section 5.6 exam 2 on 11/16 webhw due today o ce hours ...gmarple/nov10.pdfexam 2 on 11/16 8-9:30 pm...
TRANSCRIPT
Agenda
Section 5.6
Reminders
Exam 2 on 11/168-9:30 pm (Chem 1800)
WebHW due today
Office hours Tues, Thurs1-2 pm (5852 East Hall)
MathLab office hourSun 7-8 pm (MathLab)
(Gary Marple) November 10th, 2017 Math 216: Introduction to Differential Equations
§5.6 DE’s and Discontinuous Forcing Functions
Objectives
Be able to solve DE’s with discontinuous forcing terms.
(Gary Marple) November 10th, 2017 Math 216: Introduction to Differential Equations
Example
Use the Laplace transform to solve the IVP
y ′′ + y = f (t), y(0) = 5, y ′(0) = 3
where
f (t) =
{1, 0 ≤ t < π/2,0, π/2 ≤ t.
We begin by expressing f (t) using unit step functions. That is,
f (t) = 1− uπ/2(t).
Therefore, the DE can be written as
y ′′ + y = 1− uπ/2(t).
Applying the Laplace transform to both sides gives us
L{y ′′}+ L{y} = L{1} − L{uπ/2(t)
}.
(Gary Marple) November 10th, 2017 Math 216: Introduction to Differential Equations
Example
Use the Laplace transform to solve the IVP
y ′′ + y = f (t), y(0) = 5, y ′(0) = 3
where
f (t) =
{1, 0 ≤ t < π/2,0, π/2 ≤ t.
We begin by expressing f (t) using unit step functions. That is,
f (t) = 1− uπ/2(t).
Therefore, the DE can be written as
y ′′ + y = 1− uπ/2(t).
Applying the Laplace transform to both sides gives us
L{y ′′}+ L{y} = L{1} − L{uπ/2(t)
}.
(Gary Marple) November 10th, 2017 Math 216: Introduction to Differential Equations
Example
Use the Laplace transform to solve the IVP
y ′′ + y = f (t), y(0) = 5, y ′(0) = 3
where
f (t) =
{1, 0 ≤ t < π/2,0, π/2 ≤ t.
We begin by expressing f (t) using unit step functions. That is,
f (t) = 1− uπ/2(t).
Therefore, the DE can be written as
y ′′ + y = 1− uπ/2(t).
Applying the Laplace transform to both sides gives us
L{y ′′}+ L{y} = L{1} − L{uπ/2(t)
}.
(Gary Marple) November 10th, 2017 Math 216: Introduction to Differential Equations
Example
Use the Laplace transform to solve the IVP
y ′′ + y = f (t), y(0) = 5, y ′(0) = 3
where
f (t) =
{1, 0 ≤ t < π/2,0, π/2 ≤ t.
We begin by expressing f (t) using unit step functions. That is,
f (t) = 1− uπ/2(t).
Therefore, the DE can be written as
y ′′ + y = 1− uπ/2(t).
Applying the Laplace transform to both sides gives us
L{y ′′}+ L{y} = L{1} − L{uπ/2(t)
}.
(Gary Marple) November 10th, 2017 Math 216: Introduction to Differential Equations
L{y ′′}
+ L{y} = L{1} − L{uπ/2(t)
}.
Looking at the table, we see that
L{f ′′(t)
}= s2F (s)− sf (0)− f ′(0), L{1} =
1
s, L{uc (t)} =
e−cs
s.
Therefore,
s2Y (s)− sy(0)− y ′(0) + Y (s) =1
s−
e−πs/2
s.
s2Y (s)− 5s − 3 + Y (s) =1
s−
e−πs/2
s.
Keeping the terms with Y (s) on the left-hand side and moving all other terms to theright-hand side gives us
s2Y (s) + Y (s) = 5s + 3 +1
s−
e−πs/2
s.
(s2 + 1)Y (s) = 5s + 3 +1
s−
e−πs/2
s.
Y (s) = 5s
s2 + 1+ 3
1
s2 + 1+
1
s(s2 + 1)−
e−πs/2
s(s2 + 1).
(Gary Marple) November 10th, 2017 Math 216: Introduction to Differential Equations
L{y ′′}
+ L{y} = L{1} − L{uπ/2(t)
}.
Looking at the table, we see that
L{f ′′(t)
}= s2F (s)− sf (0)− f ′(0), L{1} =
1
s, L{uc (t)} =
e−cs
s.
Therefore,
s2Y (s)− sy(0)− y ′(0) + Y (s) =1
s−
e−πs/2
s.
s2Y (s)− 5s − 3 + Y (s) =1
s−
e−πs/2
s.
Keeping the terms with Y (s) on the left-hand side and moving all other terms to theright-hand side gives us
s2Y (s) + Y (s) = 5s + 3 +1
s−
e−πs/2
s.
(s2 + 1)Y (s) = 5s + 3 +1
s−
e−πs/2
s.
Y (s) = 5s
s2 + 1+ 3
1
s2 + 1+
1
s(s2 + 1)−
e−πs/2
s(s2 + 1).
(Gary Marple) November 10th, 2017 Math 216: Introduction to Differential Equations
L{y ′′}
+ L{y} = L{1} − L{uπ/2(t)
}.
Looking at the table, we see that
L{f ′′(t)
}= s2F (s)− sf (0)− f ′(0), L{1} =
1
s, L{uc (t)} =
e−cs
s.
Therefore,
s2Y (s)− sy(0)− y ′(0) + Y (s) =1
s−
e−πs/2
s.
s2Y (s)− 5s − 3 + Y (s) =1
s−
e−πs/2
s.
Keeping the terms with Y (s) on the left-hand side and moving all other terms to theright-hand side gives us
s2Y (s) + Y (s) = 5s + 3 +1
s−
e−πs/2
s.
(s2 + 1)Y (s) = 5s + 3 +1
s−
e−πs/2
s.
Y (s) = 5s
s2 + 1+ 3
1
s2 + 1+
1
s(s2 + 1)−
e−πs/2
s(s2 + 1).
(Gary Marple) November 10th, 2017 Math 216: Introduction to Differential Equations
L{y ′′}
+ L{y} = L{1} − L{uπ/2(t)
}.
Looking at the table, we see that
L{f ′′(t)
}= s2F (s)− sf (0)− f ′(0), L{1} =
1
s, L{uc (t)} =
e−cs
s.
Therefore,
s2Y (s)− sy(0)− y ′(0) + Y (s) =1
s−
e−πs/2
s.
s2Y (s)− 5s − 3 + Y (s) =1
s−
e−πs/2
s.
Keeping the terms with Y (s) on the left-hand side and moving all other terms to theright-hand side gives us
s2Y (s) + Y (s) = 5s + 3 +1
s−
e−πs/2
s.
(s2 + 1)Y (s) = 5s + 3 +1
s−
e−πs/2
s.
Y (s) = 5s
s2 + 1+ 3
1
s2 + 1+
1
s(s2 + 1)−
e−πs/2
s(s2 + 1).
(Gary Marple) November 10th, 2017 Math 216: Introduction to Differential Equations
L{y ′′}
+ L{y} = L{1} − L{uπ/2(t)
}.
Looking at the table, we see that
L{f ′′(t)
}= s2F (s)− sf (0)− f ′(0), L{1} =
1
s, L{uc (t)} =
e−cs
s.
Therefore,
s2Y (s)− sy(0)− y ′(0) + Y (s) =1
s−
e−πs/2
s.
s2Y (s)− 5s − 3 + Y (s) =1
s−
e−πs/2
s.
Keeping the terms with Y (s) on the left-hand side and moving all other terms to theright-hand side gives us
s2Y (s) + Y (s) = 5s + 3 +1
s−
e−πs/2
s.
(s2 + 1)Y (s) = 5s + 3 +1
s−
e−πs/2
s.
Y (s) = 5s
s2 + 1+ 3
1
s2 + 1+
1
s(s2 + 1)−
e−πs/2
s(s2 + 1).
(Gary Marple) November 10th, 2017 Math 216: Introduction to Differential Equations
L{y ′′}
+ L{y} = L{1} − L{uπ/2(t)
}.
Looking at the table, we see that
L{f ′′(t)
}= s2F (s)− sf (0)− f ′(0), L{1} =
1
s, L{uc (t)} =
e−cs
s.
Therefore,
s2Y (s)− sy(0)− y ′(0) + Y (s) =1
s−
e−πs/2
s.
s2Y (s)− 5s − 3 + Y (s) =1
s−
e−πs/2
s.
Keeping the terms with Y (s) on the left-hand side and moving all other terms to theright-hand side gives us
s2Y (s) + Y (s) = 5s + 3 +1
s−
e−πs/2
s.
(s2 + 1)Y (s) = 5s + 3 +1
s−
e−πs/2
s.
Y (s) = 5s
s2 + 1+ 3
1
s2 + 1+
1
s(s2 + 1)−
e−πs/2
s(s2 + 1).
(Gary Marple) November 10th, 2017 Math 216: Introduction to Differential Equations
L{y ′′}
+ L{y} = L{1} − L{uπ/2(t)
}.
Looking at the table, we see that
L{f ′′(t)
}= s2F (s)− sf (0)− f ′(0), L{1} =
1
s, L{uc (t)} =
e−cs
s.
Therefore,
s2Y (s)− sy(0)− y ′(0) + Y (s) =1
s−
e−πs/2
s.
s2Y (s)− 5s − 3 + Y (s) =1
s−
e−πs/2
s.
Keeping the terms with Y (s) on the left-hand side and moving all other terms to theright-hand side gives us
s2Y (s) + Y (s) = 5s + 3 +1
s−
e−πs/2
s.
(s2 + 1)Y (s) = 5s + 3 +1
s−
e−πs/2
s.
