − ?
2𝜋
(?)(1 −
cos ?
)
= {
0
if
? < 𝜋/2
1 − sin ?
if
𝜋/2 ≤ ? < 3𝜋/2
1 − sin ? + 3 cos ?
if
3𝜋/2 ≤ ? < 2𝜋
4 cos ? − sin ?
if
? ≥ 2𝜋
.
9.
(a)
Suppose that
? = ∑
?
𝑘
?
𝑘
+∞
𝑘=0
is a power series solution to the given ODE.
Then
?
′
= ∑ 𝑘?
𝑘
?
𝑘−1
+∞
𝑘=1
.
Substituting the series for
?
and
?
′
into the given ODE, we get
∑ 𝑘?
𝑘
?
𝑘−1
+∞
𝑘=1
= ∑ ?
𝑘
?
𝑘
+∞
𝑘=0
.
Shifting the indices of the first series we get
∑
(𝑘 + 1)?
𝑘+1
?
𝑘
+∞
𝑘=0
− ∑
?
𝑘
?
𝑘
+∞
𝑘=0
= 0
, and so
∑[(𝑘 + 1)?
𝑘+1
− ?
𝑘
]?
𝑘
+∞
𝑘=0
= 0.
Now comparing coefficients, we obtain
(𝑘 + 1)?
𝑘+1
− ?
𝑘
= 0
for every non-negative integer
𝑘,
i.e.
?
𝑘+1
=
?
𝑘
𝑘+1
for every non
-
negative integer
𝑘
.
Therefore
?
1
=
?
0
1
= ?
0
, ?
2
=
?
1
2
=
?
0
2!
, ?
3
=
?
2
3
=
?
0
3!
, …
and in general
?
𝑘
=
?
0
𝑘!
.
Therefore
? = ∑ ?
𝑘
?
𝑘
+∞
𝑘=0
= ?
0
∑
1
𝑘!
?
𝑘
+∞
𝑘=0
,
where
?
0
is an arbitrary coefficient.
MATH2350
Applied Linear Algebra and Differential Equations
Problem Set 7
L1, L2 (Fall 2023)
Page
9
of 13
(b)
Suppose that
? = ∑
?
𝑘
?
𝑘
+∞
𝑘=0
is a power series solution to the given ODE.
Then
?
′
= ∑ 𝑘?
𝑘
?
𝑘−1
+∞
𝑘=1
and
?
′′
= ∑ 𝑘(𝑘 − 1)?
𝑘
?
𝑘−2
+∞
𝑘=2
.
Substituting the series for
?
,
?
′
and
?
′
into the given ODE, we get
∑ 𝑘(𝑘 − 1)?
𝑘
?
𝑘−2
+∞
𝑘=2
− ∑ 𝑘?
𝑘
?
𝑘
+∞
𝑘=1
− ∑ ?
𝑘
?
𝑘
+∞
𝑘=0
= 0.
Shifting the indices of the first series we get
∑(𝑘 + 2)(𝑘 + 1)?
𝑘+2
?
𝑘
+∞
𝑘=0
− ∑ 𝑘?
𝑘
?
𝑘
+∞
𝑘=1
− ∑ ?
𝑘
?
𝑘
+∞
𝑘=0
= 0,
and so
(2)(1)?
2
− ?
0
+ ∑[(𝑘 + 2)(𝑘 + 1)?
𝑘+2
− 𝑘?
𝑘
− ?
𝑘
]?
𝑘
+∞
𝑘=1
= 0.
Now comparing coefficients, we obtain
?
2
=
?
0
2
and
(𝑘 + 2)(𝑘 + 1)?
𝑘+2
− (𝑘 + 1)?
𝑘
= 0
for every positive integer
𝑘,
i.e.
?
𝑘+2
=
?
𝑘
𝑘+2
for every non
-
negative integer
𝑘
.
Therefore
?
2
=
?
0
2
, ?
3
=
?
1
3
, ?
4
=
?
2
4
=
?
0
2 ⋅ 4
, ?
5
=
?
3
5
=
?
1
3 ⋅ 5
, ?
6
=
?
4
6
=
?
