The Laplace Transform of step functions (Sect. 6.3).
I Overview and notation.
I The definition of a step function.
I Piecewise discontinuous functions.
I The Laplace Transform of discontinuous functions.
I Properties of the Laplace Transform.
Overview and notation.
Overview: The Laplace Transform method can be used to solveconstant coefficients differential equations with discontinuoussource functions.
Notation:If L[f (t)] = F (s), then we denote L−1[F (s)] = f (t).
Remark: One can show that for a particular type of functions f ,that includes all functions we work with in this Section, thenotation above is well-defined.
Example
From the Laplace Transform table we know that L[eat
]=
1
s − a.
Then also holds that L−1[ 1
s − a
]= eat . C
Overview and notation.
Overview: The Laplace Transform method can be used to solveconstant coefficients differential equations with discontinuoussource functions.
Notation:If L[f (t)] = F (s), then we denote L−1[F (s)] = f (t).
Remark: One can show that for a particular type of functions f ,that includes all functions we work with in this Section, thenotation above is well-defined.
Example
From the Laplace Transform table we know that L[eat
]=
1
s − a.
Then also holds that L−1[ 1
s − a
]= eat . C
Overview and notation.
Overview: The Laplace Transform method can be used to solveconstant coefficients differential equations with discontinuoussource functions.
Notation:If L[f (t)] = F (s), then we denote L−1[F (s)] = f (t).
Remark: One can show that for a particular type of functions f ,that includes all functions we work with in this Section, thenotation above is well-defined.
Example
From the Laplace Transform table we know that L[eat
]=
1
s − a.
Then also holds that L−1[ 1
s − a
]= eat . C
Overview and notation.
Overview: The Laplace Transform method can be used to solveconstant coefficients differential equations with discontinuoussource functions.
Notation:If L[f (t)] = F (s), then we denote L−1[F (s)] = f (t).
Remark: One can show that for a particular type of functions f ,that includes all functions we work with in this Section, thenotation above is well-defined.
Example
From the Laplace Transform table we know that L[eat
]=
1
s − a.
Then also holds that L−1[ 1
s − a
]= eat . C
Overview and notation.
Overview: The Laplace Transform method can be used to solveconstant coefficients differential equations with discontinuoussource functions.
Notation:If L[f (t)] = F (s), then we denote L−1[F (s)] = f (t).
Remark: One can show that for a particular type of functions f ,that includes all functions we work with in this Section, thenotation above is well-defined.
Example
From the Laplace Transform table we know that L[eat
]=
1
s − a.
Then also holds that L−1[ 1
s − a
]= eat . C
The Laplace Transform of step functions (Sect. 6.3).
I Overview and notation.
I The definition of a step function.
I Piecewise discontinuous functions.
I The Laplace Transform of discontinuous functions.
I Properties of the Laplace Transform.
The definition of a step function.
DefinitionA function u is called a step function at t = 0 iff holds
u(t) =
{0 for t < 0,
1 for t > 0.
Example
Graph the step function values u(t) above, and the translationsu(t − c) and u(t + c) with c > 0.
Solution:u(t)
t
1
0
u(t − c)
t
1
0 c
u(t + c)
0
1
t− c
C
The definition of a step function.
DefinitionA function u is called a step function at t = 0 iff holds
u(t) =
{0 for t < 0,
1 for t > 0.
Example
Graph the step function values u(t) above, and the translationsu(t − c) and u(t + c) with c > 0.
Solution:u(t)
t
1
0
u(t − c)
t
1
0 c
u(t + c)
0
1
t− c
C
The definition of a step function.
DefinitionA function u is called a step function at t = 0 iff holds
u(t) =
{0 for t < 0,
1 for t > 0.
Example
Graph the step function values u(t) above, and the translationsu(t − c) and u(t + c) with c > 0.
Solution:u(t)
t
1
0
u(t − c)
t
1
0 c
u(t + c)
0
1
t− c
C
The definition of a step function.
DefinitionA function u is called a step function at t = 0 iff holds
u(t) =
{0 for t < 0,
1 for t > 0.
Example
Graph the step function values u(t) above, and the translationsu(t − c) and u(t + c) with c > 0.
Solution:u(t)
t
1
0
u(t − c)
t
1
0 c
u(t + c)
0
1
t− c
C
The definition of a step function.
DefinitionA function u is called a step function at t = 0 iff holds
u(t) =
{0 for t < 0,
1 for t > 0.
Example
Graph the step function values u(t) above, and the translationsu(t − c) and u(t + c) with c > 0.
Solution:u(t)
t
1
0
u(t − c)
t
1
0 c
u(t + c)
0
1
t− c
C
The definition of a step function.
Remark: Given any function values f (t) and c > 0, then f (t − c)is a right translation of f and f (t + c) is a left translation of f .
Example
f ( t ) = e
0 t
a t
1
f ( t ) f ( t ) = e
0 t
1
c
a ( t − c )f ( t )
a t
0 t
f ( t )
1
f ( t ) = u ( t ) e f ( t ) = u ( t − c ) e
0 t
1
c
a ( t − c )f ( t )
The definition of a step function.
Remark: Given any function values f (t) and c > 0, then f (t − c)is a right translation of f and f (t + c) is a left translation of f .
Example
f ( t ) = e
0 t
a t
1
f ( t )
f ( t ) = e
0 t
1
c
a ( t − c )f ( t )
a t
0 t
f ( t )
1
f ( t ) = u ( t ) e f ( t ) = u ( t − c ) e
0 t
1
c
a ( t − c )f ( t )
The definition of a step function.
Remark: Given any function values f (t) and c > 0, then f (t − c)is a right translation of f and f (t + c) is a left translation of f .
Example
f ( t ) = e
0 t
a t
1
f ( t ) f ( t ) = e
0 t
1
c
a ( t − c )f ( t )
a t
0 t
f ( t )
1
f ( t ) = u ( t ) e f ( t ) = u ( t − c ) e
0 t
1
c
a ( t − c )f ( t )
The definition of a step function.
Remark: Given any function values f (t) and c > 0, then f (t − c)is a right translation of f and f (t + c) is a left translation of f .
Example
f ( t ) = e
0 t
a t
1
f ( t ) f ( t ) = e
0 t
1
c
a ( t − c )f ( t )
a t
0 t
f ( t )
1
f ( t ) = u ( t ) e
f ( t ) = u ( t − c ) e
0 t
1
c
a ( t − c )f ( t )
The definition of a step function.
Remark: Given any function values f (t) and c > 0, then f (t − c)is a right translation of f and f (t + c) is a left translation of f .
Example
f ( t ) = e
0 t
a t
1
f ( t ) f ( t ) = e
0 t
1
c
a ( t − c )f ( t )
a t
0 t
f ( t )
1
f ( t ) = u ( t ) e f ( t ) = u ( t − c ) e
0 t
1
c
a ( t − c )f ( t )
The Laplace Transform of step functions (Sect. 6.3).
I Overview and notation.
I The definition of a step function.
I Piecewise discontinuous functions.
I The Laplace Transform of discontinuous functions.
I Properties of the Laplace Transform.
Piecewise discontinuous functions.
Example
Graph of the function b(t) = u(t − a)− u(t − b), with 0 < a < b.
Solution: The bump function b can be graphed as follows:
u(t −a)
0 a b
1
t
u(t −b)
0 a b
1
t
t0 a b
b(t)
1
C
Piecewise discontinuous functions.
Example
Graph of the function b(t) = u(t − a)− u(t − b), with 0 < a < b.
Solution: The bump function b can be graphed as follows:
u(t −a)
0 a b
1
t
u(t −b)
0 a b
1
t
t0 a b
b(t)
1
C
Piecewise discontinuous functions.
Example
Graph of the function b(t) = u(t − a)− u(t − b), with 0 < a < b.
Solution: The bump function b can be graphed as follows:
u(t −a)
0 a b
1
t
u(t −b)
0 a b
1
t
t0 a b
b(t)
1
C
Piecewise discontinuous functions.
Example
Graph of the function b(t) = u(t − a)− u(t − b), with 0 < a < b.
Solution: The bump function b can be graphed as follows:
u(t −a)
0 a b
1
t
u(t −b)
0 a b
1
t
t0 a b
b(t)
1
C
Piecewise discontinuous functions.
Example
Graph of the function f (t) = eat[u(t − 1)− u(t − 2)
].
Solution:
a t
t1 2
1
a t
f ( t ) = e [ u ( t −1 ) − u ( t −2 ) ]
[ u ( t −1 ) − u ( t −2 ) ]
e
y
Notation: The function values u(t − c) are denoted in thetextbook as uc(t).
Piecewise discontinuous functions.
Example
Graph of the function f (t) = eat[u(t − 1)− u(t − 2)
].
Solution:
a t
t1 2
1
a t
f ( t ) = e [ u ( t −1 ) − u ( t −2 ) ]
[ u ( t −1 ) − u ( t −2 ) ]
e
y
Notation: The function values u(t − c) are denoted in thetextbook as uc(t).
Piecewise discontinuous functions.
Example
Graph of the function f (t) = eat[u(t − 1)− u(t − 2)
].
Solution:
a t
t1 2
1
a t
f ( t ) = e [ u ( t −1 ) − u ( t −2 ) ]
[ u ( t −1 ) − u ( t −2 ) ]
e
y
Notation: The function values u(t − c) are denoted in thetextbook as uc(t).
The Laplace Transform of step functions (Sect. 6.3).
I Overview and notation.
I The definition of a step function.
I Piecewise discontinuous functions.
I The Laplace Transform of discontinuous functions.
I Properties of the Laplace Transform.
The Laplace Transform of discontinuous functions.
TheoremGiven any real number c > 0, the following equation holds,
L[u(t − c)] =e−cs
s, s > 0.
Proof:
L[u(t − c)] =
∫ ∞
0
e−stu(t − c) dt =
∫ ∞
ce−st dt,
L[u(t − c)] = limN→∞
−1
s
(e−Ns − e−cs
)=
e−cs
s, s > 0.
We conclude that L[u(t − c)] =e−cs
s.
The Laplace Transform of discontinuous functions.
TheoremGiven any real number c > 0, the following equation holds,
L[u(t − c)] =e−cs
s, s > 0.
Proof:
L[u(t − c)] =
∫ ∞
0
e−stu(t − c) dt
=
∫ ∞
ce−st dt,
L[u(t − c)] = limN→∞
−1
s
(e−Ns − e−cs
)=
e−cs
s, s > 0.
We conclude that L[u(t − c)] =e−cs
s.
The Laplace Transform of discontinuous functions.
TheoremGiven any real number c > 0, the following equation holds,
L[u(t − c)] =e−cs
s, s > 0.
Proof:
L[u(t − c)] =
∫ ∞
0
e−stu(t − c) dt =
∫ ∞
ce−st dt,
L[u(t − c)] = limN→∞
−1
s
(e−Ns − e−cs
)=
e−cs
s, s > 0.
We conclude that L[u(t − c)] =e−cs
s.
The Laplace Transform of discontinuous functions.
TheoremGiven any real number c > 0, the following equation holds,
L[u(t − c)] =e−cs
s, s > 0.
Proof:
L[u(t − c)] =
∫ ∞
0
e−stu(t − c) dt =
∫ ∞
ce−st dt,
L[u(t − c)] = limN→∞
−1
s
(e−Ns − e−cs
)
=e−cs
s, s > 0.
We conclude that L[u(t − c)] =e−cs
s.
The Laplace Transform of discontinuous functions.
TheoremGiven any real number c > 0, the following equation holds,
L[u(t − c)] =e−cs
s, s > 0.
Proof:
L[u(t − c)] =
∫ ∞
0
e−stu(t − c) dt =
∫ ∞
ce−st dt,
L[u(t − c)] = limN→∞
−1
s
(e−Ns − e−cs
)=
e−cs
s, s > 0.
We conclude that L[u(t − c)] =e−cs
s.
The Laplace Transform of discontinuous functions.
Example
Compute L[3u(t − 2)].
Solution: L[3u(t − 2)] = 3L[u(t − 2)] = 3e−2s
s.
We conclude: L[3u(t − 2)] =3e−2s
s. C
Example
Compute L−1[e−3s
s
].
Solution: L−1[e−3s
s
]= u(t − 3).
We conclude: L−1[e−3s
s
]= u(t − 3). C
The Laplace Transform of discontinuous functions.
Example
Compute L[3u(t − 2)].
Solution: L[3u(t − 2)] = 3L[u(t − 2)]
= 3e−2s
s.
We conclude: L[3u(t − 2)] =3e−2s
s. C
Example
Compute L−1[e−3s
s
].
Solution: L−1[e−3s
s
]= u(t − 3).
We conclude: L−1[e−3s
s
]= u(t − 3). C
The Laplace Transform of discontinuous functions.
Example
Compute L[3u(t − 2)].
Solution: L[3u(t − 2)] = 3L[u(t − 2)] = 3e−2s
s.
We conclude: L[3u(t − 2)] =3e−2s
s. C
Example
Compute L−1[e−3s
s
].
Solution: L−1[e−3s
s
]= u(t − 3).
We conclude: L−1[e−3s
s
]= u(t − 3). C
The Laplace Transform of discontinuous functions.
Example
Compute L[3u(t − 2)].
Solution: L[3u(t − 2)] = 3L[u(t − 2)] = 3e−2s
s.
We conclude: L[3u(t − 2)] =3e−2s
s. C
Example
Compute L−1[e−3s
s
].
Solution: L−1[e−3s
s
]= u(t − 3).
We conclude: L−1[e−3s
s
]= u(t − 3). C
The Laplace Transform of discontinuous functions.
Example
Compute L[3u(t − 2)].
Solution: L[3u(t − 2)] = 3L[u(t − 2)] = 3e−2s
s.
We conclude: L[3u(t − 2)] =3e−2s
s. C
Example
Compute L−1[e−3s
s
].
Solution: L−1[e−3s
s
]= u(t − 3).
We conclude: L−1[e−3s
s
]= u(t − 3). C
The Laplace Transform of discontinuous functions.
Example
Compute L[3u(t − 2)].
Solution: L[3u(t − 2)] = 3L[u(t − 2)] = 3e−2s
s.
We conclude: L[3u(t − 2)] =3e−2s
s. C
Example
Compute L−1[e−3s
s
].
Solution: L−1[e−3s
s
]
= u(t − 3).
We conclude: L−1[e−3s
s
]= u(t − 3). C
The Laplace Transform of discontinuous functions.
Example
Compute L[3u(t − 2)].
Solution: L[3u(t − 2)] = 3L[u(t − 2)] = 3e−2s
s.
We conclude: L[3u(t − 2)] =3e−2s
s. C
Example
Compute L−1[e−3s
s
].
Solution: L−1[e−3s
s
]= u(t − 3).
We conclude: L−1[e−3s
s
]= u(t − 3). C
The Laplace Transform of discontinuous functions.
Example
Compute L[3u(t − 2)].
Solution: L[3u(t − 2)] = 3L[u(t − 2)] = 3e−2s
s.
We conclude: L[3u(t − 2)] =3e−2s
s. C
Example
Compute L−1[e−3s
s
].
Solution: L−1[e−3s
s
]= u(t − 3).
We conclude: L−1[e−3s
s
]= u(t − 3). C
The Laplace Transform of step functions (Sect. 6.3).
I Overview and notation.
I The definition of a step function.
I Piecewise discontinuous functions.
I The Laplace Transform of discontinuous functions.
I Properties of the Laplace Transform.
Properties of the Laplace Transform.
Theorem (Translations)
If F (s) = L[f (t)] exists for s > a > 0 and c > 0, then holds
L[u(t − c)f (t − c)] = e−cs F (s), s > a.
Furthermore,
L[ect f (t)] = F (s − c), s > a + c .
Remark:
I L[translation (uf )
]= (exp)
(L[f ]
).
I L[(exp) (f )
]= translation
(L[f ]
).
Equivalent notation:
I L[u(t − c)f (t − c)] = e−cs L[f (t)],
I L[ect f (t)] = L[f ](s − c).
Properties of the Laplace Transform.
Theorem (Translations)
If F (s) = L[f (t)] exists for s > a > 0 and c > 0, then holds
L[u(t − c)f (t − c)] = e−cs F (s), s > a.
Furthermore,
L[ect f (t)] = F (s − c), s > a + c .
Remark:
I L[translation (uf )
]= (exp)
(L[f ]
).
I L[(exp) (f )
]= translation
(L[f ]
).
Equivalent notation:
I L[u(t − c)f (t − c)] = e−cs L[f (t)],
I L[ect f (t)] = L[f ](s − c).
Properties of the Laplace Transform.
Theorem (Translations)
If F (s) = L[f (t)] exists for s > a > 0 and c > 0, then holds
L[u(t − c)f (t − c)] = e−cs F (s), s > a.
Furthermore,
L[ect f (t)] = F (s − c), s > a + c .
Remark:
I L[translation (uf )
]= (exp)
(L[f ]
).
I L[(exp) (f )
]= translation
(L[f ]
).
Equivalent notation:
I L[u(t − c)f (t − c)] = e−cs L[f (t)],
I L[ect f (t)] = L[f ](s − c).
Properties of the Laplace Transform.
Theorem (Translations)
If F (s) = L[f (t)] exists for s > a > 0 and c > 0, then holds
L[u(t − c)f (t − c)] = e−cs F (s), s > a.
Furthermore,
L[ect f (t)] = F (s − c), s > a + c .
Remark:
I L[translation (uf )
]= (exp)
(L[f ]
).
I L[(exp) (f )
]= translation
(L[f ]
).
Equivalent notation:
I L[u(t − c)f (t − c)] = e−cs L[f (t)],
I L[ect f (t)] = L[f ](s − c).
Properties of the Laplace Transform.
Theorem (Translations)
If F (s) = L[f (t)] exists for s > a > 0 and c > 0, then holds
L[u(t − c)f (t − c)] = e−cs F (s), s > a.
Furthermore,
L[ect f (t)] = F (s − c), s > a + c .
Remark:
I L[translation (uf )
]= (exp)
(L[f ]
).
I L[(exp) (f )
]= translation
(L[f ]
).
Equivalent notation:
I L[u(t − c)f (t − c)] = e−cs L[f (t)],
I L[ect f (t)] = L[f ](s − c).
Properties of the Laplace Transform.
Example
Compute L[u(t − 2) sin(a(t − 2))
].
Solution: L[sin(at)] =a
s2 + a2, L[u(t − c)f (t − c)] = e−cs L[f (t)].
L[u(t − 2) sin(a(t − 2))
]= e−2s L[sin(at)] = e−2s a
s2 + a2.
We conclude: L[u(t − 2) sin(a(t − 2))
]= e−2s a
s2 + a2. C
Example
Compute L[e3t sin(at)
].
Solution: Recall: L[ect f (t)] = L[f ](s − c).
We conclude: L[e3t sin(at)
]=
a
(s − 3)2 + a2, with s > 3. C
Properties of the Laplace Transform.
Example
Compute L[u(t − 2) sin(a(t − 2))
].
Solution: L[sin(at)] =a
s2 + a2,
L[u(t − c)f (t − c)] = e−cs L[f (t)].
L[u(t − 2) sin(a(t − 2))
]= e−2s L[sin(at)] = e−2s a
s2 + a2.
We conclude: L[u(t − 2) sin(a(t − 2))
]= e−2s a
s2 + a2. C
Example
Compute L[e3t sin(at)
].
Solution: Recall: L[ect f (t)] = L[f ](s − c).
We conclude: L[e3t sin(at)
]=
a
(s − 3)2 + a2, with s > 3. C
Properties of the Laplace Transform.
Example
Compute L[u(t − 2) sin(a(t − 2))
].
Solution: L[sin(at)] =a
s2 + a2, L[u(t − c)f (t − c)] = e−cs L[f (t)].
L[u(t − 2) sin(a(t − 2))
]= e−2s L[sin(at)] = e−2s a
s2 + a2.
We conclude: L[u(t − 2) sin(a(t − 2))
]= e−2s a
s2 + a2. C
Example
Compute L[e3t sin(at)
].
Solution: Recall: L[ect f (t)] = L[f ](s − c).
We conclude: L[e3t sin(at)
]=
a
(s − 3)2 + a2, with s > 3. C
Properties of the Laplace Transform.
Example
Compute L[u(t − 2) sin(a(t − 2))
].
Solution: L[sin(at)] =a
s2 + a2, L[u(t − c)f (t − c)] = e−cs L[f (t)].
L[u(t − 2) sin(a(t − 2))
]= e−2s L[sin(at)]
= e−2s a
s2 + a2.
We conclude: L[u(t − 2) sin(a(t − 2))
]= e−2s a
s2 + a2. C
Example
Compute L[e3t sin(at)
].
Solution: Recall: L[ect f (t)] = L[f ](s − c).
We conclude: L[e3t sin(at)
]=
a
(s − 3)2 + a2, with s > 3. C
Properties of the Laplace Transform.
Example
Compute L[u(t − 2) sin(a(t − 2))
].
Solution: L[sin(at)] =a
s2 + a2, L[u(t − c)f (t − c)] = e−cs L[f (t)].
L[u(t − 2) sin(a(t − 2))
]= e−2s L[sin(at)] = e−2s a
s2 + a2.
We conclude: L[u(t − 2) sin(a(t − 2))
]= e−2s a
s2 + a2. C
Example
Compute L[e3t sin(at)
].
Solution: Recall: L[ect f (t)] = L[f ](s − c).
We conclude: L[e3t sin(at)
]=
a
(s − 3)2 + a2, with s > 3. C
Properties of the Laplace Transform.
Example
Compute L[u(t − 2) sin(a(t − 2))
].
Solution: L[sin(at)] =a
s2 + a2, L[u(t − c)f (t − c)] = e−cs L[f (t)].
L[u(t − 2) sin(a(t − 2))
]= e−2s L[sin(at)] = e−2s a
s2 + a2.