Y (s) = 5s
s2 + 1+ 3
1
s2 + 1+
1
s(s2 + 1)−
e−πs/2
s(s2 + 1).
(Gary Marple) November 10th, 2017 Math 216: Introduction to Differential Equations
Y (s) = 5s
s2 + 1+ 3
1
s2 + 1+
1
s(s2 + 1)−
e−πs/2
s(s2 + 1)
We need to perform completing the square on the last to terms on the right.
1
s(s2 + 1)=
A
s+
Bs + C
s2 + 1
Multiplying both sides by s(s2 + 1) give us
1 = A(s2 + 1) + (Bs + C)s.
Expanding, moving all terms to the right-hand side, and grouping them based onpowers of s gives us
0 = (A + B)s2 + Cs + (A− 1).
The only way for this equation to be satisfied for all s is if
A + B = 0, C = 0, and A− 1 = 0.
As a result, A = 1, B = −1, and C = 0. Therefore,
1
s(s2 + 1)=
1
s−
s
s2 + 1
and
Y (s) = 5s
s2 + 1+ 3
1
s2 + 1+
1
s−
s
s2 + 1− e−πs/2
(1
s−
s
s2 + 1
).
(Gary Marple) November 10th, 2017 Math 216: Introduction to Differential Equations
Y (s) = 5s
s2 + 1+ 3
1
s2 + 1+
1
s(s2 + 1)−
e−πs/2
s(s2 + 1)
We need to perform completing the square on the last to terms on the right.
1
s(s2 + 1)=
A
s+
Bs + C
s2 + 1
Multiplying both sides by s(s2 + 1) give us
1 = A(s2 + 1) + (Bs + C)s.
Expanding, moving all terms to the right-hand side, and grouping them based onpowers of s gives us
0 = (A + B)s2 + Cs + (A− 1).
The only way for this equation to be satisfied for all s is if
A + B = 0, C = 0, and A− 1 = 0.
As a result, A = 1, B = −1, and C = 0. Therefore,
1
s(s2 + 1)=
1
s−
s
s2 + 1
and
Y (s) = 5s
s2 + 1+ 3
1
s2 + 1+
1
s−
s
s2 + 1− e−πs/2
(1
s−
s
s2 + 1
).
(Gary Marple) November 10th, 2017 Math 216: Introduction to Differential Equations
Y (s) = 5s
s2 + 1+ 3
1
s2 + 1+
1
s(s2 + 1)−
e−πs/2
s(s2 + 1)
We need to perform completing the square on the last to terms on the right.
1
s(s2 + 1)=
A
s+
Bs + C
s2 + 1
Multiplying both sides by s(s2 + 1) give us
1 = A(s2 + 1) + (Bs + C)s.
Expanding, moving all terms to the right-hand side, and grouping them based onpowers of s gives us
0 = (A + B)s2 + Cs + (A− 1).
The only way for this equation to be satisfied for all s is if
A + B = 0, C = 0, and A− 1 = 0.
As a result, A = 1, B = −1, and C = 0. Therefore,
1
s(s2 + 1)=
1
s−
s
s2 + 1
and
Y (s) = 5s
s2 + 1+ 3
1
s2 + 1+
1
s−
s
s2 + 1− e−πs/2
(1
s−
s
s2 + 1
).
(Gary Marple) November 10th, 2017 Math 216: Introduction to Differential Equations
Y (s) = 5s
s2 + 1+ 3
1
s2 + 1+
1
s(s2 + 1)−
e−πs/2
s(s2 + 1)
We need to perform completing the square on the last to terms on the right.
1
s(s2 + 1)=
A
s+
Bs + C
s2 + 1
Multiplying both sides by s(s2 + 1) give us
1 = A(s2 + 1) + (Bs + C)s.
Expanding, moving all terms to the right-hand side, and grouping them based onpowers of s gives us
0 = (A + B)s2 + Cs + (A− 1).
The only way for this equation to be satisfied for all s is if
A + B = 0, C = 0, and A− 1 = 0.
As a result, A = 1, B = −1, and C = 0. Therefore,
1
s(s2 + 1)=
1
s−
s
s2 + 1
and
Y (s) = 5s
s2 + 1+ 3
1
s2 + 1+
1
s−
s
s2 + 1− e−πs/2
(1
s−
s
s2 + 1
).
(Gary Marple) November 10th, 2017 Math 216: Introduction to Differential Equations
Y (s) = 5s
s2 + 1+ 3
1
s2 + 1+
1
s(s2 + 1)−
e−πs/2
s(s2 + 1)
We need to perform completing the square on the last to terms on the right.
1
s(s2 + 1)=
A
s+
Bs + C
s2 + 1
Multiplying both sides by s(s2 + 1) give us
1 = A(s2 + 1) + (Bs + C)s.
Expanding, moving all terms to the right-hand side, and grouping them based onpowers of s gives us
0 = (A + B)s2 + Cs + (A− 1).
The only way for this equation to be satisfied for all s is if
A + B = 0, C = 0, and A− 1 = 0.
As a result, A = 1, B = −1, and C = 0. Therefore,
1
s(s2 + 1)=
1
s−
s
s2 + 1
and
Y (s) = 5s
s2 + 1+ 3
1
s2 + 1+
1
s−
s
s2 + 1− e−πs/2
(1
s−
s
s2 + 1
).
(Gary Marple) November 10th, 2017 Math 216: Introduction to Differential Equations
Y (s) = 5s
s2 + 1+ 3
1
s2 + 1+
1
s(s2 + 1)−
e−πs/2
s(s2 + 1)
We need to perform completing the square on the last to terms on the right.
1
s(s2 + 1)=
A
s+
Bs + C
s2 + 1
Multiplying both sides by s(s2 + 1) give us
1 = A(s2 + 1) + (Bs + C)s.
Expanding, moving all terms to the right-hand side, and grouping them based onpowers of s gives us
0 = (A + B)s2 + Cs + (A− 1).
The only way for this equation to be satisfied for all s is if
A + B = 0, C = 0, and A− 1 = 0.
As a result, A = 1, B = −1, and C = 0. Therefore,
1
s(s2 + 1)=
1
s−
s
s2 + 1
and
Y (s) = 5s
s2 + 1+ 3
1
s2 + 1+
1
s−
s
s2 + 1− e−πs/2
(1
s−
s
s2 + 1
).
(Gary Marple) November 10th, 2017 Math 216: Introduction to Differential Equations
Y (s) = 5s
s2 + 1+ 3
1
s2 + 1+
1
s(s2 + 1)−
e−πs/2
s(s2 + 1)
We need to perform completing the square on the last to terms on the right.
1
s(s2 + 1)=
A
s+
Bs + C
s2 + 1
Multiplying both sides by s(s2 + 1) give us
1 = A(s2 + 1) + (Bs + C)s.
Expanding, moving all terms to the right-hand side, and grouping them based onpowers of s gives us
0 = (A + B)s2 + Cs + (A− 1).
The only way for this equation to be satisfied for all s is if
A + B = 0, C = 0, and A− 1 = 0.
As a result, A = 1, B = −1, and C = 0. Therefore,
1
s(s2 + 1)=
1
s−
s
s2 + 1
and
Y (s) = 5s
s2 + 1+ 3
1
s2 + 1+
1
s−
s
s2 + 1− e−πs/2
(1
s−
s
s2 + 1
).
(Gary Marple) November 10th, 2017 Math 216: Introduction to Differential Equations
Y (s) = 5s
s2 + 1+ 3
1
s2 + 1+
1
s−
s
s2 + 1− e−πs/2
(1
s−
s
s2 + 1
)Combining the 1st and 4th terms on the right-hand side and applying the inverseLaplace operator gives us
y(t) = 4L−1
{s
s2 + 1
}+3L−1
{1
s2 + 1
}+L−1
{1
s
}−L−1
{e−πs/2
(1
s−
s
s2 + 1
)}.
By referring to a table, we see that
L{sin at} =a
s2 + a2, L{cos at} =
s
s2 + a2, L{1} =
1
s,
and L{uc (t)f (t − c)} = e−csF (s).
Therefore,
y(t) = 4 cos t + 3 sin t + 1− uπ/2(t)L−1
{1
s−
s
s2 + 1
}t→t−π/2
y(t) = 4 cos t + 3 sin t + 1− uπ/2(t) (1− cos t)t→t−π/2
y(t) = 4 cos t + 3 sin t + 1− uπ/2(t) [1− cos (t − π/2)]
y(t) = 4 cos t + 3 sin t + 1− uπ/2(t) (1− sin t)
(Gary Marple) November 10th, 2017 Math 216: Introduction to Differential Equations
Y (s) = 5s
s2 + 1+ 3
1
s2 + 1+
1
s−
s
s2 + 1− e−πs/2
(1
s−
s
s2 + 1
)Combining the 1st and 4th terms on the right-hand side and applying the inverseLaplace operator gives us
y(t) = 4L−1
{s
s2 + 1
}+3L−1
{1
s2 + 1
}+L−1
{1
s
}−L−1
{e−πs/2
(1
s−
s
s2 + 1
)}.
By referring to a table, we see that
L{sin at} =a
s2 + a2, L{cos at} =
s
s2 + a2, L{1} =
1
s,
and L{uc (t)f (t − c)} = e−csF (s).