0
2 ⋅ 4 ⋅ 6
, …
and in general
?
2𝑚
=
?
0
2 ⋅ 4 ⋅ 6 ⋯ (2? − 2)(2?)
=
?
0
2
𝑚
?!
and
?
2𝑚+1
=
?
1
3 ⋅ 5 ⋅ 7 ⋯ (2? − 1)(2? + 1)
=
?
1
2
𝑚
?!
(2? + 1)!
for every non
-
negative integer
?
.
Therefore
? = ∑ ?
𝑘
?
𝑘
+∞
𝑘=0
=
?
0
∑
1
2 ⋅ 4 ⋅ 6 ⋯ (2? − 2)(2?)
?
2𝑚
+∞
𝑚=0
+ ?
1
∑
1
3 ⋅ 5 ⋅ 7 ⋯ (2? − 1)(2? + 1)
?
2𝑚+1
+∞
𝑚=0
= ?
0
∑
1
2
𝑚
?!
?
2𝑚
+∞
𝑚=0
+ ?
1
∑
2
𝑚
?!
(2? + 1)!
?
2𝑚+1
+∞
𝑚=0
,
where
?
0
and
?
1
are arbitrary coefficients.
(c)
Suppose that
? = ∑
?
𝑘
(? − 1)
𝑘
+∞
𝑘=0
is a power series solution to the given ODE.
Then
?
′
= ∑ 𝑘?
𝑘
(? − 1)
𝑘−1
+∞
𝑘=1
and
?
′′
= ∑ 𝑘(𝑘 − 1)?
𝑘
(? − 1)
𝑘−2
+∞
𝑘=2
.
Substituting the series for
?
,
?
′
and
?
′
into the given ODE, we get
(? − 1 + 1) ∑ 𝑘(𝑘 − 1)?
𝑘
(? − 1)
𝑘−2
+∞
𝑘=2
+ ∑ 𝑘?
𝑘
(? − 1)
𝑘−1
+∞
𝑘=1
+ (? − 1 + 1) ∑ ?
𝑘
(? − 1)
𝑘
+∞
𝑘=0
= 0,
i.e.
∑ 𝑘(𝑘 − 1)?
𝑘
(? − 1)
𝑘−1
+∞
𝑘=2
+ ∑ 𝑘(𝑘 − 1)?
𝑘
(? − 1)
𝑘−2
+∞
𝑘=2
+ ∑ 𝑘?
𝑘
(? − 1)
𝑘−1
+∞
𝑘=1
+ ∑ ?
𝑘
(? − 1)
𝑘+1
+∞
𝑘=0
+ ∑ ?
𝑘
(? − 1)
𝑘
+∞
𝑘=0
= 0.
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MATH2350
Applied Linear Algebra and Differential Equations
Problem Set 7
L1, L2 (Fall 2023)
Page
10
of 13
Shifting the indices we get
∑(𝑘 + 1)(𝑘)?
𝑘+1
(? − 1)
𝑘
+∞
𝑘=1
+ ∑(𝑘 + 2)(𝑘 + 1)?
𝑘+2
(? − 1)
𝑘
+∞
𝑘=0
+ ∑(𝑘 + 1)?
𝑘+1
(? − 1)
𝑘
+∞
𝑘=0
+ ∑ ?
𝑘−1
(? − 1)
𝑘
+∞
𝑘=1
+ ∑ ?
𝑘
(? − 1)
𝑘
+∞
𝑘=0
= 0.
and so
2?
2
+ ?
1
+ ?
0
+ ∑[(𝑘 + 1)(𝑘)?
𝑘+1
+ (𝑘 + 2)(𝑘 + 1)?
𝑘+2
+ (𝑘 + 1)?
𝑘+1
+ ?
𝑘−1
+ ?
𝑘
](? − 1)
𝑘
+∞
𝑘=1
= 0.
Now comparing coefficients, we obtain
?
2
= −
?
0
2
−
?
1
2
and
(𝑘 + 2)(𝑘 + 1)?
𝑘+2
+ (𝑘 + 1)
2
?
𝑘+1
+ ?
𝑘
+ ?