We conclude: L[u(t − 2) sin(a(t − 2))
]= e−2s a
s2 + a2. C
Example
Compute L[e3t sin(at)
].
Solution: Recall: L[ect f (t)] = L[f ](s − c).
We conclude: L[e3t sin(at)
]=
a
(s − 3)2 + a2, with s > 3. C
Properties of the Laplace Transform.
Example
Compute L[u(t − 2) sin(a(t − 2))
].
Solution: L[sin(at)] =a
s2 + a2, L[u(t − c)f (t − c)] = e−cs L[f (t)].
L[u(t − 2) sin(a(t − 2))
]= e−2s L[sin(at)] = e−2s a
s2 + a2.
We conclude: L[u(t − 2) sin(a(t − 2))
]= e−2s a
s2 + a2. C
Example
Compute L[e3t sin(at)
].
Solution: Recall: L[ect f (t)] = L[f ](s − c).
We conclude: L[e3t sin(at)
]=
a
(s − 3)2 + a2, with s > 3. C
Properties of the Laplace Transform.
Example
Compute L[u(t − 2) sin(a(t − 2))
].
Solution: L[sin(at)] =a
s2 + a2, L[u(t − c)f (t − c)] = e−cs L[f (t)].
L[u(t − 2) sin(a(t − 2))
]= e−2s L[sin(at)] = e−2s a
s2 + a2.
We conclude: L[u(t − 2) sin(a(t − 2))
]= e−2s a
s2 + a2. C
Example
Compute L[e3t sin(at)
].
Solution: Recall: L[ect f (t)] = L[f ](s − c).
We conclude: L[e3t sin(at)
]=
a
(s − 3)2 + a2, with s > 3. C
Properties of the Laplace Transform.
Example
Compute L[u(t − 2) sin(a(t − 2))
].
Solution: L[sin(at)] =a
s2 + a2, L[u(t − c)f (t − c)] = e−cs L[f (t)].
L[u(t − 2) sin(a(t − 2))
]= e−2s L[sin(at)] = e−2s a
s2 + a2.
We conclude: L[u(t − 2) sin(a(t − 2))
]= e−2s a
s2 + a2. C
Example
Compute L[e3t sin(at)
].
Solution: Recall: L[ect f (t)] = L[f ](s − c).
We conclude: L[e3t sin(at)
]=
a
(s − 3)2 + a2, with s > 3. C
Properties of the Laplace Transform.
Example
Find the Laplace transform of f (t) =
{0, t < 1,
(t2 − 2t + 2), t > 1.
Solution: Using step function notation,
f (t) = u(t − 1)(t2 − 2t + 2).
Completing the square we obtain,
t2 − 2t + 2 = (t2 − 2t + 1)− 1 + 2 = (t − 1)2 + 1.
This is a parabola t2 translated to theright by 1 and up by one. This is adiscontinuous function.
1
10
f ( t )
t
Properties of the Laplace Transform.
Example
Find the Laplace transform of f (t) =
{0, t < 1,
(t2 − 2t + 2), t > 1.
Solution: Using step function notation,
f (t) = u(t − 1)(t2 − 2t + 2).
Completing the square we obtain,
t2 − 2t + 2 = (t2 − 2t + 1)− 1 + 2 = (t − 1)2 + 1.
This is a parabola t2 translated to theright by 1 and up by one. This is adiscontinuous function.
1
10
f ( t )
t
Properties of the Laplace Transform.
Example
Find the Laplace transform of f (t) =
{0, t < 1,
(t2 − 2t + 2), t > 1.
Solution: Using step function notation,
f (t) = u(t − 1)(t2 − 2t + 2).
Completing the square we obtain,
t2 − 2t + 2 = (t2 − 2t + 1)− 1 + 2 = (t − 1)2 + 1.
This is a parabola t2 translated to theright by 1 and up by one. This is adiscontinuous function.
1
10
f ( t )
t
Properties of the Laplace Transform.
Example
Find the Laplace transform of f (t) =
{0, t < 1,
(t2 − 2t + 2), t > 1.
Solution: Using step function notation,
f (t) = u(t − 1)(t2 − 2t + 2).
Completing the square we obtain,
t2 − 2t + 2 = (t2 − 2t + 1)− 1 + 2
= (t − 1)2 + 1.
This is a parabola t2 translated to theright by 1 and up by one. This is adiscontinuous function.
1
10
f ( t )
t
Properties of the Laplace Transform.
Example
Find the Laplace transform of f (t) =
{0, t < 1,
(t2 − 2t + 2), t > 1.
Solution: Using step function notation,
f (t) = u(t − 1)(t2 − 2t + 2).
Completing the square we obtain,
t2 − 2t + 2 = (t2 − 2t + 1)− 1 + 2 = (t − 1)2 + 1.
This is a parabola t2 translated to theright by 1 and up by one. This is adiscontinuous function.
1
10
f ( t )
t
Properties of the Laplace Transform.
Example
Find the Laplace transform of f (t) =
{0, t < 1,
(t2 − 2t + 2), t > 1.
Solution: Using step function notation,
f (t) = u(t − 1)(t2 − 2t + 2).
Completing the square we obtain,
t2 − 2t + 2 = (t2 − 2t + 1)− 1 + 2 = (t − 1)2 + 1.
This is a parabola t2 translated to theright by 1 and up by one. This is adiscontinuous function.
1
10
f ( t )
t
Properties of the Laplace Transform.
Example
Find the Laplace transform of f (t) =
{0, t < 1,
(t2 − 2t + 2), t > 1.
Solution: Using step function notation,
f (t) = u(t − 1)(t2 − 2t + 2).
Completing the square we obtain,
t2 − 2t + 2 = (t2 − 2t + 1)− 1 + 2 = (t − 1)2 + 1.
This is a parabola t2 translated to theright by 1 and up by one. This is adiscontinuous function.
1
10
f ( t )
t
Properties of the Laplace Transform.
Example
Find the Laplace transform of f (t) =
{0, t < 1,
(t2 − 2t + 2), t > 1.
Solution: Recall: f (t) = u(t − 1)[(t − 1)2 + 1
].
This is equivalent to
f (t) = u(t − 1) (t − 1)2 + u(t − 1).
Since L[t2] = 2/s3, and L[u(t − c)g(t − c)] = e−cs L[g(t)], then
L[f (t)] = L[u(t − 1) (t − 1)2] + L[u(t − 1)] = e−s 2
s3+ e−s 1
s.
We conclude: L[f (t)] =e−s
s3
(2 + s2
). C
Properties of the Laplace Transform.
Example
Find the Laplace transform of f (t) =
{0, t < 1,
(t2 − 2t + 2), t > 1.
Solution: Recall: f (t) = u(t − 1)[(t − 1)2 + 1
].
This is equivalent to
f (t) = u(t − 1) (t − 1)2 + u(t − 1).
Since L[t2] = 2/s3, and L[u(t − c)g(t − c)] = e−cs L[g(t)], then
L[f (t)] = L[u(t − 1) (t − 1)2] + L[u(t − 1)] = e−s 2
s3+ e−s 1
s.
We conclude: L[f (t)] =e−s
s3
(2 + s2
). C
Properties of the Laplace Transform.
Example
Find the Laplace transform of f (t) =
{0, t < 1,
(t2 − 2t + 2), t > 1.
Solution: Recall: f (t) = u(t − 1)[(t − 1)2 + 1
].
This is equivalent to
f (t) = u(t − 1) (t − 1)2 + u(t − 1).
Since L[t2] = 2/s3,
and L[u(t − c)g(t − c)] = e−cs L[g(t)], then
L[f (t)] = L[u(t − 1) (t − 1)2] + L[u(t − 1)] = e−s 2
s3+ e−s 1
s.
We conclude: L[f (t)] =e−s
s3
(2 + s2
). C
Properties of the Laplace Transform.
Example
Find the Laplace transform of f (t) =
{0, t < 1,
(t2 − 2t + 2), t > 1.
Solution: Recall: f (t) = u(t − 1)[(t − 1)2 + 1
].
This is equivalent to
f (t) = u(t − 1) (t − 1)2 + u(t − 1).
Since L[t2] = 2/s3, and L[u(t − c)g(t − c)] = e−cs L[g(t)],
then
L[f (t)] = L[u(t − 1) (t − 1)2] + L[u(t − 1)] = e−s 2
s3+ e−s 1
s.
We conclude: L[f (t)] =e−s
s3
(2 + s2
). C
Properties of the Laplace Transform.
Example
Find the Laplace transform of f (t) =
{0, t < 1,
(t2 − 2t + 2), t > 1.
Solution: Recall: f (t) = u(t − 1)[(t − 1)2 + 1
].
This is equivalent to
f (t) = u(t − 1) (t − 1)2 + u(t − 1).
Since L[t2] = 2/s3, and L[u(t − c)g(t − c)] = e−cs L[g(t)], then
L[f (t)] = L[u(t − 1) (t − 1)2] + L[u(t − 1)]
= e−s 2
s3+ e−s 1
s.
We conclude: L[f (t)] =e−s
s3
(2 + s2
). C
Properties of the Laplace Transform.
Example
Find the Laplace transform of f (t) =
{0, t < 1,
(t2 − 2t + 2), t > 1.
Solution: Recall: f (t) = u(t − 1)[(t − 1)2 + 1
].
This is equivalent to
f (t) = u(t − 1) (t − 1)2 + u(t − 1).
Since L[t2] = 2/s3, and L[u(t − c)g(t − c)] = e−cs L[g(t)], then
L[f (t)] = L[u(t − 1) (t − 1)2] + L[u(t − 1)] = e−s 2
s3+ e−s 1
s.
We conclude: L[f (t)] =e−s
s3
(2 + s2
). C
Properties of the Laplace Transform.
Example
Find the Laplace transform of f (t) =
{0, t < 1,
(t2 − 2t + 2), t > 1.
Solution: Recall: f (t) = u(t − 1)[(t − 1)2 + 1
].
This is equivalent to
f (t) = u(t − 1) (t − 1)2 + u(t − 1).
Since L[t2] = 2/s3, and L[u(t − c)g(t − c)] = e−cs L[g(t)], then
L[f (t)] = L[u(t − 1) (t − 1)2] + L[u(t − 1)] = e−s 2
s3+ e−s 1
s.
We conclude: L[f (t)] =e−s
s3
(2 + s2
). C
Properties of the Laplace Transform.
Remark: The inverse of the formulas in the Theorem above are:
L−1[e−cs F (s)
]= u(t − c) f (t − c),
L−1[F (s − c)
]= ect f (t).
Example
Find L−1[ e−4s
s2 + 9
].
Solution: L−1[ e−4s
s2 + 9
]=
1
3L−1
[e−4s 3
s2 + 9
].
Recall: L−1[ a
s2 + a2
]= sin(at). Then, we conclude that
L−1[ e−4s
s2 + 9
]=
1
3u(t − 4) sin
(3(t − 4)
). C
Properties of the Laplace Transform.
Remark: The inverse of the formulas in the Theorem above are:
L−1[e−cs F (s)
]= u(t − c) f (t − c),
L−1[F (s − c)
]= ect f (t).
Example
Find L−1[ e−4s
s2 + 9
].
Solution: L−1[ e−4s
s2 + 9
]=
1
3L−1
[e−4s 3
s2 + 9
].
Recall: L−1[ a
s2 + a2
]= sin(at). Then, we conclude that
L−1[ e−4s
s2 + 9
]=
1
3u(t − 4) sin
(3(t − 4)
). C
Properties of the Laplace Transform.
Remark: The inverse of the formulas in the Theorem above are:
L−1[e−cs F (s)
]= u(t − c) f (t − c),
L−1[F (s − c)
]= ect f (t).
Example
Find L−1[ e−4s
s2 + 9
].
Solution: L−1[ e−4s
s2 + 9
]=
1
3L−1
[e−4s 3
s2 + 9
].
Recall: L−1[ a
s2 + a2
]= sin(at). Then, we conclude that
L−1[ e−4s
s2 + 9
]=
1
3u(t − 4) sin
(3(t − 4)
). C
Properties of the Laplace Transform.
Remark: The inverse of the formulas in the Theorem above are:
L−1[e−cs F (s)
]= u(t − c) f (t − c),
L−1[F (s − c)
]= ect f (t).
Example
Find L−1[ e−4s
s2 + 9
].
Solution: L−1[ e−4s
s2 + 9
]=
1
3L−1
[e−4s 3
s2 + 9
].
Recall: L−1[ a
s2 + a2
]= sin(at). Then, we conclude that
L−1[ e−4s
s2 + 9
]=
1
3u(t − 4) sin
(3(t − 4)
). C
Properties of the Laplace Transform.
Remark: The inverse of the formulas in the Theorem above are:
L−1[e−cs F (s)
]= u(t − c) f (t − c),
L−1[F (s − c)
]= ect f (t).
Example
Find L−1[ e−4s
s2 + 9
].
Solution: L−1[ e−4s
s2 + 9
]=
1
3L−1
[e−4s 3
s2 + 9
].
Recall: L−1[ a
s2 + a2
]= sin(at).
Then, we conclude that
L−1[ e−4s
s2 + 9
]=
1
3u(t − 4) sin
(3(t − 4)
). C
Properties of the Laplace Transform.
Remark: The inverse of the formulas in the Theorem above are:
L−1[e−cs F (s)
]= u(t − c) f (t − c),
L−1[F (s − c)
]= ect f (t).
Example
Find L−1[ e−4s
s2 + 9
].
Solution: L−1[ e−4s
s2 + 9
]=
1
3L−1
[e−4s 3
s2 + 9
].
Recall: L−1[ a
s2 + a2
]= sin(at). Then, we conclude that
L−1[ e−4s
s2 + 9
]=
1
3u(t − 4) sin
(3(t − 4)
). C
Properties of the Laplace Transform.
Example
Find L−1[ (s − 2)
(s − 2)2 + 9
].
Solution: L−1[ s
s2 + a2
]= cos(at), L−1
[F (s − c)
]= ect f (t).
We conclude: L−1[ (s − 2)
(s − 2)2 + 9
]= e2t cos(3t). C
Example
Find L−1[ 2e−3s
s2 − 4
].
Solution: Recall: L−1[ a
s2 − a2
]= sinh(at)
and L−1[e−cs F (s)
]= u(t − c) f (t − c).
Properties of the Laplace Transform.
Example
Find L−1[ (s − 2)
(s − 2)2 + 9
].
Solution: L−1[ s
s2 + a2
]= cos(at),
L−1[F (s − c)
]= ect f (t).
We conclude: L−1[ (s − 2)
(s − 2)2 + 9
]= e2t cos(3t). C
Example
Find L−1[ 2e−3s
s2 − 4
].
Solution: Recall: L−1[ a
s2 − a2
]= sinh(at)
and L−1[e−cs F (s)
]= u(t − c) f (t − c).
Properties of the Laplace Transform.
Example
Find L−1[ (s − 2)
(s − 2)2 + 9
].
Solution: L−1[ s
s2 + a2
]= cos(at), L−1
[F (s − c)
]= ect f (t).
We conclude: L−1[ (s − 2)
(s − 2)2 + 9
]= e2t cos(3t). C
Example
Find L−1[ 2e−3s
s2 − 4
].
Solution: Recall: L−1[ a
s2 − a2
]= sinh(at)
and L−1[e−cs F (s)
]= u(t − c) f (t − c).
Properties of the Laplace Transform.
Example
Find L−1[ (s − 2)
(s − 2)2 + 9
].
Solution: L−1[ s
s2 + a2
]= cos(at), L−1
[F (s − c)
]= ect f (t).
We conclude: L−1[ (s − 2)
(s − 2)2 + 9
]= e2t cos(3t). C
Example
Find L−1[ 2e−3s
s2 − 4
].
Solution: Recall: L−1[ a
s2 − a2
]= sinh(at)
and L−1[e−cs F (s)
]= u(t − c) f (t − c).
Properties of the Laplace Transform.
Example
Find L−1[ (s − 2)
(s − 2)2 + 9
].
Solution: L−1[ s
s2 + a2
]= cos(at), L−1
[F (s − c)
]= ect f (t).
We conclude: L−1[ (s − 2)
(s − 2)2 + 9
]= e2t cos(3t). C
Example
Find L−1[ 2e−3s
s2 − 4
].
Solution: Recall: L−1[ a
s2 − a2
]= sinh(at)
and L−1[e−cs F (s)
]= u(t − c) f (t − c).
Properties of the Laplace Transform.
Example
Find L−1[ (s − 2)
(s − 2)2 + 9
].
Solution: L−1[ s
s2 + a2
]= cos(at), L−1
[F (s − c)
]= ect f (t).
We conclude: L−1[ (s − 2)
(s − 2)2 + 9
]= e2t cos(3t). C
Example
Find L−1[ 2e−3s
s2 − 4
].
Solution: Recall: L−1[ a
s2 − a2
]= sinh(at)
and L−1[e−cs F (s)
]= u(t − c) f (t − c).
Properties of the Laplace Transform.
Example
Find L−1[ (s − 2)
(s − 2)2 + 9
].
Solution: L−1[ s
s2 + a2
]= cos(at), L−1
[F (s − c)
]= ect f (t).
We conclude: L−1[ (s − 2)
(s − 2)2 + 9
]= e2t cos(3t). C
Example
Find L−1[ 2e−3s
s2 − 4
].
Solution: Recall: L−1[ a
s2 − a2
]= sinh(at)
and L−1[e−cs F (s)
]= u(t − c) f (t − c).
Properties of the Laplace Transform.
Example
Find L−1[ 2e−3s
s2 − 4
].
Solution: Recall:
L−1[ a
s2 − a2
]= sinh(at), L−1
[e−cs F (s)
]= u(t − c) f (t − c).
L−1[ 2e−3s
s2 − 4
]= L−1
[e−3s 2
s2 − 4
].
We conclude: L−1[ 2e−3s
s2 − 4
]= u(t − 3) sinh
(2(t − 3)
). C
Properties of the Laplace Transform.
Example
Find L−1[ 2e−3s
s2 − 4
].
Solution: Recall:
L−1[ a
s2 − a2
]= sinh(at), L−1
[e−cs F (s)
]= u(t − c) f (t − c).
L−1[ 2e−3s
s2 − 4
]= L−1
[e−3s 2
s2 − 4
].
We conclude: L−1[ 2e−3s
s2 − 4
]= u(t − 3) sinh
(2(t − 3)
). C
Properties of the Laplace Transform.
Example
Find L−1[ 2e−3s
s2 − 4
].
Solution: Recall:
L−1[ a
s2 − a2
]= sinh(at), L−1
[e−cs F (s)
]= u(t − c) f (t − c).
L−1[ 2e−3s
s2 − 4
]= L−1
[e−3s 2
s2 − 4
].
We conclude: L−1[ 2e−3s
s2 − 4
]= u(t − 3) sinh
(2(t − 3)
). C
Properties of the Laplace Transform.
Example
Find L−1[ e−2s
s2 + s − 2
].
Solution: Find the roots of the denominator:
s± =1
2
[−1±
√1 + 8
]⇒
{s+ = 1,
s− = −2.
Therefore, s2 + s − 2 = (s − 1) (s + 2).
Use partial fractions to simplify the rational function:
1
s2 + s − 2=
1
(s − 1) (s + 2)=
a
(s − 1)+
b
(s + 2),
1
s2 + s − 2= a(s + 2) + b(s − 1) =
(a + b) s + (2a− b)
(s − 1) (s + 2).
Properties of the Laplace Transform.
Example
Find L−1[ e−2s
s2 + s − 2
].
Solution: Find the roots of the denominator:
s± =1
2
[−1±
√1 + 8
]
⇒
{s+ = 1,
s− = −2.
Therefore, s2 + s − 2 = (s − 1) (s + 2).
Use partial fractions to simplify the rational function:
1
s2 + s − 2=
1
(s − 1) (s + 2)=
a
(s − 1)+
b
(s + 2),
1
s2 + s − 2= a(s + 2) + b(s − 1) =
(a + b) s + (2a− b)
(s − 1) (s + 2).
Properties of the Laplace Transform.
Example
Find L−1[ e−2s
s2 + s − 2
].
Solution: Find the roots of the denominator:
s± =1
2
[−1±
√1 + 8
]⇒
{s+ = 1,
s− = −2.
Therefore, s2 + s − 2 = (s − 1) (s + 2).
Use partial fractions to simplify the rational function:
1
s2 + s − 2=
1
(s − 1) (s + 2)=
a
(s − 1)+
b
(s + 2),
1
s2 + s − 2= a(s + 2) + b(s − 1) =
(a + b) s + (2a− b)
(s − 1) (s + 2).
Properties of the Laplace Transform.
Example
Find L−1[ e−2s
s2 + s − 2
].
Solution: Find the roots of the denominator:
s± =1
2
[−1±
√1 + 8
]⇒
{s+ = 1,
s− = −2.
Therefore, s2 + s − 2 = (s − 1) (s + 2).
Use partial fractions to simplify the rational function:
1
s2 + s − 2=
1
(s − 1) (s + 2)=
a
(s − 1)+
b
(s + 2),
1
s2 + s − 2= a(s + 2) + b(s − 1) =
(a + b) s + (2a− b)
(s − 1) (s + 2).
Properties of the Laplace Transform.
Example
Find L−1[ e−2s
s2 + s − 2
].
Solution: Find the roots of the denominator:
s± =1
2
[−1±
√1 + 8
]⇒
{s+ = 1,
s− = −2.
Therefore, s2 + s − 2 = (s − 1) (s + 2).
Use partial fractions to simplify the rational function:
1
s2 + s − 2=
1
(s − 1) (s + 2)
=a
(s − 1)+
b
(s + 2),
1
s2 + s − 2= a(s + 2) + b(s − 1) =
(a + b) s + (2a− b)
(s − 1) (s + 2).
Properties of the Laplace Transform.