Therefore,
y(t) = 4 cos t + 3 sin t + 1− uπ/2(t)L−1
{1
s−
s
s2 + 1
}t→t−π/2
y(t) = 4 cos t + 3 sin t + 1− uπ/2(t) (1− cos t)t→t−π/2
y(t) = 4 cos t + 3 sin t + 1− uπ/2(t) [1− cos (t − π/2)]
y(t) = 4 cos t + 3 sin t + 1− uπ/2(t) (1− sin t)
(Gary Marple) November 10th, 2017 Math 216: Introduction to Differential Equations
Y (s) = 5s
s2 + 1+ 3
1
s2 + 1+
1
s−
s
s2 + 1− e−πs/2
(1
s−
s
s2 + 1
)Combining the 1st and 4th terms on the right-hand side and applying the inverseLaplace operator gives us
y(t) = 4L−1
{s
s2 + 1
}+3L−1
{1
s2 + 1
}+L−1
{1
s
}−L−1
{e−πs/2
(1
s−
s
s2 + 1
)}.
By referring to a table, we see that
L{sin at} =a
s2 + a2, L{cos at} =
s
s2 + a2, L{1} =
1
s,
and L{uc (t)f (t − c)} = e−csF (s).
Therefore,
y(t) = 4 cos t + 3 sin t + 1− uπ/2(t)L−1
{1
s−
s
s2 + 1
}t→t−π/2
y(t) = 4 cos t + 3 sin t + 1− uπ/2(t) (1− cos t)t→t−π/2
y(t) = 4 cos t + 3 sin t + 1− uπ/2(t) [1− cos (t − π/2)]
y(t) = 4 cos t + 3 sin t + 1− uπ/2(t) (1− sin t)
(Gary Marple) November 10th, 2017 Math 216: Introduction to Differential Equations
Y (s) = 5s
s2 + 1+ 3
1
s2 + 1+
1
s−
s
s2 + 1− e−πs/2
(1
s−
s
s2 + 1
)Combining the 1st and 4th terms on the right-hand side and applying the inverseLaplace operator gives us
y(t) = 4L−1
{s
s2 + 1
}+3L−1
{1
s2 + 1
}+L−1
{1
s
}−L−1
{e−πs/2
(1
s−
s
s2 + 1
)}.
By referring to a table, we see that
L{sin at} =a
s2 + a2, L{cos at} =
s
s2 + a2, L{1} =
1
s,
and L{uc (t)f (t − c)} = e−csF (s).
Therefore,
y(t) = 4 cos t + 3 sin t + 1− uπ/2(t)L−1
{1
s−
s
s2 + 1
}t→t−π/2
y(t) = 4 cos t + 3 sin t + 1− uπ/2(t) (1− cos t)t→t−π/2
y(t) = 4 cos t + 3 sin t + 1− uπ/2(t) [1− cos (t − π/2)]
y(t) = 4 cos t + 3 sin t + 1− uπ/2(t) (1− sin t)
(Gary Marple) November 10th, 2017 Math 216: Introduction to Differential Equations
Y (s) = 5s
s2 + 1+ 3
1
s2 + 1+
1
s−
s
s2 + 1− e−πs/2
(1
s−
s
s2 + 1
)Combining the 1st and 4th terms on the right-hand side and applying the inverseLaplace operator gives us
y(t) = 4L−1
{s
s2 + 1
}+3L−1
{1
s2 + 1
}+L−1
{1
s
}−L−1
{e−πs/2
(1
s−
s
s2 + 1
)}.
By referring to a table, we see that
L{sin at} =a
s2 + a2, L{cos at} =
s
s2 + a2, L{1} =
1
s,
and L{uc (t)f (t − c)} = e−csF (s).
Therefore,
y(t) = 4 cos t + 3 sin t + 1− uπ/2(t)L−1
{1
s−
s
s2 + 1
}t→t−π/2
y(t) = 4 cos t + 3 sin t + 1− uπ/2(t) (1− cos t)t→t−π/2
y(t) = 4 cos t + 3 sin t + 1− uπ/2(t) [1− cos (t − π/2)]
y(t) = 4 cos t + 3 sin t + 1− uπ/2(t) (1− sin t)
(Gary Marple) November 10th, 2017 Math 216: Introduction to Differential Equations
Y (s) = 5s
s2 + 1+ 3
1
s2 + 1+
1
s−
s
s2 + 1− e−πs/2
(1
s−
s
s2 + 1
)Combining the 1st and 4th terms on the right-hand side and applying the inverseLaplace operator gives us
y(t) = 4L−1
{s
s2 + 1
}+3L−1
{1
s2 + 1
}+L−1
{1
s
}−L−1
{e−πs/2
(1
s−
s
s2 + 1
)}.
By referring to a table, we see that
L{sin at} =a
s2 + a2, L{cos at} =
s
s2 + a2, L{1} =
1
s,
and L{uc (t)f (t − c)} = e−csF (s).
Therefore,
y(t) = 4 cos t + 3 sin t + 1− uπ/2(t)L−1
{1
s−
s
s2 + 1
}t→t−π/2
y(t) = 4 cos t + 3 sin t + 1− uπ/2(t) (1− cos t)t→t−π/2
y(t) = 4 cos t + 3 sin t + 1− uπ/2(t) [1− cos (t − π/2)]
y(t) = 4 cos t + 3 sin t + 1− uπ/2(t) (1− sin t)
(Gary Marple) November 10th, 2017 Math 216: Introduction to Differential Equations
Y (s) = 5s
s2 + 1+ 3
1
s2 + 1+
1
s−
s
s2 + 1− e−πs/2
(1
s−
s
s2 + 1
)Combining the 1st and 4th terms on the right-hand side and applying the inverseLaplace operator gives us
y(t) = 4L−1
{s
s2 + 1
}+3L−1
{1
s2 + 1
}+L−1
{1
s
}−L−1
{e−πs/2
(1
s−
s
s2 + 1
)}.
By referring to a table, we see that
L{sin at} =a
s2 + a2, L{cos at} =
s
s2 + a2, L{1} =
1
s,
and L{uc (t)f (t − c)} = e−csF (s).
Therefore,
y(t) = 4 cos t + 3 sin t + 1− uπ/2(t)L−1
{1
s−
s
s2 + 1
}t→t−π/2
y(t) = 4 cos t + 3 sin t + 1− uπ/2(t) (1− cos t)t→t−π/2
y(t) = 4 cos t + 3 sin t + 1− uπ/2(t) [1− cos (t − π/2)]
y(t) = 4 cos t + 3 sin t + 1− uπ/2(t) (1− sin t).
(Gary Marple) November 10th, 2017 Math 216: Introduction to Differential Equations
y(t) = 4 cos t + 3 sin t + 1− uπ/2(t) (1− sin t)
Our solution can be expressed as piecewise defined function
y(t) =
{4 cos t + 3 sin t + 1, 0 ≤ t < π/2,4 cos t + 4 sin t, π/2 ≤ t.
(Gary Marple) November 10th, 2017 Math 216: Introduction to Differential Equations
Example
Use the Laplace transform to solve the IVP
y (4) + 5y ′′ + 4y = 1− uπ(t), y(0) = y ′(0) = y ′′(0) = y ′′′(0) = 0.
We begin by applying the Laplace transform to both sides of the DE to get
L{y (4)}
+ 5L{y ′′}
+ 4L{y} = L{1} − L{uπ(t)} .
By looking at a table, we see that
L{f ′′(t)
}= s2F (s)− sf (0)− f ′(0), L{1} =
1
s, L{uc (t)} =
e−cs
s,
andL{f (4)(t)
}= s4F (s)− s3f (0)− s2f ′(0)− sf ′′(0)− f ′′′(0).
Therefore,
s4Y (s)−s3y(0)−s2y ′(0)−sy ′′(0)−y ′′′(0)+5[s2Y (s)− sy(0)− y ′(0)
]+4Y (s) =
1
s−e−πs
s.
Plugging in the initial conditions gives us
s4Y (s) + 5s2Y (s) + 4Y (s) =1
s−
e−πs
s.
(Gary Marple) November 10th, 2017 Math 216: Introduction to Differential Equations
Example
Use the Laplace transform to solve the IVP
y (4) + 5y ′′ + 4y = 1− uπ(t), y(0) = y ′(0) = y ′′(0) = y ′′′(0) = 0.
We begin by applying the Laplace transform to both sides of the DE to get
L{y (4)}
+ 5L{y ′′}
+ 4L{y} = L{1} − L{uπ(t)} .
By looking at a table, we see that
L{f ′′(t)
}= s2F (s)− sf (0)− f ′(0), L{1} =
1
s, L{uc (t)} =
e−cs
s,
andL{f (4)(t)
}= s4F (s)− s3f (0)− s2f ′(0)− sf ′′(0)− f ′′′(0).
Therefore,
s4Y (s)−s3y(0)−s2y ′(0)−sy ′′(0)−y ′′′(0)+5[s2Y (s)− sy(0)− y ′(0)
]+4Y (s) =
1
s−e−πs
s.
Plugging in the initial conditions gives us
s4Y (s) + 5s2Y (s) + 4Y (s) =1
s−
e−πs
s.
(Gary Marple) November 10th, 2017 Math 216: Introduction to Differential Equations
Example
Use the Laplace transform to solve the IVP
y (4) + 5y ′′ + 4y = 1− uπ(t), y(0) = y ′(0) = y ′′(0) = y ′′′(0) = 0.
We begin by applying the Laplace transform to both sides of the DE to get
L{y (4)}
+ 5L{y ′′}
+ 4L{y} = L{1} − L{uπ(t)} .
By looking at a table, we see that
L{f ′′(t)
}= s2F (s)− sf (0)− f ′(0), L{1} =
1
s, L{uc (t)} =
e−cs
s,
andL{f (4)(t)
}= s4F (s)− s3f (0)− s2f ′(0)− sf ′′(0)− f ′′′(0).
Therefore,
s4Y (s)−s3y(0)−s2y ′(0)−sy ′′(0)−y ′′′(0)+5[s2Y (s)− sy(0)− y ′(0)
]+4Y (s) =
1
s−e−πs
s.