𝑘−1
= 0
for every positive integer
𝑘,
i.e.
?
𝑘+2
= −
?
𝑘−1
(𝑘+2)(𝑘+1)
−
?
𝑘
(𝑘+2)(𝑘+1)
−
(𝑘+1)?
𝑘+1
𝑘+2
for every non
-
negative integer
𝑘
.
Thus
?
3
= −
?
0
3 ⋅ 2
−
?
1
3 ⋅ 2
−
2?
2
3
=
?
0
3!
+
?
1
3!
, ?
4
= −
?
1
4 ⋅ 3
−
?
2
4 ⋅ 3
−
3?
3
4
= −
2?
0
4!
−
4?
1
4!
,
?
5
= −
?
2
5 ⋅ 4
−
?
3
5 ⋅ 4
−
4?
4
5
=
10?
0
5!
+
18?
1
5!
, …
Therefore
? = ∑ ?
𝑘
?
𝑘
+∞
𝑘=0
= ?
0
(1 −
1
2
(? − 1)
2
+
1
3!
(? − 1)
3
−
2
4!
(? − 1)
4
+
10
5!
(? − 1)
5
+ ⋯ )
+ ?
1
((? − 1) −
1
2
(? − 1)
2
+
1
3!
(? − 1)
3
−
4
4!
(? − 1)
4
+
18
5!
(? − 1)
5
+ ⋯ ) ,
where
?
0
and
?
1
are arbitrary coefficients.
10.
(a)
Suppose that
? = ∑
?
𝑘
?
𝑘
+∞
𝑘=0
is a power series solution to the given ODE.
Then
?
′
= ∑ 𝑘?
𝑘
?
𝑘−1
+∞
𝑘=1
and
?
′′
= ∑ 𝑘(𝑘 − 1)?
𝑘
?
𝑘−2
+∞
𝑘=2
.
Substituting the series for
?
,
?
′
and
?
′
into the given ODE, we get
∑ 𝑘(𝑘 − 1)?
𝑘
?
𝑘−2
+∞
𝑘=2
− ∑ 2𝑘?
𝑘
?
𝑘
+∞
𝑘=1
+ ∑ 𝜆?
𝑘
?
𝑘
+∞
𝑘=0
= 0.
Shifting the indices of the first series we get
∑(𝑘 + 2)(𝑘 + 1)?
𝑘+2
?
𝑘
+∞
𝑘=0
− ∑ 2𝑘?
𝑘
?
𝑘
+∞
𝑘=1
+ ∑ 𝜆?
𝑘
?
𝑘
+∞
𝑘=0
= 0,
and so
(2)(1)?
2
+ 𝜆?
0
+ ∑[(𝑘 + 2)(𝑘 + 1)?
𝑘+2
− 2𝑘?
𝑘
+ 𝜆?
𝑘
]?
𝑘
+∞
𝑘=1
= 0.
MATH2350
Applied Linear Algebra and Differential Equations
Problem Set 7
L1, L2 (Fall 2023)
Page
11
of 13
Now comparing coefficients, we obtain
?
2
= −
𝜆
2
?
0
and
(𝑘 + 2)(𝑘 + 1)?
𝑘+2
− (2𝑘 − 𝜆)?
𝑘
= 0
for every positive integer
𝑘,
i.e.
?
𝑘+2
=
2𝑘−𝜆
(𝑘+2)(𝑘+1)
?
𝑘
for every positive integer
𝑘
.
Therefore
?
2
= −
𝜆
2
?
0
, ?
3
=
2 − 𝜆
3 ⋅ 2
?
1
, ?
4
=
4 − 𝜆
4 ⋅ 3
?
2
=
(−𝜆)(4 − 𝜆)
4!
?
0
,
?
5
=
6 − 𝜆
5 ⋅ 4
?
3
=
(6 − 𝜆)(2 − 𝜆)
5!
?
1
, ?
6
=
8 − 𝜆
6 ⋅ 5
?
4
=
(−𝜆)(4 − 𝜆)(8 − 𝜆)
6!
?
0
, …
and in general
?