Example
Find L−1[ e−2s
s2 + s − 2
].
Solution: Find the roots of the denominator:
s± =1
2
[−1±
√1 + 8
]⇒
{s+ = 1,
s− = −2.
Therefore, s2 + s − 2 = (s − 1) (s + 2).
Use partial fractions to simplify the rational function:
1
s2 + s − 2=
1
(s − 1) (s + 2)=
a
(s − 1)+
b
(s + 2),
1
s2 + s − 2= a(s + 2) + b(s − 1) =
(a + b) s + (2a− b)
(s − 1) (s + 2).
Properties of the Laplace Transform.
Example
Find L−1[ e−2s
s2 + s − 2
].
Solution: Find the roots of the denominator:
s± =1
2
[−1±
√1 + 8
]⇒
{s+ = 1,
s− = −2.
Therefore, s2 + s − 2 = (s − 1) (s + 2).
Use partial fractions to simplify the rational function:
1
s2 + s − 2=
1
(s − 1) (s + 2)=
a
(s − 1)+
b
(s + 2),
1
s2 + s − 2= a(s + 2) + b(s − 1)
=(a + b) s + (2a− b)
(s − 1) (s + 2).
Properties of the Laplace Transform.
Example
Find L−1[ e−2s
s2 + s − 2
].
Solution: Find the roots of the denominator:
s± =1
2
[−1±
√1 + 8
]⇒
{s+ = 1,
s− = −2.
Therefore, s2 + s − 2 = (s − 1) (s + 2).
Use partial fractions to simplify the rational function:
1
s2 + s − 2=
1
(s − 1) (s + 2)=
a
(s − 1)+
b
(s + 2),
1
s2 + s − 2= a(s + 2) + b(s − 1) =
(a + b) s + (2a− b)
(s − 1) (s + 2).
Properties of the Laplace Transform.
Example
Find L−1[ e−2s
s2 + s − 2
].
Solution: Recall:1
s2 + s − 2=
(a + b) s + (2a− b)
(s − 1) (s + 2)
a + b = 0, 2a− b = 1, ⇒ a =1
3, b = −1
3.
L−1[ e−2s
s2 + s − 2
]=
1
3L−1
[e−2s 1
s − 1
]− 1
3L−1
[e−2s 1
s + 2
].
Recall: L−1[ 1
s − a
]= eat , L−1
[e−cs F (s)
]= u(t − c) f (t − c),
L−1[ e−2s
s2 + s − 2
]=
1
3u(t − 2) e(t−2) − 1
3u(t − 2) e−2(t−2).
Hence: L−1[ e−2s
s2 + s − 2
]=
1
3u(t − 2)
[e(t−2) − e−2(t−2)
]. C
Properties of the Laplace Transform.
Example
Find L−1[ e−2s
s2 + s − 2
].
Solution: Recall:1
s2 + s − 2=
(a + b) s + (2a− b)
(s − 1) (s + 2)
a + b = 0,
2a− b = 1, ⇒ a =1
3, b = −1
3.
L−1[ e−2s
s2 + s − 2
]=
1
3L−1
[e−2s 1
s − 1
]− 1
3L−1
[e−2s 1
s + 2
].
Recall: L−1[ 1
s − a
]= eat , L−1
[e−cs F (s)
]= u(t − c) f (t − c),
L−1[ e−2s
s2 + s − 2
]=
1
3u(t − 2) e(t−2) − 1
3u(t − 2) e−2(t−2).
Hence: L−1[ e−2s
s2 + s − 2
]=
1
3u(t − 2)
[e(t−2) − e−2(t−2)
]. C
Properties of the Laplace Transform.
Example
Find L−1[ e−2s
s2 + s − 2
].
Solution: Recall:1
s2 + s − 2=
(a + b) s + (2a− b)
(s − 1) (s + 2)
a + b = 0, 2a− b = 1,
⇒ a =1
3, b = −1
3.
L−1[ e−2s
s2 + s − 2
]=
1
3L−1
[e−2s 1
s − 1
]− 1
3L−1
[e−2s 1
s + 2
].
Recall: L−1[ 1
s − a
]= eat , L−1
[e−cs F (s)
]= u(t − c) f (t − c),
L−1[ e−2s
s2 + s − 2
]=
1
3u(t − 2) e(t−2) − 1
3u(t − 2) e−2(t−2).
Hence: L−1[ e−2s
s2 + s − 2
]=
1
3u(t − 2)
[e(t−2) − e−2(t−2)
]. C
Properties of the Laplace Transform.
Example
Find L−1[ e−2s
s2 + s − 2
].
Solution: Recall:1
s2 + s − 2=
(a + b) s + (2a− b)
(s − 1) (s + 2)
a + b = 0, 2a− b = 1, ⇒ a =1
3, b = −1
3.
L−1[ e−2s
s2 + s − 2
]=
1
3L−1
[e−2s 1
s − 1
]− 1
3L−1
[e−2s 1
s + 2
].
Recall: L−1[ 1
s − a
]= eat , L−1
[e−cs F (s)
]= u(t − c) f (t − c),
L−1[ e−2s
s2 + s − 2
]=
1
3u(t − 2) e(t−2) − 1
3u(t − 2) e−2(t−2).
Hence: L−1[ e−2s
s2 + s − 2
]=
1
3u(t − 2)
[e(t−2) − e−2(t−2)
]. C
Properties of the Laplace Transform.
Example
Find L−1[ e−2s
s2 + s − 2
].
Solution: Recall:1
s2 + s − 2=
(a + b) s + (2a− b)
(s − 1) (s + 2)
a + b = 0, 2a− b = 1, ⇒ a =1
3, b = −1
3.
L−1[ e−2s
s2 + s − 2
]=
1
3L−1
[e−2s 1
s − 1
]− 1
3L−1
[e−2s 1
s + 2
].
Recall: L−1[ 1
s − a
]= eat , L−1
[e−cs F (s)
]= u(t − c) f (t − c),
L−1[ e−2s
s2 + s − 2
]=
1
3u(t − 2) e(t−2) − 1
3u(t − 2) e−2(t−2).
Hence: L−1[ e−2s
s2 + s − 2
]=
1
3u(t − 2)
[e(t−2) − e−2(t−2)
]. C
Properties of the Laplace Transform.
Example
Find L−1[ e−2s
s2 + s − 2
].
Solution: Recall:1
s2 + s − 2=
(a + b) s + (2a− b)
(s − 1) (s + 2)
a + b = 0, 2a− b = 1, ⇒ a =1
3, b = −1
3.
L−1[ e−2s
s2 + s − 2
]=
1
3L−1
[e−2s 1
s − 1
]− 1
3L−1
[e−2s 1
s + 2
].
Recall: L−1[ 1
s − a
]= eat ,
L−1[e−cs F (s)
]= u(t − c) f (t − c),
L−1[ e−2s
s2 + s − 2
]=
1
3u(t − 2) e(t−2) − 1
3u(t − 2) e−2(t−2).
Hence: L−1[ e−2s
s2 + s − 2
]=
1
3u(t − 2)
[e(t−2) − e−2(t−2)
]. C
Properties of the Laplace Transform.
Example
Find L−1[ e−2s
s2 + s − 2
].
Solution: Recall:1
s2 + s − 2=
(a + b) s + (2a− b)
(s − 1) (s + 2)
a + b = 0, 2a− b = 1, ⇒ a =1
3, b = −1
3.
L−1[ e−2s
s2 + s − 2
]=
1
3L−1
[e−2s 1
s − 1
]− 1
3L−1
[e−2s 1
s + 2
].
Recall: L−1[ 1
s − a
]= eat , L−1
[e−cs F (s)
]= u(t − c) f (t − c),
L−1[ e−2s
s2 + s − 2
]=
1
3u(t − 2) e(t−2) − 1
3u(t − 2) e−2(t−2).
Hence: L−1[ e−2s
s2 + s − 2
]=
1
3u(t − 2)
[e(t−2) − e−2(t−2)
]. C
Properties of the Laplace Transform.
Example
Find L−1[ e−2s
s2 + s − 2
].
Solution: Recall:1
s2 + s − 2=
(a + b) s + (2a− b)
(s − 1) (s + 2)
a + b = 0, 2a− b = 1, ⇒ a =1
3, b = −1
3.
L−1[ e−2s
s2 + s − 2
]=
1
3L−1
[e−2s 1
s − 1
]− 1
3L−1
[e−2s 1
s + 2
].
Recall: L−1[ 1
s − a
]= eat , L−1
[e−cs F (s)
]= u(t − c) f (t − c),
L−1[ e−2s
s2 + s − 2
]=
1
3u(t − 2) e(t−2) − 1
3u(t − 2) e−2(t−2).
Hence: L−1[ e−2s
s2 + s − 2
]=
1
3u(t − 2)
[e(t−2) − e−2(t−2)
]. C
Properties of the Laplace Transform.
Example
Find L−1[ e−2s
s2 + s − 2
].
Solution: Recall:1
s2 + s − 2=
(a + b) s + (2a− b)
(s − 1) (s + 2)
a + b = 0, 2a− b = 1, ⇒ a =1
3, b = −1
3.
L−1[ e−2s
s2 + s − 2
]=
1
3L−1
[e−2s 1
s − 1
]− 1
3L−1
[e−2s 1
s + 2
].
Recall: L−1[ 1
s − a
]= eat , L−1
[e−cs F (s)
]= u(t − c) f (t − c),
L−1[ e−2s
s2 + s − 2
]=
1
3u(t − 2) e(t−2) − 1
3u(t − 2) e−2(t−2).
Hence: L−1[ e−2s
s2 + s − 2
]=
1
3u(t − 2)
[e(t−2) − e−2(t−2)
]. C
Equations with discontinuous sources (Sect. 6.4).
I Differential equations with discontinuous sources.I We solve the IVPs:
(a) Example 1:
y ′ + 2y = u(t − 4), y(0) = 3.
(b) Example 2:
y ′′ + y ′ +5
4y = b(t),
y(0) = 0,
y ′(0) = 0,b(t) =
{1, t ∈ [0, π)
0, t ∈ [π,∞).
(c) Example 3:
y ′′+y ′+5
4y = g(t),
y(0) = 0,
y ′(0) = 0,g(t) =
{sin(t), t ∈ [0, π)
0, t ∈ [π,∞).
Equations with discontinuous sources (Sect. 6.4).
I Differential equations with discontinuous sources.I We solve the IVPs:
(a) Example 1:
y ′ + 2y = u(t − 4), y(0) = 3.
(b) Example 2:
y ′′ + y ′ +5
4y = b(t),
y(0) = 0,
y ′(0) = 0,b(t) =
{1, t ∈ [0, π)
0, t ∈ [π,∞).
(c) Example 3:
y ′′+y ′+5
4y = g(t),
y(0) = 0,
y ′(0) = 0,g(t) =
{sin(t), t ∈ [0, π)
0, t ∈ [π,∞).
Differential equations with discontinuous sources.
Example
Use the Laplace transform to find the solution of the IVP
y ′ + 2y = u(t − 4), y(0) = 3.
Solution: Compute the Laplace transform of the whole equation,
L[y ′] + 2L[y ] = L[u(t − 4)] =e−4s
s.
From the previous Section we know that[s L[y ]−y(0)
]+2L[y ] =
e−4s
s⇒ (s +2)L[y ] = y(0)+
e−4s
s.
Introduce the initial condition, L[y ] =3
(s + 2)+ e−4s 1
s(s + 2),
Use the table: L[y ] = 3L[e−2t
]+ e−4s 1
s(s + 2).
Differential equations with discontinuous sources.
Example
Use the Laplace transform to find the solution of the IVP
y ′ + 2y = u(t − 4), y(0) = 3.
Solution: Compute the Laplace transform of the whole equation,
L[y ′] + 2L[y ] = L[u(t − 4)]
=e−4s
s.
From the previous Section we know that[s L[y ]−y(0)
]+2L[y ] =
e−4s
s⇒ (s +2)L[y ] = y(0)+
e−4s
s.
Introduce the initial condition, L[y ] =3
(s + 2)+ e−4s 1
s(s + 2),
Use the table: L[y ] = 3L[e−2t
]+ e−4s 1
s(s + 2).
Differential equations with discontinuous sources.
Example
Use the Laplace transform to find the solution of the IVP
y ′ + 2y = u(t − 4), y(0) = 3.
Solution: Compute the Laplace transform of the whole equation,
L[y ′] + 2L[y ] = L[u(t − 4)] =e−4s
s.
From the previous Section we know that[s L[y ]−y(0)
]+2L[y ] =
e−4s
s⇒ (s +2)L[y ] = y(0)+
e−4s
s.
Introduce the initial condition, L[y ] =3
(s + 2)+ e−4s 1
s(s + 2),
Use the table: L[y ] = 3L[e−2t
]+ e−4s 1
s(s + 2).
Differential equations with discontinuous sources.
Example
Use the Laplace transform to find the solution of the IVP
y ′ + 2y = u(t − 4), y(0) = 3.
Solution: Compute the Laplace transform of the whole equation,
L[y ′] + 2L[y ] = L[u(t − 4)] =e−4s
s.
From the previous Section we know that[s L[y ]−y(0)
]+2L[y ] =
e−4s
s
⇒ (s +2)L[y ] = y(0)+e−4s
s.
Introduce the initial condition, L[y ] =3
(s + 2)+ e−4s 1
s(s + 2),
Use the table: L[y ] = 3L[e−2t
]+ e−4s 1
s(s + 2).
Differential equations with discontinuous sources.
Example
Use the Laplace transform to find the solution of the IVP
y ′ + 2y = u(t − 4), y(0) = 3.
Solution: Compute the Laplace transform of the whole equation,
L[y ′] + 2L[y ] = L[u(t − 4)] =e−4s
s.
From the previous Section we know that[s L[y ]−y(0)
]+2L[y ] =
e−4s
s⇒ (s +2)L[y ] = y(0)+
e−4s
s.
Introduce the initial condition, L[y ] =3
(s + 2)+ e−4s 1
s(s + 2),
Use the table: L[y ] = 3L[e−2t
]+ e−4s 1
s(s + 2).
Differential equations with discontinuous sources.
Example
Use the Laplace transform to find the solution of the IVP
y ′ + 2y = u(t − 4), y(0) = 3.
Solution: Compute the Laplace transform of the whole equation,
L[y ′] + 2L[y ] = L[u(t − 4)] =e−4s
s.
From the previous Section we know that[s L[y ]−y(0)
]+2L[y ] =
e−4s
s⇒ (s +2)L[y ] = y(0)+
e−4s
s.
Introduce the initial condition,
L[y ] =3
(s + 2)+ e−4s 1
s(s + 2),
Use the table: L[y ] = 3L[e−2t
]+ e−4s 1
s(s + 2).
Differential equations with discontinuous sources.
Example
Use the Laplace transform to find the solution of the IVP
y ′ + 2y = u(t − 4), y(0) = 3.
Solution: Compute the Laplace transform of the whole equation,
L[y ′] + 2L[y ] = L[u(t − 4)] =e−4s
s.
From the previous Section we know that[s L[y ]−y(0)
]+2L[y ] =
e−4s
s⇒ (s +2)L[y ] = y(0)+
e−4s
s.
Introduce the initial condition, L[y ] =3
(s + 2)+ e−4s 1
s(s + 2),
Use the table: L[y ] = 3L[e−2t
]+ e−4s 1
s(s + 2).
Differential equations with discontinuous sources.
Example
Use the Laplace transform to find the solution of the IVP
y ′ + 2y = u(t − 4), y(0) = 3.
Solution: Compute the Laplace transform of the whole equation,
L[y ′] + 2L[y ] = L[u(t − 4)] =e−4s
s.
From the previous Section we know that[s L[y ]−y(0)
]+2L[y ] =
e−4s
s⇒ (s +2)L[y ] = y(0)+
e−4s
s.
Introduce the initial condition, L[y ] =3
(s + 2)+ e−4s 1
s(s + 2),
Use the table: L[y ] = 3L[e−2t
]+ e−4s 1
s(s + 2).
Differential equations with discontinuous sources.
Example
Use the Laplace transform to find the solution of the IVP
y ′ + 2y = u(t − 4), y(0) = 3.
Solution: Recall: L[y ] = 3L[e−2t
]+ e−4s 1
s(s + 2).
We need to invert the Laplace transform on the last term.Partial fractions:
1
s(s + 2)=
a
s+
b
(s + 2)=
a(s + 2) + bs
s(s + 2)=
(a + b) s + (2a)
s(s + 2)
We get, a + b = 0, 2a = 1. We obtain: a =1
2, b = −1
2. Hence,
1
s(s + 2)=
1
2
[1
s− 1
(s + 2)
].
Differential equations with discontinuous sources.
Example
Use the Laplace transform to find the solution of the IVP
y ′ + 2y = u(t − 4), y(0) = 3.
Solution: Recall: L[y ] = 3L[e−2t
]+ e−4s 1
s(s + 2).
We need to invert the Laplace transform on the last term.
Partial fractions:
1
s(s + 2)=
a
s+
b
(s + 2)=
a(s + 2) + bs
s(s + 2)=
(a + b) s + (2a)
s(s + 2)
We get, a + b = 0, 2a = 1. We obtain: a =1
2, b = −1
2. Hence,
1
s(s + 2)=
1
2
[1
s− 1
(s + 2)
].
Differential equations with discontinuous sources.
Example
Use the Laplace transform to find the solution of the IVP
y ′ + 2y = u(t − 4), y(0) = 3.
Solution: Recall: L[y ] = 3L[e−2t
]+ e−4s 1
s(s + 2).
We need to invert the Laplace transform on the last term.Partial fractions:
1
s(s + 2)=
a
s+
b
(s + 2)=
a(s + 2) + bs
s(s + 2)=
(a + b) s + (2a)
s(s + 2)
We get, a + b = 0, 2a = 1. We obtain: a =1
2, b = −1
2. Hence,
1
s(s + 2)=
1
2
[1
s− 1
(s + 2)
].
Differential equations with discontinuous sources.
Example
Use the Laplace transform to find the solution of the IVP
y ′ + 2y = u(t − 4), y(0) = 3.
Solution: Recall: L[y ] = 3L[e−2t
]+ e−4s 1
s(s + 2).
We need to invert the Laplace transform on the last term.Partial fractions:
1
s(s + 2)=
a
s+
b
(s + 2)
=a(s + 2) + bs
s(s + 2)=
(a + b) s + (2a)
s(s + 2)
We get, a + b = 0, 2a = 1. We obtain: a =1
2, b = −1
2. Hence,
1
s(s + 2)=
1
2
[1
s− 1
(s + 2)
].
Differential equations with discontinuous sources.
Example
Use the Laplace transform to find the solution of the IVP
y ′ + 2y = u(t − 4), y(0) = 3.
Solution: Recall: L[y ] = 3L[e−2t
]+ e−4s 1
s(s + 2).
We need to invert the Laplace transform on the last term.Partial fractions:
1
s(s + 2)=
a
s+
b
(s + 2)=
a(s + 2) + bs
s(s + 2)
=(a + b) s + (2a)
s(s + 2)
We get, a + b = 0, 2a = 1. We obtain: a =1
2, b = −1
2. Hence,
1
s(s + 2)=
1
2
[1
s− 1
(s + 2)
].
Differential equations with discontinuous sources.
Example
Use the Laplace transform to find the solution of the IVP
y ′ + 2y = u(t − 4), y(0) = 3.
Solution: Recall: L[y ] = 3L[e−2t
]+ e−4s 1
s(s + 2).
We need to invert the Laplace transform on the last term.Partial fractions:
1
s(s + 2)=
a
s+
b
(s + 2)=
a(s + 2) + bs
s(s + 2)=
(a + b) s + (2a)
s(s + 2)
We get, a + b = 0, 2a = 1. We obtain: a =1
2, b = −1
2. Hence,
1
s(s + 2)=
1
2
[1
s− 1
(s + 2)
].
Differential equations with discontinuous sources.
Example
Use the Laplace transform to find the solution of the IVP
y ′ + 2y = u(t − 4), y(0) = 3.
Solution: Recall: L[y ] = 3L[e−2t
]+ e−4s 1
s(s + 2).
We need to invert the Laplace transform on the last term.Partial fractions:
1
s(s + 2)=
a
s+
b
(s + 2)=
a(s + 2) + bs
s(s + 2)=
(a + b) s + (2a)
s(s + 2)
We get, a + b = 0, 2a = 1.
We obtain: a =1
2, b = −1
2. Hence,
1
s(s + 2)=
1
2
[1
s− 1
(s + 2)
].
Differential equations with discontinuous sources.
Example
Use the Laplace transform to find the solution of the IVP
y ′ + 2y = u(t − 4), y(0) = 3.
Solution: Recall: L[y ] = 3L[e−2t
]+ e−4s 1
s(s + 2).
We need to invert the Laplace transform on the last term.Partial fractions:
1
s(s + 2)=
a
s+
b
(s + 2)=
a(s + 2) + bs
s(s + 2)=
(a + b) s + (2a)
s(s + 2)
We get, a + b = 0, 2a = 1. We obtain: a =1
2, b = −1
2.
Hence,
1
s(s + 2)=
1
2
[1
s− 1
(s + 2)
].
Differential equations with discontinuous sources.
Example
Use the Laplace transform to find the solution of the IVP
y ′ + 2y = u(t − 4), y(0) = 3.
Solution: Recall: L[y ] = 3L[e−2t
]+ e−4s 1
s(s + 2).
We need to invert the Laplace transform on the last term.Partial fractions:
1
s(s + 2)=
a
s+
b
(s + 2)=
a(s + 2) + bs
s(s + 2)=
(a + b) s + (2a)
s(s + 2)
We get, a + b = 0, 2a = 1. We obtain: a =1
2, b = −1
2. Hence,
1
s(s + 2)=
1
2
[1
s− 1
(s + 2)
].
Differential equations with discontinuous sources.