Plugging in the initial conditions gives us
s4Y (s) + 5s2Y (s) + 4Y (s) =1
s−
e−πs
s.
(Gary Marple) November 10th, 2017 Math 216: Introduction to Differential Equations
Example
Use the Laplace transform to solve the IVP
y (4) + 5y ′′ + 4y = 1− uπ(t), y(0) = y ′(0) = y ′′(0) = y ′′′(0) = 0.
We begin by applying the Laplace transform to both sides of the DE to get
L{y (4)}
+ 5L{y ′′}
+ 4L{y} = L{1} − L{uπ(t)} .
By looking at a table, we see that
L{f ′′(t)
}= s2F (s)− sf (0)− f ′(0), L{1} =
1
s, L{uc (t)} =
e−cs
s,
andL{f (4)(t)
}= s4F (s)− s3f (0)− s2f ′(0)− sf ′′(0)− f ′′′(0).
Therefore,
s4Y (s)−s3y(0)−s2y ′(0)−sy ′′(0)−y ′′′(0)+5[s2Y (s)− sy(0)− y ′(0)
]+4Y (s) =
1
s−e−πs
s.
Plugging in the initial conditions gives us
s4Y (s) + 5s2Y (s) + 4Y (s) =1
s−
e−πs
s.
(Gary Marple) November 10th, 2017 Math 216: Introduction to Differential Equations
Example
Use the Laplace transform to solve the IVP
y (4) + 5y ′′ + 4y = 1− uπ(t), y(0) = y ′(0) = y ′′(0) = y ′′′(0) = 0.
We begin by applying the Laplace transform to both sides of the DE to get
L{y (4)}
+ 5L{y ′′}
+ 4L{y} = L{1} − L{uπ(t)} .
By looking at a table, we see that
L{f ′′(t)
}= s2F (s)− sf (0)− f ′(0), L{1} =
1
s, L{uc (t)} =
e−cs
s,
andL{f (4)(t)
}= s4F (s)− s3f (0)− s2f ′(0)− sf ′′(0)− f ′′′(0).
Therefore,
s4Y (s)−s3y(0)−s2y ′(0)−sy ′′(0)−y ′′′(0)+5[s2Y (s)− sy(0)− y ′(0)
]+4Y (s) =
1
s−e−πs
s.
Plugging in the initial conditions gives us
s4Y (s) + 5s2Y (s) + 4Y (s) =1
s−
e−πs
s.
(Gary Marple) November 10th, 2017 Math 216: Introduction to Differential Equations
Example
Use the Laplace transform to solve the IVP
y (4) + 5y ′′ + 4y = 1− uπ(t), y(0) = y ′(0) = y ′′(0) = y ′′′(0) = 0.
We begin by applying the Laplace transform to both sides of the DE to get
L{y (4)}
+ 5L{y ′′}
+ 4L{y} = L{1} − L{uπ(t)} .
By looking at a table, we see that
L{f ′′(t)
}= s2F (s)− sf (0)− f ′(0), L{1} =
1
s, L{uc (t)} =
e−cs
s,
andL{f (4)(t)
}= s4F (s)− s3f (0)− s2f ′(0)− sf ′′(0)− f ′′′(0).
Therefore,
s4Y (s)−s3y(0)−s2y ′(0)−sy ′′(0)−y ′′′(0)+5[s2Y (s)− sy(0)− y ′(0)
]+4Y (s) =
1
s−e−πs
s.
Plugging in the initial conditions gives us
s4Y (s) + 5s2Y (s) + 4Y (s) =1
s−
e−πs
s.
(Gary Marple) November 10th, 2017 Math 216: Introduction to Differential Equations
s4Y (s) + 5s2Y (s) + 4Y (s) =1
s−
e−πs
s
(s4 + 5s2 + 4)Y (s) =1
s−
e−πs
s
(s2 + 1)(s2 + 4)Y (s) =1
s−
e−πs
s
Y (s) =1
s(s2 + 1)(s2 + 4)−
e−πs
s(s2 + 1)(s2 + 4)
We need to find the partial fraction decomposition of the two terms on the right.
1
s(s2 + 1)(s2 + 4)=
A
s+
Bs + C
s2 + 1+
Ds + 2E
s2 + 4.
Multiplying both sides by s(s2 + 1)(s2 + 4) gives us
1 = A(s2 + 1)(s2 + 4) + (Bs + C)(s2 + 4)s + (Ds + 2E)(s2 + 1)s.
Expanding, moving all terms to the right-hand side, and group by power of s gives us
0 = (A + B + D)s4 + (C + 2E)s3 + (5A + 4B + D)s2 + (4C + 2E)s + (4A− 1).
(Gary Marple) November 10th, 2017 Math 216: Introduction to Differential Equations
s4Y (s) + 5s2Y (s) + 4Y (s) =1
s−
e−πs
s
(s4 + 5s2 + 4)Y (s) =1
s−
e−πs
s
(s2 + 1)(s2 + 4)Y (s) =1
s−
e−πs
s
Y (s) =1
s(s2 + 1)(s2 + 4)−
e−πs
s(s2 + 1)(s2 + 4)
We need to find the partial fraction decomposition of the two terms on the right.
1
s(s2 + 1)(s2 + 4)=
A
s+
Bs + C
s2 + 1+
Ds + 2E
s2 + 4.
Multiplying both sides by s(s2 + 1)(s2 + 4) gives us
1 = A(s2 + 1)(s2 + 4) + (Bs + C)(s2 + 4)s + (Ds + 2E)(s2 + 1)s.
Expanding, moving all terms to the right-hand side, and group by power of s gives us
0 = (A + B + D)s4 + (C + 2E)s3 + (5A + 4B + D)s2 + (4C + 2E)s + (4A− 1).
(Gary Marple) November 10th, 2017 Math 216: Introduction to Differential Equations
s4Y (s) + 5s2Y (s) + 4Y (s) =1
s−
e−πs
s
(s4 + 5s2 + 4)Y (s) =1
s−
e−πs
s
(s2 + 1)(s2 + 4)Y (s) =1
s−
e−πs
s
Y (s) =1
s(s2 + 1)(s2 + 4)−
e−πs
s(s2 + 1)(s2 + 4)
We need to find the partial fraction decomposition of the two terms on the right.
1
s(s2 + 1)(s2 + 4)=
A
s+
Bs + C
s2 + 1+
Ds + 2E
s2 + 4.
Multiplying both sides by s(s2 + 1)(s2 + 4) gives us
1 = A(s2 + 1)(s2 + 4) + (Bs + C)(s2 + 4)s + (Ds + 2E)(s2 + 1)s.
Expanding, moving all terms to the right-hand side, and group by power of s gives us
0 = (A + B + D)s4 + (C + 2E)s3 + (5A + 4B + D)s2 + (4C + 2E)s + (4A− 1).
(Gary Marple) November 10th, 2017 Math 216: Introduction to Differential Equations
s4Y (s) + 5s2Y (s) + 4Y (s) =1
s−
e−πs
s
(s4 + 5s2 + 4)Y (s) =1
s−
e−πs
s
(s2 + 1)(s2 + 4)Y (s) =1
s−
e−πs
s
Y (s) =1
s(s2 + 1)(s2 + 4)−
e−πs
s(s2 + 1)(s2 + 4)
We need to find the partial fraction decomposition of the two terms on the right.
1
s(s2 + 1)(s2 + 4)=
A
s+
Bs + C
s2 + 1+
Ds + 2E
s2 + 4.
Multiplying both sides by s(s2 + 1)(s2 + 4) gives us
1 = A(s2 + 1)(s2 + 4) + (Bs + C)(s2 + 4)s + (Ds + 2E)(s2 + 1)s.
Expanding, moving all terms to the right-hand side, and group by power of s gives us
0 = (A + B + D)s4 + (C + 2E)s3 + (5A + 4B + D)s2 + (4C + 2E)s + (4A− 1).
(Gary Marple) November 10th, 2017 Math 216: Introduction to Differential Equations
s4Y (s) + 5s2Y (s) + 4Y (s) =1
s−
e−πs
s
(s4 + 5s2 + 4)Y (s) =1
s−
e−πs
s
(s2 + 1)(s2 + 4)Y (s) =1
s−
e−πs
s
Y (s) =1
s(s2 + 1)(s2 + 4)−
e−πs
s(s2 + 1)(s2 + 4)
We need to find the partial fraction decomposition of the two terms on the right.
1
s(s2 + 1)(s2 + 4)=
A
s+
Bs + C
s2 + 1+
Ds + 2E
s2 + 4.
Multiplying both sides by s(s2 + 1)(s2 + 4) gives us
1 = A(s2 + 1)(s2 + 4) + (Bs + C)(s2 + 4)s + (Ds + 2E)(s2 + 1)s.
Expanding, moving all terms to the right-hand side, and group by power of s gives us
0 = (A + B + D)s4 + (C + 2E)s3 + (5A + 4B + D)s2 + (4C + 2E)s + (4A− 1).
(Gary Marple) November 10th, 2017 Math 216: Introduction to Differential Equations
s4Y (s) + 5s2Y (s) + 4Y (s) =1
s−
e−πs
s
(s4 + 5s2 + 4)Y (s) =1
s−
e−πs
s
(s2 + 1)(s2 + 4)Y (s) =1
s−
e−πs
s
Y (s) =1
s(s2 + 1)(s2 + 4)−
e−πs
s(s2 + 1)(s2 + 4)
We need to find the partial fraction decomposition of the two terms on the right.
1
s(s2 + 1)(s2 + 4)=
A
s+
Bs + C
s2 + 1+
Ds + 2E
s2 + 4.
Multiplying both sides by s(s2 + 1)(s2 + 4) gives us
1 = A(s2 + 1)(s2 + 4) + (Bs + C)(s2 + 4)s + (Ds + 2E)(s2 + 1)s.