2𝑚
=
(−𝜆)(4 − 𝜆)(8 − 𝜆) ⋯ (4? − 4 − 𝜆)
(2?)!
?
0
and
?
2𝑚+1
=
(2 − 𝜆)(6 − 𝜆)(10 − 𝜆) ⋯ (4? − 2 − 𝜆)
(2? + 1)!
?
1
for every non
-
negative integer
?
.
Therefore
? = ∑ ?
𝑘
?
𝑘
+∞
𝑘=0
= ?
0
∑
(−𝜆)(4 − 𝜆)(8 − 𝜆) ⋯ (4? − 4 − 𝜆)
(2?)!
?
2𝑚
+∞
𝑚=0
+ ?
1
∑
(2 − 𝜆)(6 − 𝜆)(10 − 𝜆) ⋯ (4? − 2 − 𝜆)
(2? + 1)!
?
2𝑚+1
+∞
𝑚=0
,
where
?
0
and
?
1
are arbitrary coefficients.
(b)
When
? = 0
i.e.
𝜆 = 0
, the first series terminates and only the constant term survives.
So
𝐻
0
(?) = 1.
When
? = 1
i.e.
𝜆 = 2
, the second series terminates.
We have
𝐻
1
(?) = 2?.
When
? = 2
i.e.
𝜆 = 4
, the first series terminates.
We have
𝐻
2
(?) = 2
2
2!
−4
(1 +
−4
2!
?
2
) = 4?
2
− 2.
When
? = 3
i.e.
𝜆 = 6
, the second series terminates.
We have
𝐻
3
(?) = 2
3
3!
−4
(? +
−4
3!
?
3
) = 8?
3
− 12?.
When
? = 4
i.e.
𝜆 = 8
, the first series terminates.
We have
𝐻
4
(?) = 2
4
4!
(−8)(−4)
(1 +
−8
2!
?
2
+
(−8)(−4)
4!
?
4
) = 16?
4
− 48?
2
+ 12.
When
? = 5
i.e.
𝜆 = 10
, the second series terminates.
We have
𝐻
5
(?) = 2
5
5!
(−8)(−4)
(? +
−8
3!
?
3
+
(−8)(−4)
5!
?
5
) = 32?
5
− 160?
3
+ 120?.
MATH2350
Applied Linear Algebra and Differential Equations
Problem Set 7
L1, L2 (Fall 2023)
Page
12
of 13
11.
(a)
Suppose that
? = ∑
?
𝑘
?
𝑘
+∞
𝑘=0
is a power series solution to the given ODE.
Then
?
′
= ∑ 𝑘?
𝑘
?
𝑘−1
+∞
𝑘=1
and
?
′′
= ∑ 𝑘(𝑘 − 1)?
𝑘
?
𝑘−2
+∞
𝑘=2
.
Substituting the series for
?
,
?
′
and
?
′
into the given ODE, we get
(1 − ?
2
) ∑ 𝑘(𝑘 − 1)?
𝑘
?
𝑘−2
+∞
𝑘=2
− ∑ 𝑘?
𝑘
?
𝑘
+∞
𝑘=1
+ ∑ 𝛼
2
?
𝑘
?
𝑘
+∞
𝑘=0
= 0,
i.e.
∑ 𝑘(𝑘 − 1)?
𝑘
?
𝑘−2
+∞
𝑘=2
− ∑ 𝑘(𝑘 − 1)?
𝑘
?
𝑘
+∞
𝑘=2
− ∑ 𝑘?
𝑘
?
𝑘
+∞
𝑘=1
+ ∑ 𝛼
2
?
𝑘
?
𝑘
+∞
𝑘=0
= 0.
Shifting the indices of the first series we get
∑(𝑘 + 2)(𝑘 + 1)?
𝑘+2
?
𝑘
+∞
𝑘=0
− ∑ 𝑘(𝑘 − 1)?
𝑘
?
𝑘
+∞
𝑘=2
− ∑ 𝑘?
𝑘
?
𝑘
+∞
𝑘=1
+ ∑ 𝛼
2
?
𝑘
?
𝑘
+∞
𝑘=0
= 0,
and so
(2)(1)?