Example
Use the Laplace transform to find the solution of the IVP
y ′ + 2y = u(t − 4), y(0) = 3.
Solution: Recall:1
s(s + 2)=
1
2
[1
s− 1
(s + 2)
].
The algebraic equation for L[y ] has the form,
L[y ] = 3L[e−2t
]+
1
2
[e−4s 1
s− e−4s 1
(s + 2)
].
L[y ] = 3L[e−2t
]+
1
2
(L[u(t − 4)]− L
[u(t − 4) e−2(t−4)
]).
We conclude that
y(t) = 3e−2t +1
2u(t − 4)
[1− e−2(t−4)
]. C
Differential equations with discontinuous sources.
Example
Use the Laplace transform to find the solution of the IVP
y ′ + 2y = u(t − 4), y(0) = 3.
Solution: Recall:1
s(s + 2)=
1
2
[1
s− 1
(s + 2)
].
The algebraic equation for L[y ] has the form,
L[y ] = 3L[e−2t
]+
1
2
[e−4s 1
s− e−4s 1
(s + 2)
].
L[y ] = 3L[e−2t
]+
1
2
(L[u(t − 4)]− L
[u(t − 4) e−2(t−4)
]).
We conclude that
y(t) = 3e−2t +1
2u(t − 4)
[1− e−2(t−4)
]. C
Differential equations with discontinuous sources.
Example
Use the Laplace transform to find the solution of the IVP
y ′ + 2y = u(t − 4), y(0) = 3.
Solution: Recall:1
s(s + 2)=
1
2
[1
s− 1
(s + 2)
].
The algebraic equation for L[y ] has the form,
L[y ] = 3L[e−2t
]+
1
2
[e−4s 1
s− e−4s 1
(s + 2)
].
L[y ] = 3L[e−2t
]
+1
2
(L[u(t − 4)]− L
[u(t − 4) e−2(t−4)
]).
We conclude that
y(t) = 3e−2t +1
2u(t − 4)
[1− e−2(t−4)
]. C
Differential equations with discontinuous sources.
Example
Use the Laplace transform to find the solution of the IVP
y ′ + 2y = u(t − 4), y(0) = 3.
Solution: Recall:1
s(s + 2)=
1
2
[1
s− 1
(s + 2)
].
The algebraic equation for L[y ] has the form,
L[y ] = 3L[e−2t
]+
1
2
[e−4s 1
s− e−4s 1
(s + 2)
].
L[y ] = 3L[e−2t
]+
1
2
(L[u(t − 4)]
− L[u(t − 4) e−2(t−4)
]).
We conclude that
y(t) = 3e−2t +1
2u(t − 4)
[1− e−2(t−4)
]. C
Differential equations with discontinuous sources.
Example
Use the Laplace transform to find the solution of the IVP
y ′ + 2y = u(t − 4), y(0) = 3.
Solution: Recall:1
s(s + 2)=
1
2
[1
s− 1
(s + 2)
].
The algebraic equation for L[y ] has the form,
L[y ] = 3L[e−2t
]+
1
2
[e−4s 1
s− e−4s 1
(s + 2)
].
L[y ] = 3L[e−2t
]+
1
2
(L[u(t − 4)]− L
[u(t − 4) e−2(t−4)
]).
We conclude that
y(t) = 3e−2t +1
2u(t − 4)
[1− e−2(t−4)
]. C
Differential equations with discontinuous sources.
Example
Use the Laplace transform to find the solution of the IVP
y ′ + 2y = u(t − 4), y(0) = 3.
Solution: Recall:1
s(s + 2)=
1
2
[1
s− 1
(s + 2)
].
The algebraic equation for L[y ] has the form,
L[y ] = 3L[e−2t
]+
1
2
[e−4s 1
s− e−4s 1
(s + 2)
].
L[y ] = 3L[e−2t
]+
1
2
(L[u(t − 4)]− L
[u(t − 4) e−2(t−4)
]).
We conclude that
y(t) = 3e−2t +1
2u(t − 4)
[1− e−2(t−4)
]. C
Equations with discontinuous sources (Sect. 6.4).
I Differential equations with discontinuous sources.I We solve the IVPs:
(a) Example 1:
y ′ + 2y = u(t − 4), y(0) = 3.
(b) Example 2:
y ′′ + y ′ +5
4y = b(t),
y(0) = 0,
y ′(0) = 0,b(t) =
{1, t ∈ [0, π)
0, t ∈ [π,∞).
(c) Example 3:
y ′′+y ′+5
4y = g(t),
y(0) = 0,
y ′(0) = 0,g(t) =
{sin(t), t ∈ [0, π)
0, t ∈ [π,∞).
Differential equations with discontinuous sources.
Example
Use the Laplace transform to find the solution of the IVP
y ′′ + y ′ +5
4y = b(t),
y(0) = 0,
y ′(0) = 0,b(t) =
{1, t ∈ [0, π)
0, t ∈ [π,∞).
Solution:
Rewrite the source function usingstep functions.
b ( t )
1
0 pi t
t
u ( t )
1
0 pi
u ( t − pi )
1
0 pi t
Differential equations with discontinuous sources.
Example
Use the Laplace transform to find the solution of the IVP
y ′′ + y ′ +5
4y = b(t),
y(0) = 0,
y ′(0) = 0,b(t) =
{1, t ∈ [0, π)
0, t ∈ [π,∞).
Solution:
Rewrite the source function usingstep functions.
b ( t )
1
0 pi t
t
u ( t )
1
0 pi
u ( t − pi )
1
0 pi t
Differential equations with discontinuous sources.
Example
Use the Laplace transform to find the solution of the IVP
y ′′ + y ′ +5
4y = b(t),
y(0) = 0,
y ′(0) = 0,b(t) =
{1, t ∈ [0, π)
0, t ∈ [π,∞).
Solution:
Rewrite the source function usingstep functions.
b ( t )
1
0 pi t
t
u ( t )
1
0 pi
u ( t − pi )
1
0 pi t
Differential equations with discontinuous sources.
Example
Use the Laplace transform to find the solution of the IVP
y ′′ + y ′ +5
4y = b(t),
y(0) = 0,
y ′(0) = 0,b(t) =
{1, t ∈ [0, π)
0, t ∈ [π,∞).
Solution:
Rewrite the source function usingstep functions.
b ( t )
1
0 pi t
t
u ( t )
1
0 pi
u ( t − pi )
1
0 pi t
Differential equations with discontinuous sources.
Example
Use the Laplace transform to find the solution of the IVP
y ′′ + y ′ +5
4y = b(t),
y(0) = 0,
y ′(0) = 0,b(t) =
{1, t ∈ [0, π)
0, t ∈ [π,∞).
Solution:
Rewrite the source function usingstep functions.
b ( t )
1
0 pi t
t
u ( t )
1
0 pi
u ( t − pi )
1
0 pi t
Differential equations with discontinuous sources.
Example
Use the Laplace transform to find the solution of the IVP
y ′′ + y ′ +5
4y = b(t),
y(0) = 0,
y ′(0) = 0,b(t) =
{1, t ∈ [0, π)
0, t ∈ [π,∞).
Solution: The graphs imply: b(t) = u(t)− u(t − π)
Now is simple to find L[b], since
L[b(t)] = L[u(t)]− L[u(t − π)] =1
s− e−πs
s.
So, the source is L[b(t)] =(1− e−πs
) 1
s, and the equation is
L[y ′′] + L[y ′] +5
4L[y ] =
(1− e−πs
) 1
s.
Differential equations with discontinuous sources.
Example
Use the Laplace transform to find the solution of the IVP
y ′′ + y ′ +5
4y = b(t),
y(0) = 0,
y ′(0) = 0,b(t) =
{1, t ∈ [0, π)
0, t ∈ [π,∞).
Solution: The graphs imply: b(t) = u(t)− u(t − π)
Now is simple to find L[b],
since
L[b(t)] = L[u(t)]− L[u(t − π)] =1
s− e−πs
s.
So, the source is L[b(t)] =(1− e−πs
) 1
s, and the equation is
L[y ′′] + L[y ′] +5
4L[y ] =
(1− e−πs
) 1
s.
Differential equations with discontinuous sources.
Example
Use the Laplace transform to find the solution of the IVP
y ′′ + y ′ +5
4y = b(t),
y(0) = 0,
y ′(0) = 0,b(t) =
{1, t ∈ [0, π)
0, t ∈ [π,∞).
Solution: The graphs imply: b(t) = u(t)− u(t − π)
Now is simple to find L[b], since
L[b(t)] = L[u(t)]− L[u(t − π)]
=1
s− e−πs
s.
So, the source is L[b(t)] =(1− e−πs
) 1
s, and the equation is
L[y ′′] + L[y ′] +5
4L[y ] =
(1− e−πs
) 1
s.
Differential equations with discontinuous sources.
Example
Use the Laplace transform to find the solution of the IVP
y ′′ + y ′ +5
4y = b(t),
y(0) = 0,
y ′(0) = 0,b(t) =
{1, t ∈ [0, π)
0, t ∈ [π,∞).
Solution: The graphs imply: b(t) = u(t)− u(t − π)
Now is simple to find L[b], since
L[b(t)] = L[u(t)]− L[u(t − π)] =1
s− e−πs
s.
So, the source is L[b(t)] =(1− e−πs
) 1
s, and the equation is
L[y ′′] + L[y ′] +5
4L[y ] =
(1− e−πs
) 1
s.
Differential equations with discontinuous sources.
Example
Use the Laplace transform to find the solution of the IVP
y ′′ + y ′ +5
4y = b(t),
y(0) = 0,
y ′(0) = 0,b(t) =
{1, t ∈ [0, π)
0, t ∈ [π,∞).
Solution: The graphs imply: b(t) = u(t)− u(t − π)
Now is simple to find L[b], since
L[b(t)] = L[u(t)]− L[u(t − π)] =1
s− e−πs
s.
So, the source is L[b(t)] =(1− e−πs
) 1
s,
and the equation is
L[y ′′] + L[y ′] +5
4L[y ] =
(1− e−πs
) 1
s.
Differential equations with discontinuous sources.
Example
Use the Laplace transform to find the solution of the IVP
y ′′ + y ′ +5
4y = b(t),
y(0) = 0,
y ′(0) = 0,b(t) =
{1, t ∈ [0, π)
0, t ∈ [π,∞).
Solution: The graphs imply: b(t) = u(t)− u(t − π)
Now is simple to find L[b], since
L[b(t)] = L[u(t)]− L[u(t − π)] =1
s− e−πs
s.
So, the source is L[b(t)] =(1− e−πs
) 1
s, and the equation is
L[y ′′] + L[y ′] +5
4L[y ] =
(1− e−πs
) 1
s.
Differential equations with discontinuous sources.
Example
Use the Laplace transform to find the solution of the IVP
y ′′ + y ′ +5
4y = b(t),
y(0) = 0,
y ′(0) = 0,b(t) =
{1, t ∈ [0, π)
0, t ∈ [π,∞).
Solution: So: L[y ′′] + L[y ′] +5
4L[y ] =
(1− e−πs
) 1
s.
The initial conditions imply: L[y ′′] = s2 L[y ] and L[y ′] = s L[y ].
Therefore,(s2 + s +
5
4
)L[y ] =
(1− e−πs
) 1
s.
We arrive at the expression: L[y ] =(1− e−πs
) 1
s(s2 + s + 5
4
) .
Differential equations with discontinuous sources.
Example
Use the Laplace transform to find the solution of the IVP
y ′′ + y ′ +5
4y = b(t),
y(0) = 0,
y ′(0) = 0,b(t) =
{1, t ∈ [0, π)
0, t ∈ [π,∞).
Solution: So: L[y ′′] + L[y ′] +5
4L[y ] =
(1− e−πs
) 1
s.
The initial conditions imply:
L[y ′′] = s2 L[y ] and L[y ′] = s L[y ].
Therefore,(s2 + s +
5
4
)L[y ] =
(1− e−πs
) 1
s.
We arrive at the expression: L[y ] =(1− e−πs
) 1
s(s2 + s + 5
4
) .
Differential equations with discontinuous sources.
Example
Use the Laplace transform to find the solution of the IVP
y ′′ + y ′ +5
4y = b(t),
y(0) = 0,
y ′(0) = 0,b(t) =
{1, t ∈ [0, π)
0, t ∈ [π,∞).
Solution: So: L[y ′′] + L[y ′] +5
4L[y ] =
(1− e−πs
) 1
s.
The initial conditions imply: L[y ′′] = s2 L[y ]
and L[y ′] = s L[y ].
Therefore,(s2 + s +
5
4
)L[y ] =
(1− e−πs
) 1
s.
We arrive at the expression: L[y ] =(1− e−πs
) 1
s(s2 + s + 5
4
) .
Differential equations with discontinuous sources.
Example
Use the Laplace transform to find the solution of the IVP
y ′′ + y ′ +5
4y = b(t),
y(0) = 0,
y ′(0) = 0,b(t) =
{1, t ∈ [0, π)
0, t ∈ [π,∞).
Solution: So: L[y ′′] + L[y ′] +5
4L[y ] =
(1− e−πs
) 1
s.
The initial conditions imply: L[y ′′] = s2 L[y ] and L[y ′] = s L[y ].
Therefore,(s2 + s +
5
4
)L[y ] =
(1− e−πs
) 1
s.
We arrive at the expression: L[y ] =(1− e−πs
) 1
s(s2 + s + 5
4
) .
Differential equations with discontinuous sources.
Example
Use the Laplace transform to find the solution of the IVP
y ′′ + y ′ +5
4y = b(t),
y(0) = 0,
y ′(0) = 0,b(t) =
{1, t ∈ [0, π)
0, t ∈ [π,∞).
Solution: So: L[y ′′] + L[y ′] +5
4L[y ] =
(1− e−πs
) 1
s.
The initial conditions imply: L[y ′′] = s2 L[y ] and L[y ′] = s L[y ].
Therefore,(s2 + s +
5
4
)L[y ] =
(1− e−πs
) 1
s.
We arrive at the expression: L[y ] =(1− e−πs
) 1
s(s2 + s + 5
4
) .
Differential equations with discontinuous sources.
Example
Use the Laplace transform to find the solution of the IVP
y ′′ + y ′ +5
4y = b(t),
y(0) = 0,
y ′(0) = 0,b(t) =
{1, t ∈ [0, π)
0, t ∈ [π,∞).
Solution: So: L[y ′′] + L[y ′] +5
4L[y ] =
(1− e−πs
) 1
s.
The initial conditions imply: L[y ′′] = s2 L[y ] and L[y ′] = s L[y ].
Therefore,(s2 + s +
5
4
)L[y ] =
(1− e−πs
) 1
s.
We arrive at the expression: L[y ] =(1− e−πs
) 1
s(s2 + s + 5
4
) .
Differential equations with discontinuous sources.
Example
Use the Laplace transform to find the solution of the IVP
y ′′ + y ′ +5
4y = b(t),
y(0) = 0,
y ′(0) = 0,b(t) =
{1, t ∈ [0, π)
0, t ∈ [π,∞).
Solution: Recall: L[y ] =(1− e−πs
) 1
s(s2 + s + 5
4
) .
Denoting: H(s) =1
s(s2 + s + 5
4
) ,
we obtain, L[y ] =(1− e−πs
)H(s).
In other words: y(t) = L−1[H(s)
]− L−1
[e−πs H(s)
].
Differential equations with discontinuous sources.
Example
Use the Laplace transform to find the solution of the IVP
y ′′ + y ′ +5
4y = b(t),
y(0) = 0,
y ′(0) = 0,b(t) =
{1, t ∈ [0, π)
0, t ∈ [π,∞).
Solution: Recall: L[y ] =(1− e−πs
) 1
s(s2 + s + 5
4
) .
Denoting: H(s) =1
s(s2 + s + 5
4
) ,
we obtain, L[y ] =(1− e−πs
)H(s).
In other words: y(t) = L−1[H(s)
]− L−1
[e−πs H(s)
].
Differential equations with discontinuous sources.
Example
Use the Laplace transform to find the solution of the IVP
y ′′ + y ′ +5
4y = b(t),
y(0) = 0,
y ′(0) = 0,b(t) =
{1, t ∈ [0, π)
0, t ∈ [π,∞).
Solution: Recall: L[y ] =(1− e−πs
) 1
s(s2 + s + 5
4
) .
Denoting: H(s) =1
s(s2 + s + 5
4
) ,
we obtain, L[y ] =(1− e−πs
)H(s).
In other words: y(t) = L−1[H(s)
]− L−1
[e−πs H(s)
].
Differential equations with discontinuous sources.
Example
Use the Laplace transform to find the solution of the IVP
y ′′ + y ′ +5
4y = b(t),
y(0) = 0,
y ′(0) = 0,b(t) =
{1, t ∈ [0, π)
0, t ∈ [π,∞).
Solution: Recall: L[y ] =(1− e−πs
) 1
s(s2 + s + 5
4
) .
Denoting: H(s) =1
s(s2 + s + 5
4
) ,
we obtain, L[y ] =(1− e−πs
)H(s).
In other words: y(t) = L−1[H(s)
]− L−1
[e−πs H(s)
].
Differential equations with discontinuous sources.
Example
Use the Laplace transform to find the solution of the IVP
y ′′ + y ′ +5
4y = b(t),
y(0) = 0,
y ′(0) = 0,b(t) =
{1, t ∈ [0, π)
0, t ∈ [π,∞).
Solution: Recall: y(t) = L−1[H(s)
]− L−1
[e−πs H(s)
].
Denoting: h(t) = L−1[H(s)
], the L[ ] properties imply
L−1[e−πsH(s)
]= u(t − π) h(t − π).
Therefore, the solution has the form
y(t) = h(t)− u(t − π) h(t − π).
We only need to find h(t) = L−1[ 1
s(s2 + s + 5
4
)].
Differential equations with discontinuous sources.
Example
Use the Laplace transform to find the solution of the IVP
y ′′ + y ′ +5
4y = b(t),
y(0) = 0,
y ′(0) = 0,b(t) =
{1, t ∈ [0, π)
0, t ∈ [π,∞).
Solution: Recall: y(t) = L−1[H(s)
]− L−1
[e−πs H(s)
].
Denoting: h(t) = L−1[H(s)
],
the L[ ] properties imply
L−1[e−πsH(s)
]= u(t − π) h(t − π).
Therefore, the solution has the form
y(t) = h(t)− u(t − π) h(t − π).
We only need to find h(t) = L−1[ 1
s(s2 + s + 5
4
)].
Differential equations with discontinuous sources.
Example
Use the Laplace transform to find the solution of the IVP
y ′′ + y ′ +5
4y = b(t),
y(0) = 0,
y ′(0) = 0,b(t) =
{1, t ∈ [0, π)
0, t ∈ [π,∞).
Solution: Recall: y(t) = L−1[H(s)
]− L−1
[e−πs H(s)
].
Denoting: h(t) = L−1[H(s)
], the L[ ] properties imply
L−1[e−πsH(s)
]= u(t − π) h(t − π).
Therefore, the solution has the form
y(t) = h(t)− u(t − π) h(t − π).
We only need to find h(t) = L−1[ 1
s(s2 + s + 5
4
)].
Differential equations with discontinuous sources.
Example
Use the Laplace transform to find the solution of the IVP
y ′′ + y ′ +5
4y = b(t),
y(0) = 0,
y ′(0) = 0,b(t) =
{1, t ∈ [0, π)
0, t ∈ [π,∞).
Solution: Recall: y(t) = L−1[H(s)
]− L−1
[e−πs H(s)
].
Denoting: h(t) = L−1[H(s)
], the L[ ] properties imply
L−1[e−πsH(s)
]= u(t − π) h(t − π).
Therefore, the solution has the form
y(t) = h(t)− u(t − π) h(t − π).
We only need to find h(t) = L−1[ 1
s(s2 + s + 5
4
)].
Differential equations with discontinuous sources.
Example
Use the Laplace transform to find the solution of the IVP
y ′′ + y ′ +5
4y = b(t),
y(0) = 0,
y ′(0) = 0,b(t) =
{1, t ∈ [0, π)
0, t ∈ [π,∞).
Solution: Recall: y(t) = L−1[H(s)
]− L−1
[e−πs H(s)
].
Denoting: h(t) = L−1[H(s)
], the L[ ] properties imply
L−1[e−πsH(s)
]= u(t − π) h(t − π).
Therefore, the solution has the form
y(t) = h(t)− u(t − π) h(t − π).
We only need to find h(t) = L−1[ 1
s(s2 + s + 5
4
)].
Differential equations with discontinuous sources.
Example
Use the Laplace transform to find the solution of the IVP
y ′′ + y ′ +5
4y = b(t),
y(0) = 0,
y ′(0) = 0,b(t) =
{1, t ∈ [0, π)
0, t ∈ [π,∞).
Solution: Recall: h(t) = L−1[ 1
s(s2 + s + 5
4
)].
Partial fractions: Find the zeros of the denominator,
s± =1
2
[−1±
√1− 5
]⇒ Complex roots.
The partial fraction decomposition is:
H(s) =1(
s2 + s + 54
)s
=a
s+
(bs + c)(s2 + s + 5
4
)
Differential equations with discontinuous sources.
Example
Use the Laplace transform to find the solution of the IVP
y ′′ + y ′ +5
4y = b(t),
y(0) = 0,
y ′(0) = 0,b(t) =
{1, t ∈ [0, π)
0, t ∈ [π,∞).
Solution: Recall: h(t) = L−1[ 1
s(s2 + s + 5
4
)].
Partial fractions:
Find the zeros of the denominator,
s± =1
2
[−1±
√1− 5
]⇒ Complex roots.
The partial fraction decomposition is:
H(s) =1(
s2 + s + 54
)s
=a
s+
(bs + c)(s2 + s + 5
4
)
Differential equations with discontinuous sources.