Expanding, moving all terms to the right-hand side, and group by power of s gives us
0 = (A + B + D)s4 + (C + 2E)s3 + (5A + 4B + D)s2 + (4C + 2E)s + (4A− 1).
(Gary Marple) November 10th, 2017 Math 216: Introduction to Differential Equations
s4Y (s) + 5s2Y (s) + 4Y (s) =1
s−
e−πs
s
(s4 + 5s2 + 4)Y (s) =1
s−
e−πs
s
(s2 + 1)(s2 + 4)Y (s) =1
s−
e−πs
s
Y (s) =1
s(s2 + 1)(s2 + 4)−
e−πs
s(s2 + 1)(s2 + 4)
We need to find the partial fraction decomposition of the two terms on the right.
1
s(s2 + 1)(s2 + 4)=
A
s+
Bs + C
s2 + 1+
Ds + 2E
s2 + 4.
Multiplying both sides by s(s2 + 1)(s2 + 4) gives us
1 = A(s2 + 1)(s2 + 4) + (Bs + C)(s2 + 4)s + (Ds + 2E)(s2 + 1)s.
Expanding, moving all terms to the right-hand side, and group by power of s gives us
0 = (A + B + D)s4 + (C + 2E)s3 + (5A + 4B + D)s2 + (4C + 2E)s + (4A− 1).
(Gary Marple) November 10th, 2017 Math 216: Introduction to Differential Equations
0 = (A + B + D)s4 + (C + 2E)s3 + (5A + 4B + D)s2 + (4C + 2E)s + (4A− 1)
The only way for this polynomial to be satisfied for all s is if
A + B + D = 0, C + 2E = 0, 5A + 4B + D = 0, 4C + 2E = 0, 4A− 1 = 0.
Solving the system of 5 equations gives us A = 1/4, B = −1/3, C = 0, D = 1/12,and E = 0. Recall that the partial fraction decomposition had the form
1
s(s2 + 1)(s2 + 4)=
A
s+
Bs + C
s2 + 1+
Ds + 2E
s2 + 4.
Therefore,1
s(s2 + 1)(s2 + 4)=
1
4s−
1
3
s
s2 + 1+
1
12
s
s2 + 4.
Recall that
Y (s) =1
s(s2 + 1)(s2 + 4)−
e−πs
s(s2 + 1)(s2 + 4).
Plugging in the partial fraction decomposition for the two terms on the right gives us
Y (s) =1
4s−
1
3
s
s2 + 1+
1
12
s
s2 + 4− e−πs
(1
4s−
1
3
s
s2 + 1+
1
12
s
s2 + 4
).
(Gary Marple) November 10th, 2017 Math 216: Introduction to Differential Equations
0 = (A + B + D)s4 + (C + 2E)s3 + (5A + 4B + D)s2 + (4C + 2E)s + (4A− 1)
The only way for this polynomial to be satisfied for all s is if
A + B + D = 0, C + 2E = 0, 5A + 4B + D = 0, 4C + 2E = 0, 4A− 1 = 0.
Solving the system of 5 equations gives us A = 1/4, B = −1/3, C = 0, D = 1/12,and E = 0. Recall that the partial fraction decomposition had the form
1
s(s2 + 1)(s2 + 4)=
A
s+
Bs + C
s2 + 1+
Ds + 2E
s2 + 4.
Therefore,1
s(s2 + 1)(s2 + 4)=
1
4s−
1
3
s
s2 + 1+
1
12
s
s2 + 4.
Recall that
Y (s) =1
s(s2 + 1)(s2 + 4)−
e−πs
s(s2 + 1)(s2 + 4).
Plugging in the partial fraction decomposition for the two terms on the right gives us
Y (s) =1
4s−
1
3
s
s2 + 1+
1
12
s
s2 + 4− e−πs
(1
4s−
1
3
s
s2 + 1+
1
12
s
s2 + 4
).
(Gary Marple) November 10th, 2017 Math 216: Introduction to Differential Equations
0 = (A + B + D)s4 + (C + 2E)s3 + (5A + 4B + D)s2 + (4C + 2E)s + (4A− 1)
The only way for this polynomial to be satisfied for all s is if
A + B + D = 0, C + 2E = 0, 5A + 4B + D = 0, 4C + 2E = 0, 4A− 1 = 0.
Solving the system of 5 equations gives us A = 1/4, B = −1/3, C = 0, D = 1/12,and E = 0. Recall that the partial fraction decomposition had the form
1
s(s2 + 1)(s2 + 4)=
A
s+
Bs + C
s2 + 1+
Ds + 2E
s2 + 4.
Therefore,1
s(s2 + 1)(s2 + 4)=
1
4s−
1
3
s
s2 + 1+
1
12
s
s2 + 4.
Recall that
Y (s) =1
s(s2 + 1)(s2 + 4)−
e−πs
s(s2 + 1)(s2 + 4).
Plugging in the partial fraction decomposition for the two terms on the right gives us
Y (s) =1
4s−
1
3
s
s2 + 1+
1
12
s
s2 + 4− e−πs
(1
4s−
1
3
s
s2 + 1+
1
12
s
s2 + 4
).
(Gary Marple) November 10th, 2017 Math 216: Introduction to Differential Equations
0 = (A + B + D)s4 + (C + 2E)s3 + (5A + 4B + D)s2 + (4C + 2E)s + (4A− 1)
The only way for this polynomial to be satisfied for all s is if
A + B + D = 0, C + 2E = 0, 5A + 4B + D = 0, 4C + 2E = 0, 4A− 1 = 0.
Solving the system of 5 equations gives us A = 1/4, B = −1/3, C = 0, D = 1/12,and E = 0. Recall that the partial fraction decomposition had the form
1
s(s2 + 1)(s2 + 4)=
A
s+
Bs + C
s2 + 1+
Ds + 2E
s2 + 4.
Therefore,1
s(s2 + 1)(s2 + 4)=
1
4s−
1
3
s
s2 + 1+
1
12
s
s2 + 4.
Recall that
Y (s) =1
s(s2 + 1)(s2 + 4)−
e−πs
s(s2 + 1)(s2 + 4).
Plugging in the partial fraction decomposition for the two terms on the right gives us
Y (s) =1
4s−
1
3
s
s2 + 1+
1
12
s
s2 + 4− e−πs
(1
4s−
1
3
s
s2 + 1+
1
12
s
s2 + 4
).
(Gary Marple) November 10th, 2017 Math 216: Introduction to Differential Equations
0 = (A + B + D)s4 + (C + 2E)s3 + (5A + 4B + D)s2 + (4C + 2E)s + (4A− 1)
The only way for this polynomial to be satisfied for all s is if
A + B + D = 0, C + 2E = 0, 5A + 4B + D = 0, 4C + 2E = 0, 4A− 1 = 0.
Solving the system of 5 equations gives us A = 1/4, B = −1/3, C = 0, D = 1/12,and E = 0. Recall that the partial fraction decomposition had the form
1
s(s2 + 1)(s2 + 4)=
A
s+
Bs + C
s2 + 1+
Ds + 2E
s2 + 4.
Therefore,1
s(s2 + 1)(s2 + 4)=
1
4s−
1
3
s
s2 + 1+
1
12
s
s2 + 4.
Recall that
Y (s) =1
s(s2 + 1)(s2 + 4)−
e−πs
s(s2 + 1)(s2 + 4).
Plugging in the partial fraction decomposition for the two terms on the right gives us
Y (s) =1
4s−
1
3
s
s2 + 1+
1
12
s
s2 + 4− e−πs
(1
4s−
1
3
s
s2 + 1+
1
12
s
s2 + 4
).
(Gary Marple) November 10th, 2017 Math 216: Introduction to Differential Equations
0 = (A + B + D)s4 + (C + 2E)s3 + (5A + 4B + D)s2 + (4C + 2E)s + (4A− 1)
The only way for this polynomial to be satisfied for all s is if
A + B + D = 0, C + 2E = 0, 5A + 4B + D = 0, 4C + 2E = 0, 4A− 1 = 0.
Solving the system of 5 equations gives us A = 1/4, B = −1/3, C = 0, D = 1/12,and E = 0. Recall that the partial fraction decomposition had the form
1
s(s2 + 1)(s2 + 4)=
A
s+
Bs + C
s2 + 1+
Ds + 2E
s2 + 4.
Therefore,1
s(s2 + 1)(s2 + 4)=
1
4s−
1
3
s
s2 + 1+
1
12
s
s2 + 4.
Recall that
Y (s) =1
s(s2 + 1)(s2 + 4)−
e−πs
s(s2 + 1)(s2 + 4).
Plugging in the partial fraction decomposition for the two terms on the right gives us
Y (s) =1
4s−
1
3
s
s2 + 1+
1
12
s
s2 + 4− e−πs
(1
4s−
1
3
s
s2 + 1+
1
12
s
s2 + 4
).
(Gary Marple) November 10th, 2017 Math 216: Introduction to Differential Equations
Y (s) =1
4s−
1
3
s
s2 + 1+
1
12
s
s2 + 4− e−πs
(1
4s−
1
3
s
s2 + 1+
1
12
s
s2 + 4
)Applying the inverse Laplace transform to both sides gives us
y(t) =1
4L−1
{1
s
}−
1
3L−1
{s
s2 + 1
}+
1
12L−1
{s
s2 + 4
}− L−1
{e−πs
(1
4s−
1
3
s
s2 + 1+
1
12
s
s2 + 4
)}.
By referring to a table, we see that
L{cos at} =s
s2 + a2, L{1} =
1
s, L{uc (t)f (t − c)} = e−csF (s).