2
+ 𝛼
2
?
0
+ (3)(2)?
3
? − ?
1
? + 𝛼
2
?
1
? + ∑[(𝑘 + 2)(𝑘 + 1)?
𝑘+2
− 𝑘(𝑘 − 1)?
𝑘
− 𝑘?
𝑘
+ 𝛼
2
?
𝑘
]?
𝑘
+∞
𝑘=2
= 0.
Now comparing coefficients, we obtain
?
2
= −
𝛼
2
2
?
0
and
?
3
=
1−𝛼
2
6
?
1
and
(𝑘 + 2)(𝑘 + 1)?
𝑘+2
− (𝑘
2
− 𝛼
2
)?
𝑘
= 0
for every integer
𝑘 ≥ 2,
i.e.
?
𝑘+2
=
𝑘
2
−𝛼
2
(𝑘+2)(𝑘+1)
?
𝑘
for every positive integer
𝑘
.
Therefore
?
4
=
2
2
− 𝛼
2
4 ⋅ 3
?
2
=
(−𝛼
2
)(2
2
− 𝛼
2
)
4!
?
0
, ?
5
=
3
2
− 𝛼
2
5 ⋅ 4
?
3
=
(1
2
− 𝛼
2
)(3
2
− 𝛼
2
)
5!
?
1
,
?
6
=
4
2
− 𝛼
2
6 ⋅ 5
?
4
=
(−𝛼
2
)(2
2
− 𝛼
2
)(4
2
− 𝛼
2
)
6!
?
0
, …
and in general
?
2𝑚
=
(−𝛼
2
)(2
2
− 𝛼
2
)(4
2
− 𝛼
2
) ⋯ ((2? − 2)
2
− 𝛼
2
)
(2?)!
?
0
and
?
2𝑚+1
=
(1
2
− 𝛼
2
)(3
2
− 𝛼
2
)(5
2
− 𝛼
2
) ⋯ ((2? − 1)
2
− 𝛼
2
)
(2? + 1)!
?
1
for every non
-
negative integer
?
.
Therefore
? = ∑ ?
𝑘
?
𝑘
+∞
𝑘=0
= ?
0
∑
(−𝛼
2
)(2
2
− 𝛼
2
)(4
2
− 𝛼
2
) ⋯ ((2? − 2)
2
− 𝛼
2
)
(2?)!
?
2𝑚
+∞
𝑚=0
+ ?
1
∑
(1
2
− 𝛼
2
)(3
2
− 𝛼
2
)(5
2
− 𝛼
2
) ⋯ ((2? − 1)
2
− 𝛼
2
)
(2? + 1)!
?
2𝑚+1
+∞
𝑚=0
,
where
?
0
and
?
1
are arbitrary coefficients.
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MATH2350
Applied Linear Algebra and Differential Equations
Problem Set 7
L1, L2 (Fall 2023)
Page
13
of 13
(b)
When
𝛼 = ? = 1
, the second series terminates.
We have
𝑇
1
(?) = ?.
When
𝛼 = ? = 2
, the first series terminates.
We have
𝑇
2
(?) = 2
2!
−2
2
(1 +
−2
2
2!
?
2
) = 2?
2
− 1.
When
𝛼 = ? = 3
, the second series terminates.
We have
𝑇
3
(?) = 2
2
3!
1
2
− 3
2
(? +
1
2
− 3
2
3!
?
3
) = 4?
3
− 3?.
When
𝛼 = ? = 4
, the first series terminates.
We have
𝑇
4
(?) = 2
3
4!
(−4
2
)(2
2
− 4
2
)
(1 +
−4
2
2!
?
2
+
(−4
2
)(2
2
− 4
2
)
4!
?
4
) = 8?
4
− 8?
2
+ 1.
When
𝛼 = ? = 5
, the second series terminates.
We have
𝑇
5
(?) = 2
4
5!
(1
2
− 5
2
)(3
2
− 5
2
)
(? +
1
2
− 5
2
3!
?
3
+
(1
2
− 5
2
)(3
2
− 5
2
)
5!
?
5
) = 16?
5
− 40?
3
+ 5?.