Example
Use the Laplace transform to find the solution of the IVP
y ′′ + y ′ +5
4y = b(t),
y(0) = 0,
y ′(0) = 0,b(t) =
{1, t ∈ [0, π)
0, t ∈ [π,∞).
Solution: Recall: h(t) = L−1[ 1
s(s2 + s + 5
4
)].
Partial fractions: Find the zeros of the denominator,
s± =1
2
[−1±
√1− 5
]⇒ Complex roots.
The partial fraction decomposition is:
H(s) =1(
s2 + s + 54
)s
=a
s+
(bs + c)(s2 + s + 5
4
)
Differential equations with discontinuous sources.
Example
Use the Laplace transform to find the solution of the IVP
y ′′ + y ′ +5
4y = b(t),
y(0) = 0,
y ′(0) = 0,b(t) =
{1, t ∈ [0, π)
0, t ∈ [π,∞).
Solution: Recall: h(t) = L−1[ 1
s(s2 + s + 5
4
)].
Partial fractions: Find the zeros of the denominator,
s± =1
2
[−1±
√1− 5
]
⇒ Complex roots.
The partial fraction decomposition is:
H(s) =1(
s2 + s + 54
)s
=a
s+
(bs + c)(s2 + s + 5
4
)
Differential equations with discontinuous sources.
Example
Use the Laplace transform to find the solution of the IVP
y ′′ + y ′ +5
4y = b(t),
y(0) = 0,
y ′(0) = 0,b(t) =
{1, t ∈ [0, π)
0, t ∈ [π,∞).
Solution: Recall: h(t) = L−1[ 1
s(s2 + s + 5
4
)].
Partial fractions: Find the zeros of the denominator,
s± =1
2
[−1±
√1− 5
]⇒ Complex roots.
The partial fraction decomposition is:
H(s) =1(
s2 + s + 54
)s
=a
s+
(bs + c)(s2 + s + 5
4
)
Differential equations with discontinuous sources.
Example
Use the Laplace transform to find the solution of the IVP
y ′′ + y ′ +5
4y = b(t),
y(0) = 0,
y ′(0) = 0,b(t) =
{1, t ∈ [0, π)
0, t ∈ [π,∞).
Solution: Recall: h(t) = L−1[ 1
s(s2 + s + 5
4
)].
Partial fractions: Find the zeros of the denominator,
s± =1
2
[−1±
√1− 5
]⇒ Complex roots.
The partial fraction decomposition is:
H(s) =1(
s2 + s + 54
)s
=a
s+
(bs + c)(s2 + s + 5
4
)
Differential equations with discontinuous sources.
Example
Use the Laplace transform to find the solution of the IVP
y ′′ + y ′ +5
4y = b(t),
y(0) = 0,
y ′(0) = 0,b(t) =
{1, t ∈ [0, π)
0, t ∈ [π,∞).
Solution: Recall: h(t) = L−1[ 1
s(s2 + s + 5
4
)].
Partial fractions: Find the zeros of the denominator,
s± =1
2
[−1±
√1− 5
]⇒ Complex roots.
The partial fraction decomposition is:
H(s) =1(
s2 + s + 54
)s
=a
s+
(bs + c)(s2 + s + 5
4
)
Differential equations with discontinuous sources.
Example
Use the Laplace transform to find the solution of the IVP
y ′′ + y ′ +5
4y = b(t),
y(0) = 0,
y ′(0) = 0,b(t) =
{1, t ∈ [0, π)
0, t ∈ [π,∞).
Solution: Recall: h(t) = L−1[ 1
s(s2 + s + 5
4
)].
Partial fractions: Find the zeros of the denominator,
s± =1
2
[−1±
√1− 5
]⇒ Complex roots.
The partial fraction decomposition is:
H(s) =1(
s2 + s + 54
)s
=a
s+
(bs + c)(s2 + s + 5
4
)
Differential equations with discontinuous sources.
Example
Use the Laplace transform to find the solution of the IVP
y ′′ + y ′ +5
4y = b(t),
y(0) = 0,
y ′(0) = 0,b(t) =
{1, t ∈ [0, π)
0, t ∈ [π,∞).
Solution: Recall: H(s) =1(
s2 + s + 54
)s
=a
s+
(bs + c)(s2 + s + 5
4
) .
The partial fraction decomposition is:
1 = a(s2 + s +
5
4
)+ s (bs + c) = (a + b) s2 + (a + c) s +
5
4a.
This equation implies that a, b, and c , are solutions of
a + b = 0, a + c = 0,5
4a = 1.
Differential equations with discontinuous sources.
Example
Use the Laplace transform to find the solution of the IVP
y ′′ + y ′ +5
4y = b(t),
y(0) = 0,
y ′(0) = 0,b(t) =
{1, t ∈ [0, π)
0, t ∈ [π,∞).
Solution: Recall: H(s) =1(
s2 + s + 54
)s
=a
s+
(bs + c)(s2 + s + 5
4
) .
The partial fraction decomposition is:
1 = a(s2 + s +
5
4
)+ s (bs + c) = (a + b) s2 + (a + c) s +
5
4a.
This equation implies that a, b, and c , are solutions of
a + b = 0, a + c = 0,5
4a = 1.
Differential equations with discontinuous sources.
Example
Use the Laplace transform to find the solution of the IVP
y ′′ + y ′ +5
4y = b(t),
y(0) = 0,
y ′(0) = 0,b(t) =
{1, t ∈ [0, π)
0, t ∈ [π,∞).
Solution: Recall: H(s) =1(
s2 + s + 54
)s
=a
s+
(bs + c)(s2 + s + 5
4
) .
The partial fraction decomposition is:
1 = a(s2 + s +
5
4
)+ s (bs + c)
= (a + b) s2 + (a + c) s +5
4a.
This equation implies that a, b, and c , are solutions of
a + b = 0, a + c = 0,5
4a = 1.
Differential equations with discontinuous sources.
Example
Use the Laplace transform to find the solution of the IVP
y ′′ + y ′ +5
4y = b(t),
y(0) = 0,
y ′(0) = 0,b(t) =
{1, t ∈ [0, π)
0, t ∈ [π,∞).
Solution: Recall: H(s) =1(
s2 + s + 54
)s
=a
s+
(bs + c)(s2 + s + 5
4
) .
The partial fraction decomposition is:
1 = a(s2 + s +
5
4
)+ s (bs + c) = (a + b) s2 + (a + c) s +
5
4a.
This equation implies that a, b, and c , are solutions of
a + b = 0, a + c = 0,5
4a = 1.
Differential equations with discontinuous sources.
Example
Use the Laplace transform to find the solution of the IVP
y ′′ + y ′ +5
4y = b(t),
y(0) = 0,
y ′(0) = 0,b(t) =
{1, t ∈ [0, π)
0, t ∈ [π,∞).
Solution: Recall: H(s) =1(
s2 + s + 54
)s
=a
s+
(bs + c)(s2 + s + 5
4
) .
The partial fraction decomposition is:
1 = a(s2 + s +
5
4
)+ s (bs + c) = (a + b) s2 + (a + c) s +
5
4a.
This equation implies that a, b, and c , are solutions of
a + b = 0, a + c = 0,5
4a = 1.
Differential equations with discontinuous sources.
Example
Use the Laplace transform to find the solution of the IVP
y ′′ + y ′ +5
4y = b(t),
y(0) = 0,
y ′(0) = 0,b(t) =
{1, t ∈ [0, π)
0, t ∈ [π,∞).
Solution: So: a =4
5, b = −4
5, c = −4
5.
Hence, we have found that,
H(s) =1(
s2 + s + 54
)s
=4
5
[1
s− (s + 1)(
s2 + s + 54
)]We have to compute the inverse Laplace Transform
h(t) =4
5L−1
[1
s− (s + 1)(
s2 + s + 54
)]
Differential equations with discontinuous sources.
Example
Use the Laplace transform to find the solution of the IVP
y ′′ + y ′ +5
4y = b(t),
y(0) = 0,
y ′(0) = 0,b(t) =
{1, t ∈ [0, π)
0, t ∈ [π,∞).
Solution: So: a =4
5, b = −4
5, c = −4
5.
Hence, we have found that,
H(s) =1(
s2 + s + 54
)s
=4
5
[1
s− (s + 1)(
s2 + s + 54
)]We have to compute the inverse Laplace Transform
h(t) =4
5L−1
[1
s− (s + 1)(
s2 + s + 54
)]
Differential equations with discontinuous sources.
Example
Use the Laplace transform to find the solution of the IVP
y ′′ + y ′ +5
4y = b(t),
y(0) = 0,
y ′(0) = 0,b(t) =
{1, t ∈ [0, π)
0, t ∈ [π,∞).
Solution: So: a =4
5, b = −4
5, c = −4
5.
Hence, we have found that,
H(s) =1(
s2 + s + 54
)s
=4
5
[1
s− (s + 1)(
s2 + s + 54
)]
We have to compute the inverse Laplace Transform
h(t) =4
5L−1
[1
s− (s + 1)(
s2 + s + 54
)]
Differential equations with discontinuous sources.
Example
Use the Laplace transform to find the solution of the IVP
y ′′ + y ′ +5
4y = b(t),
y(0) = 0,
y ′(0) = 0,b(t) =
{1, t ∈ [0, π)
0, t ∈ [π,∞).
Solution: So: a =4
5, b = −4
5, c = −4
5.
Hence, we have found that,
H(s) =1(
s2 + s + 54
)s
=4
5
[1
s− (s + 1)(
s2 + s + 54
)]We have to compute the inverse Laplace Transform
h(t) =4
5L−1
[1
s− (s + 1)(
s2 + s + 54
)]
Differential equations with discontinuous sources.
Example
Use the Laplace transform to find the solution of the IVP
y ′′ + y ′ +5
4y = b(t),
y(0) = 0,
y ′(0) = 0,b(t) =
{1, t ∈ [0, π)
0, t ∈ [π,∞).
Solution: So: a =4
5, b = −4
5, c = −4
5.
Hence, we have found that,
H(s) =1(
s2 + s + 54
)s
=4
5
[1
s− (s + 1)(
s2 + s + 54
)]We have to compute the inverse Laplace Transform
h(t) =4
5L−1
[1
s− (s + 1)(
s2 + s + 54
)]
Differential equations with discontinuous sources.
Example
Use the Laplace transform to find the solution of the IVP
y ′′ + y ′ +5
4y = b(t),
y(0) = 0,
y ′(0) = 0,b(t) =
{1, t ∈ [0, π)
0, t ∈ [π,∞).
Solution: Recall: h(t) =4
5L−1
[1
s− (s + 1)(
s2 + s + 54
)].
In this case we complete the square in the denominator,
s2 + s +5
4=
[s2 + 2
(1
2
)s +
1
4
]− 1
4+
5
4=
(s +
1
2
)2+ 1.
So: h(t) =4
5L−1
[1
s− (s + 1)[(
s + 12
)2+ 1
]].
That is, h(t) =4
5L−1
[1
s
]− 4
5L−1
[ (s + 1
2
)+ 1
2[(s + 1
2
)2+ 1
]].
Differential equations with discontinuous sources.
Example
Use the Laplace transform to find the solution of the IVP
y ′′ + y ′ +5
4y = b(t),
y(0) = 0,
y ′(0) = 0,b(t) =
{1, t ∈ [0, π)
0, t ∈ [π,∞).
Solution: Recall: h(t) =4
5L−1
[1
s− (s + 1)(
s2 + s + 54
)].
In this case we complete the square in the denominator,
s2 + s +5
4=
[s2 + 2
(1
2
)s +
1
4
]− 1
4+
5
4=
(s +
1
2
)2+ 1.
So: h(t) =4
5L−1
[1
s− (s + 1)[(
s + 12
)2+ 1
]].
That is, h(t) =4
5L−1
[1
s
]− 4
5L−1
[ (s + 1
2
)+ 1
2[(s + 1
2
)2+ 1
]].
Differential equations with discontinuous sources.
Example
Use the Laplace transform to find the solution of the IVP
y ′′ + y ′ +5
4y = b(t),
y(0) = 0,
y ′(0) = 0,b(t) =
{1, t ∈ [0, π)
0, t ∈ [π,∞).
Solution: Recall: h(t) =4
5L−1
[1
s− (s + 1)(
s2 + s + 54
)].
In this case we complete the square in the denominator,
s2 + s +5
4=
[s2 + 2
(1
2
)s +
1
4
]− 1
4+
5
4
=(s +
1
2
)2+ 1.
So: h(t) =4
5L−1
[1
s− (s + 1)[(
s + 12
)2+ 1
]].
That is, h(t) =4
5L−1
[1
s
]− 4
5L−1
[ (s + 1
2
)+ 1
2[(s + 1
2
)2+ 1
]].
Differential equations with discontinuous sources.
Example
Use the Laplace transform to find the solution of the IVP
y ′′ + y ′ +5
4y = b(t),
y(0) = 0,
y ′(0) = 0,b(t) =
{1, t ∈ [0, π)
0, t ∈ [π,∞).
Solution: Recall: h(t) =4
5L−1
[1
s− (s + 1)(
s2 + s + 54
)].
In this case we complete the square in the denominator,
s2 + s +5
4=
[s2 + 2
(1
2
)s +
1
4
]− 1
4+
5
4=
(s +
1
2
)2+ 1.
So: h(t) =4
5L−1
[1
s− (s + 1)[(
s + 12
)2+ 1
]].
That is, h(t) =4
5L−1
[1
s
]− 4
5L−1
[ (s + 1
2
)+ 1
2[(s + 1
2
)2+ 1
]].
Differential equations with discontinuous sources.
Example
Use the Laplace transform to find the solution of the IVP
y ′′ + y ′ +5
4y = b(t),
y(0) = 0,
y ′(0) = 0,b(t) =
{1, t ∈ [0, π)
0, t ∈ [π,∞).
Solution: Recall: h(t) =4
5L−1
[1
s− (s + 1)(
s2 + s + 54
)].
In this case we complete the square in the denominator,
s2 + s +5
4=
[s2 + 2
(1
2
)s +
1
4
]− 1
4+
5
4=
(s +
1
2
)2+ 1.
So: h(t) =4
5L−1
[1
s− (s + 1)[(
s + 12
)2+ 1
]].
That is, h(t) =4
5L−1
[1
s
]− 4
5L−1
[ (s + 1
2
)+ 1
2[(s + 1
2
)2+ 1
]].
Differential equations with discontinuous sources.
Example
Use the Laplace transform to find the solution of the IVP
y ′′ + y ′ +5
4y = b(t),
y(0) = 0,
y ′(0) = 0,b(t) =
{1, t ∈ [0, π)
0, t ∈ [π,∞).
Solution: Recall: h(t) =4
5L−1
[1
s− (s + 1)(
s2 + s + 54
)].
In this case we complete the square in the denominator,
s2 + s +5
4=
[s2 + 2
(1
2
)s +
1
4
]− 1
4+
5
4=
(s +
1
2
)2+ 1.
So: h(t) =4
5L−1
[1
s− (s + 1)[(
s + 12
)2+ 1
]].
That is, h(t) =4
5L−1
[1
s
]− 4
5L−1
[ (s + 1
2
)+ 1
2[(s + 1
2
)2+ 1
]].
Differential equations with discontinuous sources.
Example
Use the Laplace transform to find the solution of the IVP
y ′′ + y ′ +5
4y = b(t),
y(0) = 0,
y ′(0) = 0,b(t) =
{1, t ∈ [0, π)
0, t ∈ [π,∞).
Solution: Recall: h(t) =4
5L−1
[1
s
]− 4
5L−1
[ (s + 1
2
)+ 1
2[(s + 1
2
)2+ 1
]].
h(t) =4
5L−1
[1
s
]− 4
5L−1
[ (s + 1
2
)[(
s + 12
)2+ 1
]]− 2
5L−1
[ 1[(s + 1
2
)2+ 1
]].
Recall: L−1[F (s − c)
]= ect f (t). Hence,
h(t) =4
5
[1− e−t/2 cos(t)− 1
2e−t/2 sin(t)
].
We conclude: y(t) = h(t) + u(t − π)h(t − π). C
Differential equations with discontinuous sources.
Example
Use the Laplace transform to find the solution of the IVP
y ′′ + y ′ +5
4y = b(t),
y(0) = 0,
y ′(0) = 0,b(t) =
{1, t ∈ [0, π)
0, t ∈ [π,∞).
Solution: Recall: h(t) =4
5L−1
[1
s
]− 4
5L−1
[ (s + 1
2
)+ 1
2[(s + 1
2
)2+ 1
]].
h(t) =4
5L−1
[1
s
]− 4
5L−1
[ (s + 1
2
)[(
s + 12
)2+ 1
]]− 2
5L−1
[ 1[(s + 1
2
)2+ 1
]].
Recall: L−1[F (s − c)
]= ect f (t). Hence,
h(t) =4
5
[1− e−t/2 cos(t)− 1
2e−t/2 sin(t)
].
We conclude: y(t) = h(t) + u(t − π)h(t − π). C
Differential equations with discontinuous sources.
Example
Use the Laplace transform to find the solution of the IVP
y ′′ + y ′ +5
4y = b(t),
y(0) = 0,
y ′(0) = 0,b(t) =
{1, t ∈ [0, π)
0, t ∈ [π,∞).
Solution: Recall: h(t) =4
5L−1
[1
s
]− 4
5L−1
[ (s + 1
2
)+ 1
2[(s + 1
2
)2+ 1
]].
h(t) =4
5L−1
[1
s
]− 4
5L−1
[ (s + 1
2
)[(
s + 12
)2+ 1
]]− 2
5L−1
[ 1[(s + 1
2
)2+ 1
]].
Recall: L−1[F (s − c)
]= ect f (t).
Hence,
h(t) =4
5
[1− e−t/2 cos(t)− 1
2e−t/2 sin(t)
].
We conclude: y(t) = h(t) + u(t − π)h(t − π). C
Differential equations with discontinuous sources.
Example
Use the Laplace transform to find the solution of the IVP
y ′′ + y ′ +5
4y = b(t),
y(0) = 0,
y ′(0) = 0,b(t) =
{1, t ∈ [0, π)
0, t ∈ [π,∞).
Solution: Recall: h(t) =4
5L−1
[1
s
]− 4
5L−1
[ (s + 1
2
)+ 1
2[(s + 1
2
)2+ 1
]].
h(t) =4
5L−1
[1
s
]− 4
5L−1
[ (s + 1
2
)[(
s + 12
)2+ 1
]]− 2
5L−1
[ 1[(s + 1
2
)2+ 1
]].
Recall: L−1[F (s − c)
]= ect f (t). Hence,
h(t) =4
5
[1− e−t/2 cos(t)− 1
2e−t/2 sin(t)
].
We conclude: y(t) = h(t) + u(t − π)h(t − π). C
Differential equations with discontinuous sources.
Example
Use the Laplace transform to find the solution of the IVP
y ′′ + y ′ +5
4y = b(t),
y(0) = 0,
y ′(0) = 0,b(t) =
{1, t ∈ [0, π)
0, t ∈ [π,∞).
Solution: Recall: h(t) =4
5L−1
[1
s
]− 4
5L−1
[ (s + 1
2
)+ 1
2[(s + 1
2
)2+ 1
]].
h(t) =4
5L−1
[1
s
]− 4
5L−1
[ (s + 1
2
)[(
s + 12
)2+ 1
]]− 2
5L−1
[ 1[(s + 1
2
)2+ 1
]].
Recall: L−1[F (s − c)
]= ect f (t). Hence,
h(t) =4
5
[1− e−t/2 cos(t)− 1
2e−t/2 sin(t)
].
We conclude: y(t) = h(t) + u(t − π)h(t − π). C
Equations with discontinuous sources (Sect. 6.4).
I Differential equations with discontinuous sources.I We solve the IVPs:
(a) Example 1:
y ′ + 2y = u(t − 4), y(0) = 3.
(b) Example 2:
y ′′ + y ′ +5
4y = b(t),
y(0) = 0,
y ′(0) = 0,b(t) =
{1, t ∈ [0, π)
0, t ∈ [π,∞).
(c) Example 3:
y ′′+y ′+5
4y = g(t),
y(0) = 0,
y ′(0) = 0,g(t) =
{sin(t), t ∈ [0, π)
0, t ∈ [π,∞).
Differential equations with discontinuous sources.
Example
Use the Laplace transform to find the solution of the IVP
y ′′ + y ′ +5
4y = g(t),
y(0) = 0,
y ′(0) = 0,g(t) =
{sin(t) t ∈ [0, π)
0 t ∈ [π,∞).
Solution:
Rewrite the source function usingstep functions.
g ( t )
0 tpi
1
y
sin ( t )
1
0 pi t
y y
0
1
u ( t ) − u ( t − pi )
pi t
Differential equations with discontinuous sources.
Example
Use the Laplace transform to find the solution of the IVP
y ′′ + y ′ +5
4y = g(t),
y(0) = 0,
y ′(0) = 0,g(t) =
{sin(t) t ∈ [0, π)
0 t ∈ [π,∞).
Solution:
Rewrite the source function usingstep functions.
g ( t )
0 tpi
1
y
sin ( t )
1
0 pi t
y y
0
1
u ( t ) − u ( t − pi )
pi t
Differential equations with discontinuous sources.
Example
Use the Laplace transform to find the solution of the IVP
y ′′ + y ′ +5
4y = g(t),
y(0) = 0,
y ′(0) = 0,g(t) =
{sin(t) t ∈ [0, π)
0 t ∈ [π,∞).
Solution:
Rewrite the source function usingstep functions.
g ( t )
0 tpi
1
y
sin ( t )
1
0 pi t
y y
0
1
u ( t ) − u ( t − pi )
pi t
Differential equations with discontinuous sources.
Example
Use the Laplace transform to find the solution of the IVP
y ′′ + y ′ +5
4y = g(t),
y(0) = 0,
y ′(0) = 0,g(t) =
{sin(t) t ∈ [0, π)
0 t ∈ [π,∞).