Therefore,
y(t) =1
4−
1
3cos t +
1
12cos 2t − uπ(t)L−1
{(1
4s−
1
3
s
s2 + 1+
1
12
s
s2 + 4
)}t→t−π
y(t) =1
4−
1
3cos t +
1
12cos 2t − uπ(t)
(1
4−
1
3cos t +
1
12cos 2t
)t→t−π
y(t) =1
4−
1
3cos t +
1
12cos 2t − uπ(t)
[1
4−
1
3cos (t − π) +
1
12cos (2(t − π))
].
(Gary Marple) November 10th, 2017 Math 216: Introduction to Differential Equations
Y (s) =1
4s−
1
3
s
s2 + 1+
1
12
s
s2 + 4− e−πs
(1
4s−
1
3
s
s2 + 1+
1
12
s
s2 + 4
)Applying the inverse Laplace transform to both sides gives us
y(t) =1
4L−1
{1
s
}−
1
3L−1
{s
s2 + 1
}+
1
12L−1
{s
s2 + 4
}− L−1
{e−πs
(1
4s−
1
3
s
s2 + 1+
1
12
s
s2 + 4
)}.
By referring to a table, we see that
L{cos at} =s
s2 + a2, L{1} =
1
s, L{uc (t)f (t − c)} = e−csF (s).
Therefore,
y(t) =1
4−
1
3cos t +
1
12cos 2t − uπ(t)L−1
{(1
4s−
1
3
s
s2 + 1+
1
12
s
s2 + 4
)}t→t−π
y(t) =1
4−
1
3cos t +
1
12cos 2t − uπ(t)
(1
4−
1
3cos t +
1
12cos 2t
)t→t−π
y(t) =1
4−
1
3cos t +
1
12cos 2t − uπ(t)
[1
4−
1
3cos (t − π) +
1
12cos (2(t − π))
].
(Gary Marple) November 10th, 2017 Math 216: Introduction to Differential Equations
Y (s) =1
4s−
1
3
s
s2 + 1+
1
12
s
s2 + 4− e−πs
(1
4s−
1
3
s
s2 + 1+
1
12
s
s2 + 4
)Applying the inverse Laplace transform to both sides gives us
y(t) =1
4L−1
{1
s
}−
1
3L−1
{s
s2 + 1
}+
1
12L−1
{s
s2 + 4
}− L−1
{e−πs
(1
4s−
1
3
s
s2 + 1+
1
12
s
s2 + 4
)}.
By referring to a table, we see that
L{cos at} =s
s2 + a2, L{1} =
1
s, L{uc (t)f (t − c)} = e−csF (s).
Therefore,
y(t) =1
4−
1
3cos t +
1
12cos 2t − uπ(t)L−1
{(1
4s−
1
3
s
s2 + 1+
1
12
s
s2 + 4
)}t→t−π
y(t) =1
4−
1
3cos t +
1
12cos 2t − uπ(t)
(1
4−
1
3cos t +
1
12cos 2t
)t→t−π
y(t) =1
4−
1
3cos t +
1
12cos 2t − uπ(t)
[1
4−
1
3cos (t − π) +
1
12cos (2(t − π))
].
(Gary Marple) November 10th, 2017 Math 216: Introduction to Differential Equations
Y (s) =1
4s−
1
3
s
s2 + 1+
1
12
s
s2 + 4− e−πs
(1
4s−
1
3
s
s2 + 1+
1
12
s
s2 + 4
)Applying the inverse Laplace transform to both sides gives us
y(t) =1
4L−1
{1
s
}−
1
3L−1
{s
s2 + 1
}+
1
12L−1
{s
s2 + 4
}− L−1
{e−πs
(1
4s−
1
3
s
s2 + 1+
1
12
s
s2 + 4
)}.
By referring to a table, we see that
L{cos at} =s
s2 + a2, L{1} =
1
s, L{uc (t)f (t − c)} = e−csF (s).
Therefore,
y(t) =1
4−
1
3cos t +
1
12cos 2t − uπ(t)L−1
{(1
4s−
1
3
s
s2 + 1+
1
12
s
s2 + 4
)}t→t−π
y(t) =1
4−
1
3cos t +
1
12cos 2t − uπ(t)
(1
4−
1
3cos t +
1
12cos 2t
)t→t−π
y(t) =1
4−
1
3cos t +
1
12cos 2t − uπ(t)
[1
4−
1
3cos (t − π) +
1
12cos (2(t − π))
].
(Gary Marple) November 10th, 2017 Math 216: Introduction to Differential Equations
Y (s) =1
4s−
1
3
s
s2 + 1+
1
12
s
s2 + 4− e−πs
(1
4s−
1
3
s
s2 + 1+
1
12
s
s2 + 4
)Applying the inverse Laplace transform to both sides gives us
y(t) =1
4L−1
{1
s
}−
1
3L−1
{s
s2 + 1
}+
1
12L−1
{s
s2 + 4
}− L−1
{e−πs
(1
4s−
1
3
s
s2 + 1+
1
12
s
s2 + 4
)}.
By referring to a table, we see that
L{cos at} =s
s2 + a2, L{1} =
1
s, L{uc (t)f (t − c)} = e−csF (s).
Therefore,
y(t) =1
4−
1
3cos t +
1
12cos 2t − uπ(t)L−1
{(1
4s−
1
3
s
s2 + 1+
1
12
s
s2 + 4
)}t→t−π
y(t) =1
4−
1
3cos t +
1
12cos 2t − uπ(t)
(1
4−
1
3cos t +
1
12cos 2t
)t→t−π
y(t) =1
4−
1
3cos t +
1
12cos 2t − uπ(t)
[1
4−
1
3cos (t − π) +
1
12cos (2(t − π))
].
(Gary Marple) November 10th, 2017 Math 216: Introduction to Differential Equations
Y (s) =1
4s−
1
3
s
s2 + 1+
1
12
s
s2 + 4− e−πs
(1
4s−
1
3
s
s2 + 1+
1
12
s
s2 + 4
)Applying the inverse Laplace transform to both sides gives us
y(t) =1
4L−1
{1
s
}−
1
3L−1
{s
s2 + 1
}+
1
12L−1
{s
s2 + 4
}− L−1
{e−πs
(1
4s−
1
3
s
s2 + 1+
1
12
s
s2 + 4
)}.
By referring to a table, we see that
L{cos at} =s
s2 + a2, L{1} =
1
s, L{uc (t)f (t − c)} = e−csF (s).
Therefore,
y(t) =1
4−
1
3cos t +
1
12cos 2t − uπ(t)L−1
{(1
4s−
1
3
s
s2 + 1+
1
12
s
s2 + 4
)}t→t−π
y(t) =1
4−
1
3cos t +
1
12cos 2t − uπ(t)
(1
4−
1
3cos t +
1
12cos 2t
)t→t−π
y(t) =1
4−
1
3cos t +
1
12cos 2t − uπ(t)
[1
4−
1
3cos (t − π) +
1
12cos (2(t − π))
].
(Gary Marple) November 10th, 2017 Math 216: Introduction to Differential Equations
y(t) =1
4−
1
3cos t +
1
12cos 2t − uπ(t)
[1
4−
1
3cos (t − π) +
1
12cos (2(t − π))
]y(t) =
1
4−
1
3cos t +
1
12cos 2t − uπ(t)
(1
4+
1
3cos t +
1
12cos 2t
)y(t) =
{(3− 4 cos t + cos 2t) /12 0 ≤ t < π−2(cos t)/3 π ≤ t
(Gary Marple) November 10th, 2017 Math 216: Introduction to Differential Equations
Example
Solve the IVPy ′′ + y = f (t), y(0) = 0, y ′(0) = 0,
where
f (t) = 1 + 215∑k=1
(−1)kukπ(t).
We begin by applying the Laplace transform to both sides of the DE
L{y ′′}
+ L{y} = L{1}+ 215∑k=1
(−1)kL{ukπ(t)}.
By looking in a table, we’ll see that
L{f ′′(t)
}= s2F (s)− sy(0)− y ′(0), L{1} =
1
s, L{uc (t)} =
e−cs
s.
Therefore,
s2Y (s)− sy(0)− y ′(0) + Y (s) =1
s+ 2
15∑k=1
(−1)ke−kπs
s.
s2Y (s) + Y (s) =1
s+ 2
15∑k=1
(−1)ke−kπs
s.
(Gary Marple) November 10th, 2017 Math 216: Introduction to Differential Equations
Example
Solve the IVPy ′′ + y = f (t), y(0) = 0, y ′(0) = 0,
where
f (t) = 1 + 215∑k=1
(−1)kukπ(t).
We begin by applying the Laplace transform to both sides of the DE
L{y ′′}
+ L{y} = L{1}+ 215∑k=1
(−1)kL{ukπ(t)}.
By looking in a table, we’ll see that
L{f ′′(t)
}= s2F (s)− sy(0)− y ′(0), L{1} =
1
s, L{uc (t)} =
e−cs
s.
Therefore,
s2Y (s)− sy(0)− y ′(0) + Y (s) =1
s+ 2
15∑k=1
(−1)ke−kπs
s.
s2Y (s) + Y (s) =1
s+ 2
15∑k=1
(−1)ke−kπs
s.
(Gary Marple) November 10th, 2017 Math 216: Introduction to Differential Equations
Example
Solve the IVPy ′′ + y = f (t), y(0) = 0, y ′(0) = 0,
where
f (t) = 1 + 215∑k=1
(−1)kukπ(t).
We begin by applying the Laplace transform to both sides of the DE
L{y ′′}
+ L{y} = L{1}+ 215∑k=1
(−1)kL{ukπ(t)}.
By looking in a table, we’ll see that
L{f ′′(t)
}= s2F (s)− sy(0)− y ′(0), L{1} =
1
s, L{uc (t)} =
e−cs
s.
Therefore,
s2Y (s)− sy(0)− y ′(0) + Y (s) =1
s+ 2
15∑k=1
(−1)ke−kπs
s.
s2Y (s) + Y (s) =1
s+ 2
15∑k=1
(−1)ke−kπs
s.
(Gary Marple) November 10th, 2017 Math 216: Introduction to Differential Equations
Example
Solve the IVPy ′′ + y = f (t), y(0) = 0, y ′(0) = 0,
where
f (t) = 1 + 215∑k=1
(−1)kukπ(t).
We begin by applying the Laplace transform to both sides of the DE
L{y ′′}
+ L{y} = L{1}+ 215∑k=1
(−1)kL{ukπ(t)}.
By looking in a table, we’ll see that
L{f ′′(t)
}= s2F (s)− sy(0)− y ′(0), L{1} =
1
s, L{uc (t)} =
e−cs
s.
Therefore,
s2Y (s)− sy(0)− y ′(0) + Y (s) =1
s+ 2
15∑k=1
(−1)ke−kπs
s.
s2Y (s) + Y (s) =1
s+ 2
15∑k=1
(−1)ke−kπs
s.
(Gary Marple) November 10th, 2017 Math 216: Introduction to Differential Equations
Example
Solve the IVPy ′′ + y = f (t), y(0) = 0, y ′(0) = 0,
where
f (t) = 1 + 215∑k=1
(−1)kukπ(t).
We begin by applying the Laplace transform to both sides of the DE
L{y ′′}
+ L{y} = L{1}+ 215∑k=1
(−1)kL{ukπ(t)}.
By looking in a table, we’ll see that
L{f ′′(t)
}= s2F (s)− sy(0)− y ′(0), L{1} =
1
s, L{uc (t)} =
e−cs
s.
Therefore,
s2Y (s)− sy(0)− y ′(0) + Y (s) =1
s+ 2
15∑k=1
(−1)ke−kπs
s.
s2Y (s) + Y (s) =1
s+ 2
15∑k=1
(−1)ke−kπs
s.
(Gary Marple) November 10th, 2017 Math 216: Introduction to Differential Equations
s2Y (s) + Y (s) =1
s+ 2
15∑k=1
(−1)ke−kπs
s
(s2 + 1)Y (s) =1
s+ 2
15∑k=1
(−1)ke−kπs
s
Y (s) =1
s(s2 + 1)+ 2
15∑k=1
(−1)ke−kπs
s(s2 + 1)
We need to perform a partial fraction expansion.
1
s(s2 + 1)=
A
s+
Bs + C
s2 + 1
Multiplying both sides by s(s2 + 1) gives us
1 = A(s2 + 1) + (Bs + C)s.
Expanding, moving all terms to the right-hand side, and grouping terms by powers ofs gives us
0 = (A + B)s2 + Cs + (A− 1).
The only way for this equation to be satisfied for all s is if
A + B = 0, C = 0, and A− 1 = 0.
(Gary Marple) November 10th, 2017 Math 216: Introduction to Differential Equations
s2Y (s) + Y (s) =1
s+ 2
15∑k=1
(−1)ke−kπs
s
(s2 + 1)Y (s) =1
s+ 2
15∑k=1
(−1)ke−kπs
s
Y (s) =1
s(s2 + 1)+ 2
15∑k=1
(−1)ke−kπs
s(s2 + 1)
We need to perform a partial fraction expansion.
1
s(s2 + 1)=
A
s+
Bs + C
s2 + 1
Multiplying both sides by s(s2 + 1) gives us
1 = A(s2 + 1) + (Bs + C)s.
Expanding, moving all terms to the right-hand side, and grouping terms by powers ofs gives us
0 = (A + B)s2 + Cs + (A− 1).
The only way for this equation to be satisfied for all s is if
A + B = 0, C = 0, and A− 1 = 0.
(Gary Marple) November 10th, 2017 Math 216: Introduction to Differential Equations
s2Y (s) + Y (s) =1
s+ 2
15∑k=1
(−1)ke−kπs
s
(s2 + 1)Y (s) =1
s+ 2
15∑k=1
(−1)ke−kπs
s
Y (s) =1
s(s2 + 1)+ 2
15∑k=1
(−1)ke−kπs
s(s2 + 1)
We need to perform a partial fraction expansion.
1
s(s2 + 1)=
A
s+
Bs + C
s2 + 1
Multiplying both sides by s(s2 + 1) gives us
1 = A(s2 + 1) + (Bs + C)s.
Expanding, moving all terms to the right-hand side, and grouping terms by powers ofs gives us
0 = (A + B)s2 + Cs + (A− 1).
The only way for this equation to be satisfied for all s is if
A + B = 0, C = 0, and A− 1 = 0.
(Gary Marple) November 10th, 2017 Math 216: Introduction to Differential Equations
s2Y (s) + Y (s) =1
s+ 2
15∑k=1
(−1)ke−kπs
s
(s2 + 1)Y (s) =1
s+ 2
15∑k=1
(−1)ke−kπs
s
Y (s) =1
s(s2 + 1)+ 2
15∑k=1
(−1)ke−kπs
s(s2 + 1)
We need to perform a partial fraction expansion.
1
s(s2 + 1)=
A
s+
Bs + C
s2 + 1
Multiplying both sides by s(s2 + 1) gives us
1 = A(s2 + 1) + (Bs + C)s.
Expanding, moving all terms to the right-hand side, and grouping terms by powers ofs gives us
0 = (A + B)s2 + Cs + (A− 1).
The only way for this equation to be satisfied for all s is if
A + B = 0, C = 0, and A− 1 = 0.
(Gary Marple) November 10th, 2017 Math 216: Introduction to Differential Equations
s2Y (s) + Y (s) =1
s+ 2
15∑k=1
(−1)ke−kπs
s
(s2 + 1)Y (s) =1
s+ 2
15∑k=1
(−1)ke−kπs
s
Y (s) =1
s(s2 + 1)+ 2
15∑k=1
(−1)ke−kπs
s(s2 + 1)
We need to perform a partial fraction expansion.
1
s(s2 + 1)=
A
s+
Bs + C
s2 + 1
Multiplying both sides by s(s2 + 1) gives us
1 = A(s2 + 1) + (Bs + C)s.
Expanding, moving all terms to the right-hand side, and grouping terms by powers ofs gives us
0 = (A + B)s2 + Cs + (A− 1).
The only way for this equation to be satisfied for all s is if
A + B = 0, C = 0, and A− 1 = 0.
(Gary Marple) November 10th, 2017 Math 216: Introduction to Differential Equations
s2Y (s) + Y (s) =1
s+ 2
15∑k=1
(−1)ke−kπs
s
(s2 + 1)Y (s) =1
s+ 2
15∑k=1
(−1)ke−kπs
s
Y (s) =1
s(s2 + 1)+ 2
15∑k=1
(−1)ke−kπs
s(s2 + 1)
We need to perform a partial fraction expansion.
1
s(s2 + 1)=
A
s+
Bs + C
s2 + 1
Multiplying both sides by s(s2 + 1) gives us
1 = A(s2 + 1) + (Bs + C)s.
Expanding, moving all terms to the right-hand side, and grouping terms by powers ofs gives us
0 = (A + B)s2 + Cs + (A− 1).
The only way for this equation to be satisfied for all s is if
A + B = 0, C = 0, and A− 1 = 0.
(Gary Marple) November 10th, 2017 Math 216: Introduction to Differential Equations
s2Y (s) + Y (s) =1
s+ 2
15∑k=1
(−1)ke−kπs
s
(s2 + 1)Y (s) =1
s+ 2
15∑k=1
(−1)ke−kπs
s
Y (s) =1
s(s2 + 1)+ 2
15∑k=1
(−1)ke−kπs
s(s2 + 1)
We need to perform a partial fraction expansion.
1
s(s2 + 1)=
A
s+
Bs + C
s2 + 1
Multiplying both sides by s(s2 + 1) gives us
1 = A(s2 + 1) + (Bs + C)s.
Expanding, moving all terms to the right-hand side, and grouping terms by powers ofs gives us
0 = (A + B)s2 + Cs + (A− 1).
The only way for this equation to be satisfied for all s is if
A + B = 0, C = 0, and A− 1 = 0.
(Gary Marple) November 10th, 2017 Math 216: Introduction to Differential Equations
A + B = 0, C = 0, and A− 1 = 0.
Solving the system tells us that A = 1, B = −1, and C = 0. Recall that the form ofthe partial fraction decomposition was
1
s(s2 + 1)=
A
s+
Bs + C
s2 + 1.
Plugging in our values for A, B, and C gives us
1
s(s2 + 1)=
1
s−
s
s2 + 1.
Recall that
Y (s) =1
s(s2 + 1)+ 2
15∑k=1
(−1)ke−kπs
s(s2 + 1).
Plugging in the result from our partial fraction decomposition gives us
Y (s) =1
s−
s
s2 + 1+ 2
15∑k=1
(−1)ke−kπs
(1
s−
s
s2 + 1
).
(Gary Marple) November 10th, 2017 Math 216: Introduction to Differential Equations
A + B = 0, C = 0, and A− 1 = 0.
Solving the system tells us that A = 1, B = −1, and C = 0. Recall that the form ofthe partial fraction decomposition was
1
s(s2 + 1)=
A
s+
Bs + C
s2 + 1.
Plugging in our values for A, B, and C gives us
1
s(s2 + 1)=
1
s−
s
s2 + 1.
Recall that
Y (s) =1
s(s2 + 1)+ 2
15∑k=1
(−1)ke−kπs
s(s2 + 1).
Plugging in the result from our partial fraction decomposition gives us
Y (s) =1
s−
s
s2 + 1+ 2
15∑k=1
(−1)ke−kπs
(1
s−
s
s2 + 1
).