Solution:
Rewrite the source function usingstep functions.
g ( t )
0 tpi
1
y
sin ( t )
1
0 pi t
y
y
0
1
u ( t ) − u ( t − pi )
pi t
Differential equations with discontinuous sources.
Example
Use the Laplace transform to find the solution of the IVP
y ′′ + y ′ +5
4y = g(t),
y(0) = 0,
y ′(0) = 0,g(t) =
{sin(t) t ∈ [0, π)
0 t ∈ [π,∞).
Solution:
Rewrite the source function usingstep functions.
g ( t )
0 tpi
1
y
sin ( t )
1
0 pi t
y y
0
1
u ( t ) − u ( t − pi )
pi t
Differential equations with discontinuous sources.
Example
Use the Laplace transform to find the solution of the IVP
y ′′ + y ′ +5
4y = g(t),
y(0) = 0,
y ′(0) = 0,g(t) =
{sin(t) t ∈ [0, π)
0 t ∈ [π,∞).
Solution: The graphs imply: g(t) =[u(t)− u(t − π)
]sin(t).
Recall the identity: sin(t) = − sin(t − π). Then,
g(t) = u(t) sin(t)− u(t − π) sin(t),
g(t) = u(t) sin(t) + u(t − π) sin(t − π).
Now is simple to find L[g ], since
L[g(t)] = L[u(t) sin(t)] + L[u(t − π) sin(t − π)].
Differential equations with discontinuous sources.
Example
Use the Laplace transform to find the solution of the IVP
y ′′ + y ′ +5
4y = g(t),
y(0) = 0,
y ′(0) = 0,g(t) =
{sin(t) t ∈ [0, π)
0 t ∈ [π,∞).
Solution: The graphs imply: g(t) =[u(t)− u(t − π)
]sin(t).
Recall the identity: sin(t) = − sin(t − π).
Then,
g(t) = u(t) sin(t)− u(t − π) sin(t),
g(t) = u(t) sin(t) + u(t − π) sin(t − π).
Now is simple to find L[g ], since
L[g(t)] = L[u(t) sin(t)] + L[u(t − π) sin(t − π)].
Differential equations with discontinuous sources.
Example
Use the Laplace transform to find the solution of the IVP
y ′′ + y ′ +5
4y = g(t),
y(0) = 0,
y ′(0) = 0,g(t) =
{sin(t) t ∈ [0, π)
0 t ∈ [π,∞).
Solution: The graphs imply: g(t) =[u(t)− u(t − π)
]sin(t).
Recall the identity: sin(t) = − sin(t − π). Then,
g(t) = u(t) sin(t)− u(t − π) sin(t),
g(t) = u(t) sin(t) + u(t − π) sin(t − π).
Now is simple to find L[g ], since
L[g(t)] = L[u(t) sin(t)] + L[u(t − π) sin(t − π)].
Differential equations with discontinuous sources.
Example
Use the Laplace transform to find the solution of the IVP
y ′′ + y ′ +5
4y = g(t),
y(0) = 0,
y ′(0) = 0,g(t) =
{sin(t) t ∈ [0, π)
0 t ∈ [π,∞).
Solution: The graphs imply: g(t) =[u(t)− u(t − π)
]sin(t).
Recall the identity: sin(t) = − sin(t − π). Then,
g(t) = u(t) sin(t)− u(t − π) sin(t),
g(t) = u(t) sin(t) + u(t − π) sin(t − π).
Now is simple to find L[g ], since
L[g(t)] = L[u(t) sin(t)] + L[u(t − π) sin(t − π)].
Differential equations with discontinuous sources.
Example
Use the Laplace transform to find the solution of the IVP
y ′′ + y ′ +5
4y = g(t),
y(0) = 0,
y ′(0) = 0,g(t) =
{sin(t) t ∈ [0, π)
0 t ∈ [π,∞).
Solution: The graphs imply: g(t) =[u(t)− u(t − π)
]sin(t).
Recall the identity: sin(t) = − sin(t − π). Then,
g(t) = u(t) sin(t)− u(t − π) sin(t),
g(t) = u(t) sin(t) + u(t − π) sin(t − π).
Now is simple to find L[g ],
since
L[g(t)] = L[u(t) sin(t)] + L[u(t − π) sin(t − π)].
Differential equations with discontinuous sources.
Example
Use the Laplace transform to find the solution of the IVP
y ′′ + y ′ +5
4y = g(t),
y(0) = 0,
y ′(0) = 0,g(t) =
{sin(t) t ∈ [0, π)
0 t ∈ [π,∞).
Solution: The graphs imply: g(t) =[u(t)− u(t − π)
]sin(t).
Recall the identity: sin(t) = − sin(t − π). Then,
g(t) = u(t) sin(t)− u(t − π) sin(t),
g(t) = u(t) sin(t) + u(t − π) sin(t − π).
Now is simple to find L[g ], since
L[g(t)] = L[u(t) sin(t)] + L[u(t − π) sin(t − π)].
Differential equations with discontinuous sources.
Example
Use the Laplace transform to find the solution of the IVP
y ′′ + y ′ +5
4y = g(t),
y(0) = 0,
y ′(0) = 0,g(t) =
{sin(t) t ∈ [0, π)
0 t ∈ [π,∞).
Solution: So: L[g(t)] = L[u(t) sin(t)] + L[u(t − π) sin(t − π)].
L[g(t)] =1
(s2 + 1)+ e−πs 1
(s2 + 1).
Recall the Laplace transform of the differential equation
L[y ′′] + L[y ′] +5
4L[y ] = L[g ].
The initial conditions imply: L[y ′′] = s2 L[y ] and L[y ′] = s L[y ].
Therefore,(s2 + s +
5
4
)L[y ] =
(1 + e−πs
) 1
(s2 + 1).
Differential equations with discontinuous sources.
Example
Use the Laplace transform to find the solution of the IVP
y ′′ + y ′ +5
4y = g(t),
y(0) = 0,
y ′(0) = 0,g(t) =
{sin(t) t ∈ [0, π)
0 t ∈ [π,∞).
Solution: So: L[g(t)] = L[u(t) sin(t)] + L[u(t − π) sin(t − π)].
L[g(t)] =1
(s2 + 1)+ e−πs 1
(s2 + 1).
Recall the Laplace transform of the differential equation
L[y ′′] + L[y ′] +5
4L[y ] = L[g ].
The initial conditions imply: L[y ′′] = s2 L[y ] and L[y ′] = s L[y ].
Therefore,(s2 + s +
5
4
)L[y ] =
(1 + e−πs
) 1
(s2 + 1).
Differential equations with discontinuous sources.
Example
Use the Laplace transform to find the solution of the IVP
y ′′ + y ′ +5
4y = g(t),
y(0) = 0,
y ′(0) = 0,g(t) =
{sin(t) t ∈ [0, π)
0 t ∈ [π,∞).
Solution: So: L[g(t)] = L[u(t) sin(t)] + L[u(t − π) sin(t − π)].
L[g(t)] =1
(s2 + 1)+ e−πs 1
(s2 + 1).
Recall the Laplace transform of the differential equation
L[y ′′] + L[y ′] +5
4L[y ] = L[g ].
The initial conditions imply: L[y ′′] = s2 L[y ] and L[y ′] = s L[y ].
Therefore,(s2 + s +
5
4
)L[y ] =
(1 + e−πs
) 1
(s2 + 1).
Differential equations with discontinuous sources.
Example
Use the Laplace transform to find the solution of the IVP
y ′′ + y ′ +5
4y = g(t),
y(0) = 0,
y ′(0) = 0,g(t) =
{sin(t) t ∈ [0, π)
0 t ∈ [π,∞).
Solution: So: L[g(t)] = L[u(t) sin(t)] + L[u(t − π) sin(t − π)].
L[g(t)] =1
(s2 + 1)+ e−πs 1
(s2 + 1).
Recall the Laplace transform of the differential equation
L[y ′′] + L[y ′] +5
4L[y ] = L[g ].
The initial conditions imply:
L[y ′′] = s2 L[y ] and L[y ′] = s L[y ].
Therefore,(s2 + s +
5
4
)L[y ] =
(1 + e−πs
) 1
(s2 + 1).
Differential equations with discontinuous sources.
Example
Use the Laplace transform to find the solution of the IVP
y ′′ + y ′ +5
4y = g(t),
y(0) = 0,
y ′(0) = 0,g(t) =
{sin(t) t ∈ [0, π)
0 t ∈ [π,∞).
Solution: So: L[g(t)] = L[u(t) sin(t)] + L[u(t − π) sin(t − π)].
L[g(t)] =1
(s2 + 1)+ e−πs 1
(s2 + 1).
Recall the Laplace transform of the differential equation
L[y ′′] + L[y ′] +5
4L[y ] = L[g ].
The initial conditions imply: L[y ′′] = s2 L[y ]
and L[y ′] = s L[y ].
Therefore,(s2 + s +
5
4
)L[y ] =
(1 + e−πs
) 1
(s2 + 1).
Differential equations with discontinuous sources.
Example
Use the Laplace transform to find the solution of the IVP
y ′′ + y ′ +5
4y = g(t),
y(0) = 0,
y ′(0) = 0,g(t) =
{sin(t) t ∈ [0, π)
0 t ∈ [π,∞).
Solution: So: L[g(t)] = L[u(t) sin(t)] + L[u(t − π) sin(t − π)].
L[g(t)] =1
(s2 + 1)+ e−πs 1
(s2 + 1).
Recall the Laplace transform of the differential equation
L[y ′′] + L[y ′] +5
4L[y ] = L[g ].
The initial conditions imply: L[y ′′] = s2 L[y ] and L[y ′] = s L[y ].
Therefore,(s2 + s +
5
4
)L[y ] =
(1 + e−πs
) 1
(s2 + 1).
Differential equations with discontinuous sources.
Example
Use the Laplace transform to find the solution of the IVP
y ′′ + y ′ +5
4y = g(t),
y(0) = 0,
y ′(0) = 0,g(t) =
{sin(t) t ∈ [0, π)
0 t ∈ [π,∞).
Solution: So: L[g(t)] = L[u(t) sin(t)] + L[u(t − π) sin(t − π)].
L[g(t)] =1
(s2 + 1)+ e−πs 1
(s2 + 1).
Recall the Laplace transform of the differential equation
L[y ′′] + L[y ′] +5
4L[y ] = L[g ].
The initial conditions imply: L[y ′′] = s2 L[y ] and L[y ′] = s L[y ].
Therefore,(s2 + s +
5
4
)L[y ] =
(1 + e−πs
) 1
(s2 + 1).
Differential equations with discontinuous sources.
Example
Use the Laplace transform to find the solution of the IVP
y ′′ + y ′ +5
4y = g(t),
y(0) = 0,
y ′(0) = 0,g(t) =
{sin(t) t ∈ [0, π)
0 t ∈ [π,∞).
Solution: Recall:(s2 + s +
5
4
)L[y ] =
(1 + e−πs
) 1
(s2 + 1).
L[y ] =(1 + e−πs
) 1(s2 + s + 5
4
)(s2 + 1)
.
Introduce the function H(s) =1(
s2 + s + 54
)(s2 + 1)
.
Then, y(t) = L−1[H(s)] + L−1[e−πs H(s)].
Differential equations with discontinuous sources.
Example
Use the Laplace transform to find the solution of the IVP
y ′′ + y ′ +5
4y = g(t),
y(0) = 0,
y ′(0) = 0,g(t) =
{sin(t) t ∈ [0, π)
0 t ∈ [π,∞).
Solution: Recall:(s2 + s +
5
4
)L[y ] =
(1 + e−πs
) 1
(s2 + 1).
L[y ] =(1 + e−πs
) 1(s2 + s + 5
4
)(s2 + 1)
.
Introduce the function H(s) =1(
s2 + s + 54
)(s2 + 1)
.
Then, y(t) = L−1[H(s)] + L−1[e−πs H(s)].
Differential equations with discontinuous sources.
Example
Use the Laplace transform to find the solution of the IVP
y ′′ + y ′ +5
4y = g(t),
y(0) = 0,
y ′(0) = 0,g(t) =
{sin(t) t ∈ [0, π)
0 t ∈ [π,∞).
Solution: Recall:(s2 + s +
5
4
)L[y ] =
(1 + e−πs
) 1
(s2 + 1).
L[y ] =(1 + e−πs
) 1(s2 + s + 5
4
)(s2 + 1)
.
Introduce the function H(s) =1(
s2 + s + 54
)(s2 + 1)
.
Then, y(t) = L−1[H(s)] + L−1[e−πs H(s)].
Differential equations with discontinuous sources.
Example
Use the Laplace transform to find the solution of the IVP
y ′′ + y ′ +5
4y = g(t),
y(0) = 0,
y ′(0) = 0,g(t) =
{sin(t) t ∈ [0, π)
0 t ∈ [π,∞).
Solution: Recall:(s2 + s +
5
4
)L[y ] =
(1 + e−πs
) 1
(s2 + 1).
L[y ] =(1 + e−πs
) 1(s2 + s + 5
4
)(s2 + 1)
.
Introduce the function H(s) =1(
s2 + s + 54
)(s2 + 1)
.
Then, y(t) = L−1[H(s)] + L−1[e−πs H(s)].
Differential equations with discontinuous sources.Example
Use the Laplace transform to find the solution of the IVP
y ′′ + y ′ +5
4y = g(t),
y(0) = 0,
y ′(0) = 0,g(t) =
{sin(t) t ∈ [0, π)
0 t ∈ [π,∞).
Solution: Recall: y(t) = L−1[H(s)] + L−1[e−πs H(s)], and
H(s) =1(
s2 + s + 54
)(s2 + 1)
.
Partial fractions: Find the zeros of the denominator,
s± =1
2
[−1±
√1− 5
]⇒ Complex roots.
The partial fraction decomposition is:
1(s2 + s + 5
4
)(s2 + 1)
=(as + b)(
s2 + s + 54
) +(cs + d)
(s2 + 1).
Differential equations with discontinuous sources.Example
Use the Laplace transform to find the solution of the IVP
y ′′ + y ′ +5
4y = g(t),
y(0) = 0,
y ′(0) = 0,g(t) =
{sin(t) t ∈ [0, π)
0 t ∈ [π,∞).
Solution: Recall: y(t) = L−1[H(s)] + L−1[e−πs H(s)], and
H(s) =1(
s2 + s + 54
)(s2 + 1)
.
Partial fractions:
Find the zeros of the denominator,
s± =1
2
[−1±
√1− 5
]⇒ Complex roots.
The partial fraction decomposition is:
1(s2 + s + 5
4
)(s2 + 1)
=(as + b)(
s2 + s + 54
) +(cs + d)
(s2 + 1).
Differential equations with discontinuous sources.Example
Use the Laplace transform to find the solution of the IVP
y ′′ + y ′ +5
4y = g(t),
y(0) = 0,
y ′(0) = 0,g(t) =
{sin(t) t ∈ [0, π)
0 t ∈ [π,∞).
Solution: Recall: y(t) = L−1[H(s)] + L−1[e−πs H(s)], and
H(s) =1(
s2 + s + 54
)(s2 + 1)
.
Partial fractions: Find the zeros of the denominator,
s± =1
2
[−1±
√1− 5
]⇒ Complex roots.
The partial fraction decomposition is:
1(s2 + s + 5
4
)(s2 + 1)
=(as + b)(
s2 + s + 54
) +(cs + d)
(s2 + 1).
Differential equations with discontinuous sources.Example
Use the Laplace transform to find the solution of the IVP
y ′′ + y ′ +5
4y = g(t),
y(0) = 0,
y ′(0) = 0,g(t) =
{sin(t) t ∈ [0, π)
0 t ∈ [π,∞).
Solution: Recall: y(t) = L−1[H(s)] + L−1[e−πs H(s)], and
H(s) =1(
s2 + s + 54
)(s2 + 1)
.
Partial fractions: Find the zeros of the denominator,
s± =1
2
[−1±
√1− 5
]
⇒ Complex roots.
The partial fraction decomposition is:
1(s2 + s + 5
4
)(s2 + 1)
=(as + b)(
s2 + s + 54
) +(cs + d)
(s2 + 1).
Differential equations with discontinuous sources.Example
Use the Laplace transform to find the solution of the IVP
y ′′ + y ′ +5
4y = g(t),
y(0) = 0,
y ′(0) = 0,g(t) =
{sin(t) t ∈ [0, π)
0 t ∈ [π,∞).
Solution: Recall: y(t) = L−1[H(s)] + L−1[e−πs H(s)], and
H(s) =1(
s2 + s + 54
)(s2 + 1)
.
Partial fractions: Find the zeros of the denominator,
s± =1
2
[−1±
√1− 5
]⇒ Complex roots.
The partial fraction decomposition is:
1(s2 + s + 5
4
)(s2 + 1)
=(as + b)(
s2 + s + 54
) +(cs + d)
(s2 + 1).
Differential equations with discontinuous sources.Example
Use the Laplace transform to find the solution of the IVP
y ′′ + y ′ +5
4y = g(t),
y(0) = 0,
y ′(0) = 0,g(t) =
{sin(t) t ∈ [0, π)
0 t ∈ [π,∞).
Solution: Recall: y(t) = L−1[H(s)] + L−1[e−πs H(s)], and
H(s) =1(
s2 + s + 54
)(s2 + 1)
.
Partial fractions: Find the zeros of the denominator,
s± =1
2
[−1±
√1− 5
]⇒ Complex roots.
The partial fraction decomposition is:
1(s2 + s + 5
4
)(s2 + 1)
=(as + b)(
s2 + s + 54
) +(cs + d)
(s2 + 1).
Differential equations with discontinuous sources.Example
Use the Laplace transform to find the solution of the IVP
y ′′ + y ′ +5
4y = g(t),
y(0) = 0,
y ′(0) = 0,g(t) =
{sin(t) t ∈ [0, π)
0 t ∈ [π,∞).
Solution: Recall: y(t) = L−1[H(s)] + L−1[e−πs H(s)], and
H(s) =1(
s2 + s + 54
)(s2 + 1)
.
Partial fractions: Find the zeros of the denominator,
s± =1
2
[−1±
√1− 5
]⇒ Complex roots.
The partial fraction decomposition is:
1(s2 + s + 5
4
)(s2 + 1)
=(as + b)(
s2 + s + 54
) +(cs + d)
(s2 + 1).
Differential equations with discontinuous sources.
Example
Use the Laplace transform to find the solution of the IVP
y ′′ + y ′ +5
4y = g(t),
y(0) = 0,
y ′(0) = 0,g(t) =
{sin(t) t ∈ [0, π)
0 t ∈ [π,∞).
Solution: So:1(
s2 + s + 54
)(s2 + 1)
=(as + b)(
s2 + s + 54
) +(cs + d)
(s2 + 1).
Therefore, we get
1 = (as + b)(s2 + 1) + (cs + d)(s2 + s +
5
4
),
1 = (a + c) s3 + (b + c + d) s2 +(a +
5
4c + d
)s +
(b +
5
4d).
This equation implies that a, b, c , and d , are solutions of
a + c = 0, b + c + d = 0, a +5
4c + d = 0, b +
5
4d = 1.
Differential equations with discontinuous sources.
Example
Use the Laplace transform to find the solution of the IVP
y ′′ + y ′ +5
4y = g(t),
y(0) = 0,
y ′(0) = 0,g(t) =
{sin(t) t ∈ [0, π)
0 t ∈ [π,∞).
Solution: So:1(
s2 + s + 54
)(s2 + 1)
=(as + b)(
s2 + s + 54
) +(cs + d)
(s2 + 1).
Therefore, we get
1 = (as + b)(s2 + 1) + (cs + d)(s2 + s +
5
4
),
1 = (a + c) s3 + (b + c + d) s2 +(a +
5
4c + d
)s +
(b +
5
4d).
This equation implies that a, b, c , and d , are solutions of
a + c = 0, b + c + d = 0, a +5
4c + d = 0, b +
5
4d = 1.
Differential equations with discontinuous sources.
Example
Use the Laplace transform to find the solution of the IVP
y ′′ + y ′ +5
4y = g(t),
y(0) = 0,
y ′(0) = 0,g(t) =
{sin(t) t ∈ [0, π)
0 t ∈ [π,∞).
Solution: So:1(
s2 + s + 54
)(s2 + 1)
=(as + b)(
s2 + s + 54
) +(cs + d)
(s2 + 1).
Therefore, we get
1 = (as + b)(s2 + 1) + (cs + d)(s2 + s +
5
4
),
1 = (a + c) s3 + (b + c + d) s2 +(a +
5
4c + d
)s +
(b +
5
4d).
This equation implies that a, b, c , and d , are solutions of
a + c = 0, b + c + d = 0, a +5
4c + d = 0, b +
5
4d = 1.
Differential equations with discontinuous sources.
Example
Use the Laplace transform to find the solution of the IVP
y ′′ + y ′ +5
4y = g(t),
y(0) = 0,
y ′(0) = 0,g(t) =
{sin(t) t ∈ [0, π)
0 t ∈ [π,∞).
Solution: So:1(
s2 + s + 54
)(s2 + 1)
=(as + b)(
s2 + s + 54
) +(cs + d)
(s2 + 1).
Therefore, we get
1 = (as + b)(s2 + 1) + (cs + d)(s2 + s +
5
4
),
1 = (a + c) s3 + (b + c + d) s2 +(a +
5
4c + d
)s +
(b +
5
4d).
This equation implies that a, b, c , and d , are solutions of
a + c = 0, b + c + d = 0, a +5
4c + d = 0, b +
5
4d = 1.
Differential equations with discontinuous sources.
Example
Use the Laplace transform to find the solution of the IVP
y ′′ + y ′ +5
4y = g(t),
y(0) = 0,
y ′(0) = 0,g(t) =
{sin(t) t ∈ [0, π)
0 t ∈ [π,∞).