(Gary Marple) November 10th, 2017 Math 216: Introduction to Differential Equations
A + B = 0, C = 0, and A− 1 = 0.
Solving the system tells us that A = 1, B = −1, and C = 0. Recall that the form ofthe partial fraction decomposition was
1
s(s2 + 1)=
A
s+
Bs + C
s2 + 1.
Plugging in our values for A, B, and C gives us
1
s(s2 + 1)=
1
s−
s
s2 + 1.
Recall that
Y (s) =1
s(s2 + 1)+ 2
15∑k=1
(−1)ke−kπs
s(s2 + 1).
Plugging in the result from our partial fraction decomposition gives us
Y (s) =1
s−
s
s2 + 1+ 2
15∑k=1
(−1)ke−kπs
(1
s−
s
s2 + 1
).
(Gary Marple) November 10th, 2017 Math 216: Introduction to Differential Equations
A + B = 0, C = 0, and A− 1 = 0.
Solving the system tells us that A = 1, B = −1, and C = 0. Recall that the form ofthe partial fraction decomposition was
1
s(s2 + 1)=
A
s+
Bs + C
s2 + 1.
Plugging in our values for A, B, and C gives us
1
s(s2 + 1)=
1
s−
s
s2 + 1.
Recall that
Y (s) =1
s(s2 + 1)+ 2
15∑k=1
(−1)ke−kπs
s(s2 + 1).
Plugging in the result from our partial fraction decomposition gives us
Y (s) =1
s−
s
s2 + 1+ 2
15∑k=1
(−1)ke−kπs
(1
s−
s
s2 + 1
).
(Gary Marple) November 10th, 2017 Math 216: Introduction to Differential Equations
A + B = 0, C = 0, and A− 1 = 0.
Solving the system tells us that A = 1, B = −1, and C = 0. Recall that the form ofthe partial fraction decomposition was
1
s(s2 + 1)=
A
s+
Bs + C
s2 + 1.
Plugging in our values for A, B, and C gives us
1
s(s2 + 1)=
1
s−
s
s2 + 1.
Recall that
Y (s) =1
s(s2 + 1)+ 2
15∑k=1
(−1)ke−kπs
s(s2 + 1).
Plugging in the result from our partial fraction decomposition gives us
Y (s) =1
s−
s
s2 + 1+ 2
15∑k=1
(−1)ke−kπs
(1
s−
s
s2 + 1
).
(Gary Marple) November 10th, 2017 Math 216: Introduction to Differential Equations
Y (s) =1
s−
s
s2 + 1+ 2
15∑k=1
(−1)ke−kπs
(1
s−
s
s2 + 1
)Applying the inverse Laplace transform gives us
y(t) = L−1
{1
s
}− L−1
{s
s2 + 1
}+ 2
15∑k=1
(−1)kL−1
{e−kπs
(1
s−
s
s2 + 1
)}By referring to a table, we see that
L{cos at} =s
s2 + a2, L{1} =
1
s, L{uc (t)f (t − c)} = e−csF (s).
Therefore,
y(t) = 1− cos t + 215∑k=1
(−1)kukπ(t)L−1
{(1
s−
s
s2 + 1
)}t→t−kπ
y(t) = 1− cos t + 215∑k=1
(−1)kukπ(t) (1− cos t)t→t−kπ
y(t) = 1− cos t + 215∑k=1
(−1)kukπ(t) [1− cos (t − kπ)].
(Gary Marple) November 10th, 2017 Math 216: Introduction to Differential Equations
Y (s) =1
s−
s
s2 + 1+ 2
15∑k=1
(−1)ke−kπs
(1
s−
s
s2 + 1
)Applying the inverse Laplace transform gives us
y(t) = L−1
{1
s
}− L−1
{s
s2 + 1
}+ 2
15∑k=1
(−1)kL−1
{e−kπs
(1
s−
s
s2 + 1
)}By referring to a table, we see that
L{cos at} =s
s2 + a2, L{1} =
1
s, L{uc (t)f (t − c)} = e−csF (s).
Therefore,
y(t) = 1− cos t + 215∑k=1
(−1)kukπ(t)L−1
{(1
s−
s
s2 + 1
)}t→t−kπ
y(t) = 1− cos t + 215∑k=1
(−1)kukπ(t) (1− cos t)t→t−kπ
y(t) = 1− cos t + 215∑k=1
(−1)kukπ(t) [1− cos (t − kπ)].
(Gary Marple) November 10th, 2017 Math 216: Introduction to Differential Equations
Y (s) =1
s−
s
s2 + 1+ 2
15∑k=1
(−1)ke−kπs
(1
s−
s
s2 + 1
)Applying the inverse Laplace transform gives us
y(t) = L−1
{1
s
}− L−1
{s
s2 + 1
}+ 2
15∑k=1
(−1)kL−1
{e−kπs
(1
s−
s
s2 + 1
)}By referring to a table, we see that
L{cos at} =s
s2 + a2, L{1} =
1
s, L{uc (t)f (t − c)} = e−csF (s).
Therefore,
y(t) = 1− cos t + 215∑k=1
(−1)kukπ(t)L−1
{(1
s−
s
s2 + 1
)}t→t−kπ
y(t) = 1− cos t + 215∑k=1
(−1)kukπ(t) (1− cos t)t→t−kπ
y(t) = 1− cos t + 215∑k=1
(−1)kukπ(t) [1− cos (t − kπ)].
(Gary Marple) November 10th, 2017 Math 216: Introduction to Differential Equations
Y (s) =1
s−
s
s2 + 1+ 2
15∑k=1
(−1)ke−kπs
(1
s−
s
s2 + 1
)Applying the inverse Laplace transform gives us
y(t) = L−1
{1
s
}− L−1
{s
s2 + 1
}+ 2
15∑k=1
(−1)kL−1
{e−kπs
(1
s−
s
s2 + 1
)}By referring to a table, we see that
L{cos at} =s
s2 + a2, L{1} =
1
s, L{uc (t)f (t − c)} = e−csF (s).
Therefore,
y(t) = 1− cos t + 215∑k=1
(−1)kukπ(t)L−1
{(1
s−
s
s2 + 1
)}t→t−kπ
y(t) = 1− cos t + 215∑k=1
(−1)kukπ(t) (1− cos t)t→t−kπ
y(t) = 1− cos t + 215∑k=1
(−1)kukπ(t) [1− cos (t − kπ)].
(Gary Marple) November 10th, 2017 Math 216: Introduction to Differential Equations
Y (s) =1
s−
s
s2 + 1+ 2
15∑k=1
(−1)ke−kπs
(1
s−
s
s2 + 1
)Applying the inverse Laplace transform gives us
y(t) = L−1
{1
s
}− L−1
{s
s2 + 1
}+ 2
15∑k=1
(−1)kL−1
{e−kπs
(1
s−
s
s2 + 1
)}By referring to a table, we see that
L{cos at} =s
s2 + a2, L{1} =
1
s, L{uc (t)f (t − c)} = e−csF (s).
Therefore,
y(t) = 1− cos t + 215∑k=1
(−1)kukπ(t)L−1
{(1
s−
s
s2 + 1
)}t→t−kπ
y(t) = 1− cos t + 215∑k=1
(−1)kukπ(t) (1− cos t)t→t−kπ
y(t) = 1− cos t + 215∑k=1
(−1)kukπ(t) [1− cos (t − kπ)].
(Gary Marple) November 10th, 2017 Math 216: Introduction to Differential Equations
Y (s) =1
s−
s
s2 + 1+ 2
15∑k=1
(−1)ke−kπs
(1
s−
s
s2 + 1
)Applying the inverse Laplace transform gives us
y(t) = L−1
{1
s
}− L−1
{s
s2 + 1
}+ 2
15∑k=1
(−1)kL−1
{e−kπs
(1
s−
s
s2 + 1
)}By referring to a table, we see that
L{cos at} =s
s2 + a2, L{1} =
1
s, L{uc (t)f (t − c)} = e−csF (s).
Therefore,
y(t) = 1− cos t + 215∑k=1
(−1)kukπ(t)L−1
{(1
s−
s
s2 + 1
)}t→t−kπ
y(t) = 1− cos t + 215∑k=1
(−1)kukπ(t) (1− cos t)t→t−kπ
y(t) = 1− cos t + 215∑k=1
(−1)kukπ(t) [1− cos (t − kπ)].
(Gary Marple) November 10th, 2017 Math 216: Introduction to Differential Equations
y(t) = 1− cos t + 215∑k=1
(−1)kukπ(t) [1− cos (t − kπ)].
(Gary Marple) November 10th, 2017 Math 216: Introduction to Differential Equations
Example
Use the Laplace transform to solve the IVP
y ′′ + 3y ′ + 2y = f (t), y(0) = 0, y ′(0) = 0,
where
f (t) =
{1, 0 ≤ t < 10,0, t ≥ 10.
(Gary Marple) November 10th, 2017 Math 216: Introduction to Differential Equations
Example
Use the Laplace transform to solve the IVP
y ′′ + y = u3π(t), y(0) = 1, y ′(0) = 0.
(Gary Marple) November 10th, 2017 Math 216: Introduction to Differential Equations
Example
Use the Laplace transform to solve the IVP
y ′′ + y = g(t), y(0) = 6, y ′(0) = 8,
where
g(t) =
{t/2, 0 ≤ t < 6,3, t ≥ 6.
(Gary Marple) November 10th, 2017 Math 216: Introduction to Differential Equations
Example
Use the Laplace transform to solve the IVP
y ′′ + 4y = uπ(t)− u3π(t), y(0) = 3, y ′(0) = 7.
(Gary Marple) November 10th, 2017 Math 216: Introduction to Differential Equations