Solution: So: a =16
17, b =
12
17, c = −16
17, d =
4
17.
We have found: H(s) =4
17
[ (4s + 3)(s2 + s + 5
4
) +(−4s + 1)
(s2 + 1)
].
Complete the square in the denominator,
s2 + s +5
4=
[s2 + 2
(1
2
)s +
1
4
]− 1
4+
5
4=
(s +
1
2
)2+ 1.
H(s) =4
17
[ (4s + 3)[(s + 1
2
)2+ 1
] +(−4s + 1)
(s2 + 1)
].
Differential equations with discontinuous sources.
Example
Use the Laplace transform to find the solution of the IVP
y ′′ + y ′ +5
4y = g(t),
y(0) = 0,
y ′(0) = 0,g(t) =
{sin(t) t ∈ [0, π)
0 t ∈ [π,∞).
Solution: So: a =16
17, b =
12
17, c = −16
17, d =
4
17.
We have found: H(s) =4
17
[ (4s + 3)(s2 + s + 5
4
) +(−4s + 1)
(s2 + 1)
].
Complete the square in the denominator,
s2 + s +5
4=
[s2 + 2
(1
2
)s +
1
4
]− 1
4+
5
4=
(s +
1
2
)2+ 1.
H(s) =4
17
[ (4s + 3)[(s + 1
2
)2+ 1
] +(−4s + 1)
(s2 + 1)
].
Differential equations with discontinuous sources.
Example
Use the Laplace transform to find the solution of the IVP
y ′′ + y ′ +5
4y = g(t),
y(0) = 0,
y ′(0) = 0,g(t) =
{sin(t) t ∈ [0, π)
0 t ∈ [π,∞).
Solution: So: a =16
17, b =
12
17, c = −16
17, d =
4
17.
We have found: H(s) =4
17
[ (4s + 3)(s2 + s + 5
4
) +(−4s + 1)
(s2 + 1)
].
Complete the square in the denominator,
s2 + s +5
4=
[s2 + 2
(1
2
)s +
1
4
]− 1
4+
5
4=
(s +
1
2
)2+ 1.
H(s) =4
17
[ (4s + 3)[(s + 1
2
)2+ 1
] +(−4s + 1)
(s2 + 1)
].
Differential equations with discontinuous sources.
Example
Use the Laplace transform to find the solution of the IVP
y ′′ + y ′ +5
4y = g(t),
y(0) = 0,
y ′(0) = 0,g(t) =
{sin(t) t ∈ [0, π)
0 t ∈ [π,∞).
Solution: So: a =16
17, b =
12
17, c = −16
17, d =
4
17.
We have found: H(s) =4
17
[ (4s + 3)(s2 + s + 5
4
) +(−4s + 1)
(s2 + 1)
].
Complete the square in the denominator,
s2 + s +5
4=
[s2 + 2
(1
2
)s +
1
4
]− 1
4+
5
4
=(s +
1
2
)2+ 1.
H(s) =4
17
[ (4s + 3)[(s + 1
2
)2+ 1
] +(−4s + 1)
(s2 + 1)
].
Differential equations with discontinuous sources.
Example
Use the Laplace transform to find the solution of the IVP
y ′′ + y ′ +5
4y = g(t),
y(0) = 0,
y ′(0) = 0,g(t) =
{sin(t) t ∈ [0, π)
0 t ∈ [π,∞).
Solution: So: a =16
17, b =
12
17, c = −16
17, d =
4
17.
We have found: H(s) =4
17
[ (4s + 3)(s2 + s + 5
4
) +(−4s + 1)
(s2 + 1)
].
Complete the square in the denominator,
s2 + s +5
4=
[s2 + 2
(1
2
)s +
1
4
]− 1
4+
5
4=
(s +
1
2
)2+ 1.
H(s) =4
17
[ (4s + 3)[(s + 1
2
)2+ 1
] +(−4s + 1)
(s2 + 1)
].
Differential equations with discontinuous sources.
Example
Use the Laplace transform to find the solution of the IVP
y ′′ + y ′ +5
4y = g(t),
y(0) = 0,
y ′(0) = 0,g(t) =
{sin(t) t ∈ [0, π)
0 t ∈ [π,∞).
Solution: So: a =16
17, b =
12
17, c = −16
17, d =
4
17.
We have found: H(s) =4
17
[ (4s + 3)(s2 + s + 5
4
) +(−4s + 1)
(s2 + 1)
].
Complete the square in the denominator,
s2 + s +5
4=
[s2 + 2
(1
2
)s +
1
4
]− 1
4+
5
4=
(s +
1
2
)2+ 1.
H(s) =4
17
[ (4s + 3)[(s + 1
2
)2+ 1
] +(−4s + 1)
(s2 + 1)
].
Differential equations with discontinuous sources.
Example
Use the Laplace transform to find the solution of the IVP
y ′′ + y ′ +5
4y = g(t),
y(0) = 0,
y ′(0) = 0,g(t) =
{sin(t) t ∈ [0, π)
0 t ∈ [π,∞).
Solution: So: H(s) =4
17
[ (4s + 3)[(s + 1
2
)2+ 1
] +(−4s + 1)
(s2 + 1)
].
Rewrite the polynomial in the numerator,
(4s + 3) = 4(s +
1
2− 1
2
)+ 3 = 4
(s +
1
2
)+ 1,
H(s) =4
17
[4
(s + 1
2
)[(s + 1
2
)2+ 1
] +1[(
s + 12
)2+ 1
] − 4s
(s2 + 1)+
1
(s2 + 1)
],
Differential equations with discontinuous sources.
Example
Use the Laplace transform to find the solution of the IVP
y ′′ + y ′ +5
4y = g(t),
y(0) = 0,
y ′(0) = 0,g(t) =
{sin(t) t ∈ [0, π)
0 t ∈ [π,∞).
Solution: So: H(s) =4
17
[ (4s + 3)[(s + 1
2
)2+ 1
] +(−4s + 1)
(s2 + 1)
].
Rewrite the polynomial in the numerator,
(4s + 3) = 4(s +
1
2− 1
2
)+ 3 = 4
(s +
1
2
)+ 1,
H(s) =4
17
[4
(s + 1
2
)[(s + 1
2
)2+ 1
] +1[(
s + 12
)2+ 1
] − 4s
(s2 + 1)+
1
(s2 + 1)
],
Differential equations with discontinuous sources.
Example
Use the Laplace transform to find the solution of the IVP
y ′′ + y ′ +5
4y = g(t),
y(0) = 0,
y ′(0) = 0,g(t) =
{sin(t) t ∈ [0, π)
0 t ∈ [π,∞).
Solution: So: H(s) =4
17
[ (4s + 3)[(s + 1
2
)2+ 1
] +(−4s + 1)
(s2 + 1)
].
Rewrite the polynomial in the numerator,
(4s + 3) = 4(s +
1
2− 1
2
)+ 3
= 4(s +
1
2
)+ 1,
H(s) =4
17
[4
(s + 1
2
)[(s + 1
2
)2+ 1
] +1[(
s + 12
)2+ 1
] − 4s
(s2 + 1)+
1
(s2 + 1)
],
Differential equations with discontinuous sources.
Example
Use the Laplace transform to find the solution of the IVP
y ′′ + y ′ +5
4y = g(t),
y(0) = 0,
y ′(0) = 0,g(t) =
{sin(t) t ∈ [0, π)
0 t ∈ [π,∞).
Solution: So: H(s) =4
17
[ (4s + 3)[(s + 1
2
)2+ 1
] +(−4s + 1)
(s2 + 1)
].
Rewrite the polynomial in the numerator,
(4s + 3) = 4(s +
1
2− 1
2
)+ 3 = 4
(s +
1
2
)+ 1,
H(s) =4
17
[4
(s + 1
2
)[(s + 1
2
)2+ 1
] +1[(
s + 12
)2+ 1
] − 4s
(s2 + 1)+
1
(s2 + 1)
],
Differential equations with discontinuous sources.
Example
Use the Laplace transform to find the solution of the IVP
y ′′ + y ′ +5
4y = g(t),
y(0) = 0,
y ′(0) = 0,g(t) =
{sin(t) t ∈ [0, π)
0 t ∈ [π,∞).
Solution: So: H(s) =4
17
[ (4s + 3)[(s + 1
2
)2+ 1
] +(−4s + 1)
(s2 + 1)
].
Rewrite the polynomial in the numerator,
(4s + 3) = 4(s +
1
2− 1
2
)+ 3 = 4
(s +
1
2
)+ 1,
H(s) =4
17
[4
(s + 1
2
)[(s + 1
2
)2+ 1
] +1[(
s + 12
)2+ 1
] − 4s
(s2 + 1)+
1
(s2 + 1)
],
Differential equations with discontinuous sources.Example
Use the Laplace transform to find the solution of the IVP
y ′′ + y ′ +5
4y = g(t),
y(0) = 0,
y ′(0) = 0,g(t) =
{sin(t) t ∈ [0, π)
0 t ∈ [π,∞).
Solution:
H(s) =4
17
[4
(s + 1
2
)[(s + 1
2
)2+ 1
] +1[(
s + 12
)2+ 1
] − 4s
(s2 + 1)+
1
(s2 + 1)
],
Use the Laplace Transform table to get H(s) equal to
H(s) =4
17
[4L
[e−t/2 cos(t)
]+L
[e−t/2 sin(t)
]− 4L[cos(t)]+L[sin(t)]
].
H(s) = L[ 4
17
(4e−t/2 cos(t) + e−t/2 sin(t)− 4 cos(t) + sin(t)
)].
Differential equations with discontinuous sources.Example
Use the Laplace transform to find the solution of the IVP
y ′′ + y ′ +5
4y = g(t),
y(0) = 0,
y ′(0) = 0,g(t) =
{sin(t) t ∈ [0, π)
0 t ∈ [π,∞).
Solution:
H(s) =4
17
[4
(s + 1
2
)[(s + 1
2
)2+ 1
] +1[(
s + 12
)2+ 1
] − 4s
(s2 + 1)+
1
(s2 + 1)
],
Use the Laplace Transform table to get H(s) equal to
H(s) =4
17
[4L
[e−t/2 cos(t)
]+L
[e−t/2 sin(t)
]− 4L[cos(t)]+L[sin(t)]
].
H(s) = L[ 4
17
(4e−t/2 cos(t) + e−t/2 sin(t)− 4 cos(t) + sin(t)
)].
Differential equations with discontinuous sources.Example
Use the Laplace transform to find the solution of the IVP
y ′′ + y ′ +5
4y = g(t),
y(0) = 0,
y ′(0) = 0,g(t) =
{sin(t) t ∈ [0, π)
0 t ∈ [π,∞).
Solution:
H(s) =4
17
[4
(s + 1
2
)[(s + 1
2
)2+ 1
] +1[(
s + 12
)2+ 1
] − 4s
(s2 + 1)+
1
(s2 + 1)
],
Use the Laplace Transform table to get H(s) equal to
H(s) =4
17
[4L
[e−t/2 cos(t)
]+L
[e−t/2 sin(t)
]− 4L[cos(t)]+L[sin(t)]
].
H(s) = L[ 4
17
(4e−t/2 cos(t) + e−t/2 sin(t)− 4 cos(t) + sin(t)
)].
Differential equations with discontinuous sources.
Example
Use the Laplace transform to find the solution of the IVP
y ′′ + y ′ +5
4y = g(t),
y(0) = 0,
y ′(0) = 0,g(t) =
{sin(t) t ∈ [0, π)
0 t ∈ [π,∞).
Solution: Recall:
H(s) = L[ 4
17
(4e−t/2 cos(t) + e−t/2 sin(t)− 4 cos(t) + sin(t)
)].
Denote:
h(t) =4
17
[4e−t/2 cos(t) + e−t/2 sin(t)− 4 cos(t) + sin(t)
].
Then, H(s) = L[h(t)]. Recalling: L[y(t)] = H(s) + e−πs H(s),
L[y(t)] = L[h(t)] + e−πs L[h(t)].
We conclude: y(t) = h(t) + u(t − π)h(t − π). C
Differential equations with discontinuous sources.
Example
Use the Laplace transform to find the solution of the IVP
y ′′ + y ′ +5
4y = g(t),
y(0) = 0,
y ′(0) = 0,g(t) =
{sin(t) t ∈ [0, π)
0 t ∈ [π,∞).
Solution: Recall:
H(s) = L[ 4
17
(4e−t/2 cos(t) + e−t/2 sin(t)− 4 cos(t) + sin(t)
)].
Denote:
h(t) =4
17
[4e−t/2 cos(t) + e−t/2 sin(t)− 4 cos(t) + sin(t)
].
Then, H(s) = L[h(t)]. Recalling: L[y(t)] = H(s) + e−πs H(s),
L[y(t)] = L[h(t)] + e−πs L[h(t)].
We conclude: y(t) = h(t) + u(t − π)h(t − π). C
Differential equations with discontinuous sources.
Example
Use the Laplace transform to find the solution of the IVP
y ′′ + y ′ +5
4y = g(t),
y(0) = 0,
y ′(0) = 0,g(t) =
{sin(t) t ∈ [0, π)
0 t ∈ [π,∞).
Solution: Recall:
H(s) = L[ 4
17
(4e−t/2 cos(t) + e−t/2 sin(t)− 4 cos(t) + sin(t)
)].
Denote:
h(t) =4
17
[4e−t/2 cos(t) + e−t/2 sin(t)− 4 cos(t) + sin(t)
].
Then, H(s) = L[h(t)].
Recalling: L[y(t)] = H(s) + e−πs H(s),
L[y(t)] = L[h(t)] + e−πs L[h(t)].
We conclude: y(t) = h(t) + u(t − π)h(t − π). C
Differential equations with discontinuous sources.
Example
Use the Laplace transform to find the solution of the IVP
y ′′ + y ′ +5
4y = g(t),
y(0) = 0,
y ′(0) = 0,g(t) =
{sin(t) t ∈ [0, π)
0 t ∈ [π,∞).
Solution: Recall:
H(s) = L[ 4
17
(4e−t/2 cos(t) + e−t/2 sin(t)− 4 cos(t) + sin(t)
)].
Denote:
h(t) =4
17
[4e−t/2 cos(t) + e−t/2 sin(t)− 4 cos(t) + sin(t)
].
Then, H(s) = L[h(t)]. Recalling: L[y(t)] = H(s) + e−πs H(s),
L[y(t)] = L[h(t)] + e−πs L[h(t)].
We conclude: y(t) = h(t) + u(t − π)h(t − π). C
Differential equations with discontinuous sources.
Example
Use the Laplace transform to find the solution of the IVP
y ′′ + y ′ +5
4y = g(t),
y(0) = 0,
y ′(0) = 0,g(t) =
{sin(t) t ∈ [0, π)
0 t ∈ [π,∞).
Solution: Recall:
H(s) = L[ 4
17
(4e−t/2 cos(t) + e−t/2 sin(t)− 4 cos(t) + sin(t)
)].
Denote:
h(t) =4
17
[4e−t/2 cos(t) + e−t/2 sin(t)− 4 cos(t) + sin(t)
].
Then, H(s) = L[h(t)]. Recalling: L[y(t)] = H(s) + e−πs H(s),
L[y(t)] = L[h(t)] + e−πs L[h(t)].
We conclude: y(t) = h(t) + u(t − π)h(t − π). C
Differential equations with discontinuous sources.
Example
Use the Laplace transform to find the solution of the IVP
y ′′ + y ′ +5
4y = g(t),
y(0) = 0,
y ′(0) = 0,g(t) =
{sin(t) t ∈ [0, π)
0 t ∈ [π,∞).
Solution: Recall:
H(s) = L[ 4
17
(4e−t/2 cos(t) + e−t/2 sin(t)− 4 cos(t) + sin(t)
)].
Denote:
h(t) =4
17
[4e−t/2 cos(t) + e−t/2 sin(t)− 4 cos(t) + sin(t)
].
Then, H(s) = L[h(t)]. Recalling: L[y(t)] = H(s) + e−πs H(s),
L[y(t)] = L[h(t)] + e−πs L[h(t)].
We conclude: y(t) = h(t) + u(t − π)h(t − π). C
Generalized sources (Sect. 6.5).
I The Dirac delta generalized function.
I Properties of Dirac’s delta.
I Relation between deltas and steps.
I Dirac’s delta in Physics.
I The Laplace Transform of Dirac’s delta.
I Differential equations with Dirac’s delta sources.
Generalized sources (Sect. 6.5).
I The Dirac delta generalized function.
I Properties of Dirac’s delta.
I Relation between deltas and steps.
I Dirac’s delta in Physics.
I The Laplace Transform of Dirac’s delta.
I Differential equations with Dirac’s delta sources.
The Dirac delta generalized function.
DefinitionConsider the sequence of functions for n > 1,
δn(t) =
0, t < 0
n, 0 6 t 61
n
0, t >1
n.
d
0
1
2
3
2
1
1 t
3
nd
d
d
The Dirac delta generalized function is given by
limn→∞
δn(t) = δ(t), t ∈ R.
Remarks:
(a) There exist infinitely many sequences δn that define the samegeneralized function δ.
(b) For example, compare with the sequence δn in the textbook.
The Dirac delta generalized function.
DefinitionConsider the sequence of functions for n > 1,
δn(t) =
0, t < 0
n, 0 6 t 61
n
0, t >1
n. d
0
1
2
3
2
1
1 t
3
nd
d
d
The Dirac delta generalized function is given by
limn→∞
δn(t) = δ(t), t ∈ R.
Remarks:
(a) There exist infinitely many sequences δn that define the samegeneralized function δ.
(b) For example, compare with the sequence δn in the textbook.
The Dirac delta generalized function.
DefinitionConsider the sequence of functions for n > 1,
δn(t) =
0, t < 0
n, 0 6 t 61
n
0, t >1
n. d
0
1
2
3
2
1
1 t
3
nd
d
d
The Dirac delta generalized function is given by
limn→∞
δn(t) = δ(t), t ∈ R.
Remarks:
(a) There exist infinitely many sequences δn that define the samegeneralized function δ.
(b) For example, compare with the sequence δn in the textbook.
The Dirac delta generalized function.
DefinitionConsider the sequence of functions for n > 1,
δn(t) =
0, t < 0
n, 0 6 t 61
n
0, t >1
n. d
0
1
2
3
2
1
1 t
3
nd
d
d
The Dirac delta generalized function is given by
limn→∞
δn(t) = δ(t), t ∈ R.
Remarks:
(a) There exist infinitely many sequences δn that define the samegeneralized function δ.
(b) For example, compare with the sequence δn in the textbook.
The Dirac delta generalized function.
DefinitionConsider the sequence of functions for n > 1,
δn(t) =
0, t < 0
n, 0 6 t 61
n
0, t >1
n. d
0
1
2
3
2
1
1 t
3
nd
d
d
The Dirac delta generalized function is given by
limn→∞
δn(t) = δ(t), t ∈ R.
Remarks:
(a) There exist infinitely many sequences δn that define the samegeneralized function δ.
(b) For example, compare with the sequence δn in the textbook.
The Dirac delta generalized function.
d
0
1
2
3
2
1
1 t
3
nd
d
d
R − { 0 }
0 t
delta
Remarks:
(a) The Dirac δ is a function on the domain R− {0}, andδ(t) = 0 for t ∈ R− {0}.
(b) δ at t = 0 is not defined, since δ(0) = limn→∞ n = +∞.
(c) δ is not a function on R.
The Dirac delta generalized function.
d
0
1
2
3
2
1
1 t
3
nd
d
d
R − { 0 }
0 t
delta
Remarks:
(a) The Dirac δ is a function on the domain R− {0}, andδ(t) = 0 for t ∈ R− {0}.
(b) δ at t = 0 is not defined, since δ(0) = limn→∞ n = +∞.
(c) δ is not a function on R.
The Dirac delta generalized function.
d
0
1
2
3
2
1
1 t
3
nd
d
d
R − { 0 }
0 t
delta
Remarks:
(a) The Dirac δ is a function on the domain R− {0}, andδ(t) = 0 for t ∈ R− {0}.
(b) δ at t = 0 is not defined, since δ(0) = limn→∞ n = +∞.
(c) δ is not a function on R.
The Dirac delta generalized function.
d
0
1
2
3
2
1
1 t
3
nd
d
d
R − { 0 }
0 t
delta
Remarks:
(a) The Dirac δ is a function on the domain R− {0}, andδ(t) = 0 for t ∈ R− {0}.
(b) δ at t = 0 is not defined, since δ(0) = limn→∞ n = +∞.
(c) δ is not a function on R.
The Dirac delta generalized function.
d
0
1
2
3
2
1
1 t
3
nd
d
d
R − { 0 }
0 t
delta
Remarks:
(a) The Dirac δ is a function on the domain R− {0}, andδ(t) = 0 for t ∈ R− {0}.
(b) δ at t = 0 is not defined, since δ(0) = limn→∞ n = +∞.
(c) δ is not a function on R.
Generalized sources (Sect. 6.5).
I The Dirac delta generalized function.
I Properties of Dirac’s delta.
I Relation between deltas and steps.
I Dirac’s delta in Physics.
I The Laplace Transform of Dirac’s delta.
I Differential equations with Dirac’s delta sources.
Properties of Dirac’s delta.
Remark: The Dirac δ is not a function.
We define operations on Dirac’s δ as limits n →∞ of theoperation on the sequence elements δn.
Definition
δ(t − c) = limn→∞
δn(t − c),
a δ(t) + b δ(t) = limn→∞
[a δn(t) + b δn(t)
],
f (t) δ(t) = limn→∞
[f (t) δn(t)
],∫ b
aδ(t) dt = lim
n→∞
∫ b
aδn(t) dt,
L[δ] = limn→∞
L[δn].
Properties of Dirac’s delta.
Remark: The Dirac δ is not a function.
We define operations on Dirac’s δ as limits n →∞ of theoperation on the sequence elements δn.
Definition
δ(t − c) = limn→∞
δn(t − c),
a δ(t) + b δ(t) = limn→∞
[a δn(t) + b δn(t)
],
f (t) δ(t) = limn→∞
[f (t) δn(t)
],∫ b
aδ(t) dt = lim
n→∞
∫ b
aδn(t) dt,
L[δ] = limn→∞
L[δn].
Properties of Dirac’s delta.
Remark: The Dirac δ is not a function.
We define operations on Dirac’s δ as limits n →∞ of theoperation on the sequence elements δn.
Definition
δ(t − c) = limn→∞
δn(t − c),
a δ(t) + b δ(t) = limn→∞
[a δn(t) + b δn(t)
],
f (t) δ(t) = limn→∞
[f (t) δn(t)
],∫ b
aδ(t) dt = lim
n→∞
∫ b
aδn(t) dt,
L[δ] = limn→∞
L[δn].
Properties of Dirac’s delta.
Remark: The Dirac δ is not a function.
We define operations on Dirac’s δ as limits n →∞ of theoperation on the sequence elements δn.
Definition
δ(t − c) = limn→∞
δn(t − c),
a δ(t) + b δ(t) = limn→∞
[a δn(t) + b δn(t)
],
f (t) δ(t) = limn→∞
[f (t) δn(t)
],∫ b
aδ(t) dt = lim
n→∞
∫ b
aδn(t) dt,
L[δ] = limn→∞
L[δn].
Properties of Dirac’s delta.
Remark: The Dirac δ is not a function.
We define operations on Dirac’s δ as limits n →∞ of theoperation on the sequence elements δn.
Definition
δ(t − c) = limn→∞
δn(t − c),
a δ(t) + b δ(t) = limn→∞
[a δn(t) + b δn(t)
],
f (t) δ(t) = limn→∞
[f (t) δn(t)
],
∫ b
aδ(t) dt = lim
n→∞
∫ b
aδn(t) dt,
L[δ] = limn→∞
L[δn].
Properties of Dirac’s delta.
Remark: The Dirac δ is not a function.
We define operations on Dirac’s δ as limits n →∞ of theoperation on the sequence elements δn.
Definition
δ(t − c) = limn→∞
δn(t − c),
a δ(t) + b δ(t) = limn→∞
[a δn(t) + b δn(t)
],
f (t) δ(t) = limn→∞
[f (t) δn(t)
],∫ b
aδ(t) dt = lim
n→∞
∫ b
aδn(t) dt,
L[δ] = limn→∞
L[δn].
Properties of Dirac’s delta.
Remark: The Dirac δ is not a function.
We define operations on Dirac’s δ as limits n →∞ of theoperation on the sequence elements δn.
Definition
δ(t − c) = limn→∞
δn(t − c),
a δ(t) + b δ(t) = limn→∞
[a δn(t) + b δn(t)
],
f (t) δ(t) = limn→∞
[f (t) δn(t)
],∫ b
aδ(t) dt = lim
n→∞
∫ b
aδn(t) dt,
L[δ] = limn→∞
L[δn].
Properties of Dirac’s delta.
Theorem ∫ a
−aδ(t) dt = 1, a > 0.
Proof: ∫ a
−aδ(t) dt = lim
n→∞
∫ a
−aδn(t) dt = lim
n→∞
∫ 1/n
0n dt
∫ a
−aδ(t) dt = lim
n→∞
[n(t∣∣∣1/n
0
)]= lim
n→∞
[n
1
n
].
We conclude:
∫ a
−aδ(t) dt = 1.
Properties of Dirac’s delta.
Theorem ∫ a
−aδ(t) dt = 1, a > 0.
Proof: ∫ a
−aδ(t) dt
= limn→∞
∫ a
−aδn(t) dt = lim
n→∞
∫ 1/n
0n dt
∫ a
−aδ(t) dt = lim
n→∞
[n(t∣∣∣1/n
0
)]= lim
n→∞
[n
1
n
].
We conclude:
∫ a
−aδ(t) dt = 1.
Properties of Dirac’s delta.
Theorem ∫ a
−aδ(t) dt = 1, a > 0.
Proof: ∫ a
−aδ(t) dt = lim
n→∞
∫ a
−aδn(t) dt
= limn→∞
∫ 1/n
0n dt
∫ a
−aδ(t) dt = lim
n→∞
[n(t∣∣∣1/n
0
)]= lim
n→∞
[n
1
n
].
We conclude:
∫ a
−aδ(t) dt = 1.
Properties of Dirac’s delta.
Theorem ∫ a
−aδ(t) dt = 1, a > 0.
Proof: ∫ a
−aδ(t) dt = lim
n→∞
∫ a
−aδn(t) dt = lim
n→∞
∫ 1/n
0n dt
∫ a
−aδ(t) dt = lim
n→∞
[n(t∣∣∣1/n
0
)]= lim
n→∞
[n
1
n
].
We conclude:
∫ a
−aδ(t) dt = 1.
Properties of Dirac’s delta.
Theorem ∫ a
−aδ(t) dt = 1, a > 0.
Proof: ∫ a
−aδ(t) dt = lim
n→∞
∫ a
−aδn(t) dt = lim
n→∞
∫ 1/n
0n dt
∫ a
−aδ(t) dt = lim
n→∞
[n(t∣∣∣1/n
0
)]
= limn→∞
[n
1
n
].
We conclude:
∫ a
−aδ(t) dt = 1.
Properties of Dirac’s delta.
Theorem ∫ a
−aδ(t) dt = 1, a > 0.
Proof: ∫ a
−aδ(t) dt = lim
n→∞
∫ a
−aδn(t) dt = lim
n→∞
∫ 1/n
0n dt
∫ a
−aδ(t) dt = lim
n→∞
[n(t∣∣∣1/n
0
)]= lim
n→∞
[n
1
n
].
We conclude:
∫ a
−aδ(t) dt = 1.
Properties of Dirac’s delta.
Theorem ∫ a
−aδ(t) dt = 1, a > 0.
Proof: ∫ a
−aδ(t) dt = lim
n→∞
∫ a
−aδn(t) dt = lim
n→∞
∫ 1/n
0n dt
∫ a
−aδ(t) dt = lim
n→∞
[n(t∣∣∣1/n
0
)]= lim
n→∞
[n
1
n
].
We conclude:
∫ a
−aδ(t) dt = 1.
Properties of Dirac’s delta.
TheoremIf f : R → R is continuous, t0 ∈ R and a > 0, then∫ t0+a
t0−aδ(t − t0) f (t) dt = f (t0).
Proof: Introduce the change of variable τ = t − t0,
I =
∫ t0+a
t0−aδ(t − t0) f (t) dt =
∫ a
−aδ(τ) f (τ + t0) dτ,
I = limn→∞
∫ a
−aδn(τ) f (τ + t0) dτ = lim
n→∞
∫ 1/n
0n f (τ + t0) dτ
Therefore, I = limn→∞
n
∫ 1/n
0F ′(τ + t0) dτ , where we introduced the
primitive F (t) =
∫f (t) dt, that is, f (t) = F ′(t).
Properties of Dirac’s delta.
TheoremIf f : R → R is continuous, t0 ∈ R and a > 0, then∫ t0+a
t0−aδ(t − t0) f (t) dt = f (t0).
Proof: Introduce the change of variable τ = t − t0,
I =
∫ t0+a
t0−aδ(t − t0) f (t) dt
=
∫ a
−aδ(τ) f (τ + t0) dτ,
I = limn→∞
∫ a
−aδn(τ) f (τ + t0) dτ = lim
n→∞
∫ 1/n
0n f (τ + t0) dτ
Therefore, I = limn→∞
n
∫ 1/n
0F ′(τ + t0) dτ , where we introduced the
primitive F (t) =
∫f (t) dt, that is, f (t) = F ′(t).
Properties of Dirac’s delta.
TheoremIf f : R → R is continuous, t0 ∈ R and a > 0, then∫ t0+a
t0−aδ(t − t0) f (t) dt = f (t0).
Proof: Introduce the change of variable τ = t − t0,
I =
∫ t0+a
t0−aδ(t − t0) f (t) dt =
∫ a
−aδ(τ) f (τ + t0) dτ,
I = limn→∞
∫ a
−aδn(τ) f (τ + t0) dτ = lim
n→∞
∫ 1/n
0n f (τ + t0) dτ
Therefore, I = limn→∞
n
∫ 1/n
0F ′(τ + t0) dτ , where we introduced the
primitive F (t) =
∫f (t) dt, that is, f (t) = F ′(t).
Properties of Dirac’s delta.
TheoremIf f : R → R is continuous, t0 ∈ R and a > 0, then∫ t0+a
t0−aδ(t − t0) f (t) dt = f (t0).
Proof: Introduce the change of variable τ = t − t0,
I =
∫ t0+a
t0−aδ(t − t0) f (t) dt =
∫ a
−aδ(τ) f (τ + t0) dτ,
I = limn→∞
∫ a
−aδn(τ) f (τ + t0) dτ
= limn→∞
∫ 1/n
0n f (τ + t0) dτ
Therefore, I = limn→∞
n
∫ 1/n
0F ′(τ + t0) dτ , where we introduced the
primitive F (t) =
∫f (t) dt, that is, f (t) = F ′(t).
Properties of Dirac’s delta.
TheoremIf f : R → R is continuous, t0 ∈ R and a > 0, then∫ t0+a
t0−aδ(t − t0) f (t) dt = f (t0).
Proof: Introduce the change of variable τ = t − t0,
I =
∫ t0+a
t0−aδ(t − t0) f (t) dt =
∫ a
−aδ(τ) f (τ + t0) dτ,
I = limn→∞
∫ a
−aδn(τ) f (τ + t0) dτ = lim
n→∞
∫ 1/n
0n f (τ + t0) dτ
Therefore, I = limn→∞
n
∫ 1/n
0F ′(τ + t0) dτ , where we introduced the
primitive F (t) =
∫f (t) dt, that is, f (t) = F ′(t).
Properties of Dirac’s delta.
TheoremIf f : R → R is continuous, t0 ∈ R and a > 0, then∫ t0+a
t0−aδ(t − t0) f (t) dt = f (t0).
Proof: Introduce the change of variable τ = t − t0,
I =
∫ t0+a
t0−aδ(t − t0) f (t) dt =
∫ a
−aδ(τ) f (τ + t0) dτ,
I = limn→∞
∫ a
−aδn(τ) f (τ + t0) dτ = lim
n→∞
∫ 1/n
0n f (τ + t0) dτ
Therefore, I = limn→∞
n
∫ 1/n
0F ′(τ + t0) dτ ,
where we introduced the
primitive F (t) =
∫f (t) dt, that is, f (t) = F ′(t).
Properties of Dirac’s delta.
TheoremIf f : R → R is continuous, t0 ∈ R and a > 0, then∫ t0+a
t0−aδ(t − t0) f (t) dt = f (t0).
Proof: Introduce the change of variable τ = t − t0,
I =
∫ t0+a
t0−aδ(t − t0) f (t) dt =
∫ a
−aδ(τ) f (τ + t0) dτ,
I = limn→∞
∫ a
−aδn(τ) f (τ + t0) dτ = lim
n→∞
∫ 1/n
0n f (τ + t0) dτ
Therefore, I = limn→∞
n
∫ 1/n
0F ′(τ + t0) dτ , where we introduced the
primitive F (t) =
∫f (t) dt, that is, f (t) = F ′(t).
Properties of Dirac’s delta.
TheoremIf f : R → R is continuous, t0 ∈ R and a > 0, then∫ t0+a
t0−aδ(t − t0) f (t) dt = f (t0).
Proof: So, I = limn→∞
n
∫ 1/n
0F ′(τ + t0) dτ , with f (t) = F ′(t).
I = limn→∞
n[F (τ + t0)
∣∣∣1/n
0
]= lim
n→∞n
[F
(t0 +
1
n
)− F (t0)
].
I = limn→∞
[F
(t0 + 1
n
)− F (t0)
]1n
= F ′(t0) = f (t0).
We conclude:
∫ t0+a
t0−aδ(t − t0) f (t) dt = f (t0).
Properties of Dirac’s delta.
TheoremIf f : R → R is continuous, t0 ∈ R and a > 0, then∫ t0+a
t0−aδ(t − t0) f (t) dt = f (t0).
Proof: So, I = limn→∞
n
∫ 1/n
0F ′(τ + t0) dτ , with f (t) = F ′(t).
I = limn→∞
n[F (τ + t0)
∣∣∣1/n
0
]
= limn→∞
n[F
(t0 +
1
n
)− F (t0)
].
I = limn→∞
[F
(t0 + 1
n
)− F (t0)
]1n
= F ′(t0) = f (t0).
We conclude:
∫ t0+a
t0−aδ(t − t0) f (t) dt = f (t0).
Properties of Dirac’s delta.
TheoremIf f : R → R is continuous, t0 ∈ R and a > 0, then∫ t0+a
t0−aδ(t − t0) f (t) dt = f (t0).
Proof: So, I = limn→∞
n
∫ 1/n
0F ′(τ + t0) dτ , with f (t) = F ′(t).
I = limn→∞
n[F (τ + t0)
∣∣∣1/n
0
]= lim
n→∞n
[F
(t0 +
1
n
)− F (t0)
].
I = limn→∞
[F
(t0 + 1
n
)− F (t0)
]1n
= F ′(t0) = f (t0).
We conclude:
∫ t0+a
t0−aδ(t − t0) f (t) dt = f (t0).
Properties of Dirac’s delta.
TheoremIf f : R → R is continuous, t0 ∈ R and a > 0, then∫ t0+a
t0−aδ(t − t0) f (t) dt = f (t0).
Proof: So, I = limn→∞
n
∫ 1/n
0F ′(τ + t0) dτ , with f (t) = F ′(t).
I = limn→∞
n[F (τ + t0)
∣∣∣1/n
0
]= lim
n→∞n
[F
(t0 +
1
n
)− F (t0)
].
I = limn→∞
[F
(t0 + 1
n
)− F (t0)
]1n
= F ′(t0) = f (t0).
We conclude:
∫ t0+a
t0−aδ(t − t0) f (t) dt = f (t0).
Properties of Dirac’s delta.
TheoremIf f : R → R is continuous, t0 ∈ R and a > 0, then∫ t0+a
t0−aδ(t − t0) f (t) dt = f (t0).
Proof: So, I = limn→∞
n
∫ 1/n
0F ′(τ + t0) dτ , with f (t) = F ′(t).
I = limn→∞
n[F (τ + t0)
∣∣∣1/n
0
]= lim
n→∞n
[F
(t0 +
1
n
)− F (t0)
].
I = limn→∞
[F
(t0 + 1
n
)− F (t0)
]1n
= F ′(t0)
= f (t0).
We conclude:
∫ t0+a
t0−aδ(t − t0) f (t) dt = f (t0).
Properties of Dirac’s delta.
TheoremIf f : R → R is continuous, t0 ∈ R and a > 0, then∫ t0+a
t0−aδ(t − t0) f (t) dt = f (t0).
Proof: So, I = limn→∞
n
∫ 1/n
0F ′(τ + t0) dτ , with f (t) = F ′(t).
I = limn→∞
n[F (τ + t0)
∣∣∣1/n
0
]= lim
n→∞n
[F
(t0 +
1
n
)− F (t0)
].
I = limn→∞
[F
(t0 + 1
n
)− F (t0)
]1n
= F ′(t0) = f (t0).
We conclude:
∫ t0+a
t0−aδ(t − t0) f (t) dt = f (t0).
Properties of Dirac’s delta.
TheoremIf f : R → R is continuous, t0 ∈ R and a > 0, then∫ t0+a
t0−aδ(t − t0) f (t) dt = f (t0).
Proof: So, I = limn→∞
n
∫ 1/n
0F ′(τ + t0) dτ , with f (t) = F ′(t).
I = limn→∞
n[F (τ + t0)
∣∣∣1/n
0
]= lim
n→∞n
[F
(t0 +
1
n
)− F (t0)
].
I = limn→∞
[F
(t0 + 1
n
)− F (t0)
]1n
= F ′(t0) = f (t0).
We conclude:
∫ t0+a
t0−aδ(t − t0) f (t) dt = f (t0).
Generalized sources (Sect. 6.5).
I The Dirac delta generalized function.
I Properties of Dirac’s delta.
I Relation between deltas and steps.
I Dirac’s delta in Physics.
I The Laplace Transform of Dirac’s delta.
I Differential equations with Dirac’s delta sources.
Relation between deltas and steps.
TheoremThe sequence of functions for n > 1,
un(t) =
0, t < 0
nt, 0 6 t 61
n
1, t >1
n.
u
t0 1/3 1/2 1
3 2 1u u u
1
n
satisfies, for t ∈ (−∞, 0) ∪ (0, 1/n) ∪ (1/n,∞), both equations,
u′n(t) = δn(t), limn→∞
un(t) = u(t), t ∈ R.
Remark:
I If we generalize the notion of derivative asu′(t) = lim
n→∞δn(t), then holds u′(t) = δ(t).
I Dirac’s delta is a generalized derivative of the step function.
Relation between deltas and steps.
TheoremThe sequence of functions for n > 1,
un(t) =
0, t < 0
nt, 0 6 t 61
n
1, t >1
n.
u
t0 1/3 1/2 1
3 2 1u u u
1
n
satisfies, for t ∈ (−∞, 0) ∪ (0, 1/n) ∪ (1/n,∞), both equations,
u′n(t) = δn(t), limn→∞
un(t) = u(t), t ∈ R.
Remark:
I If we generalize the notion of derivative asu′(t) = lim
n→∞δn(t), then holds u′(t) = δ(t).
I Dirac’s delta is a generalized derivative of the step function.
Relation between deltas and steps.
TheoremThe sequence of functions for n > 1,
un(t) =
0, t < 0
nt, 0 6 t 61
n
1, t >1
n.
u
t0 1/3 1/2 1
3 2 1u u u
1
n
satisfies, for t ∈ (−∞, 0) ∪ (0, 1/n) ∪ (1/n,∞), both equations,
u′n(t) = δn(t), limn→∞
un(t) = u(t), t ∈ R.
Remark:
I If we generalize the notion of derivative asu′(t) = lim
n→∞δn(t), then holds u′(t) = δ(t).
I Dirac’s delta is a generalized derivative of the step function.
Relation between deltas and steps.
TheoremThe sequence of functions for n > 1,
un(t) =
0, t < 0
nt, 0 6 t 61
n
1, t >1
n.
u
t0 1/3 1/2 1
3 2 1u u u
1
n
satisfies, for t ∈ (−∞, 0) ∪ (0, 1/n) ∪ (1/n,∞), both equations,
u′n(t) = δn(t), limn→∞
un(t) = u(t), t ∈ R.
Remark:
I If we generalize the notion of derivative asu′(t) = lim
n→∞δn(t), then holds u′(t) = δ(t).
I Dirac’s delta is a generalized derivative of the step function.
Relation between deltas and steps.
TheoremThe sequence of functions for n > 1,
un(t) =
0, t < 0
nt, 0 6 t 61
n
1, t >1
n.
u
t0 1/3 1/2 1
3 2 1u u u
1
n
satisfies, for t ∈ (−∞, 0) ∪ (0, 1/n) ∪ (1/n,∞), both equations,
u′n(t) = δn(t), limn→∞
un(t) = u(t), t ∈ R.
Remark:
I If we generalize the notion of derivative asu′(t) = lim
n→∞δn(t), then holds u′(t) = δ(t).
I Dirac’s delta is a generalized derivative of the step function.
Generalized sources (Sect. 6.5).
I The Dirac delta generalized function.
I Properties of Dirac’s delta.
I Relation between deltas and steps.
I Dirac’s delta in Physics.
I The Laplace Transform of Dirac’s delta.
I Differential equations with Dirac’s delta sources.
Dirac’s delta in Physics.
Remarks:
(a) Dirac’s delta generalized function is useful to describeimpulsive forces in mechanical systems.
(b) An impulsive force transmits a finite momentum in aninfinitely short time.
(c) For example: The momentum transmitted to a pendulumwhen hit by a hammer. Newton’s law of motion says,
m v ′(t) = F (t), with F (t) = F0 δ(t − t0).
The momentum transfer is:
∆I = lim∆t→0
mv(t)∣∣∣t0+∆t
t0−∆t= lim
∆t→0
∫ t0+∆t
t0−∆tF (t) dt = F0.
That is, ∆I = F0.
Dirac’s delta in Physics.
Remarks:
(a) Dirac’s delta generalized function is useful to describeimpulsive forces in mechanical systems.
(b) An impulsive force transmits a finite momentum in aninfinitely short time.
(c) For example: The momentum transmitted to a pendulumwhen hit by a hammer. Newton’s law of motion says,
m v ′(t) = F (t), with F (t) = F0 δ(t − t0).
The momentum transfer is:
∆I = lim∆t→0
mv(t)∣∣∣t0+∆t
t0−∆t= lim
∆t→0
∫ t0+∆t
t0−∆tF (t) dt = F0.
That is, ∆I = F0.
Dirac’s delta in Physics.
Remarks:
(a) Dirac’s delta generalized function is useful to describeimpulsive forces in mechanical systems.
(b) An impulsive force transmits a finite momentum in aninfinitely short time.
(c) For example: The momentum transmitted to a pendulumwhen hit by a hammer.
Newton’s law of motion says,
m v ′(t) = F (t), with F (t) = F0 δ(t − t0).
The momentum transfer is:
∆I = lim∆t→0
mv(t)∣∣∣t0+∆t
t0−∆t= lim
∆t→0
∫ t0+∆t
t0−∆tF (t) dt = F0.
That is, ∆I = F0.
Dirac’s delta in Physics.
Remarks:
(a) Dirac’s delta generalized function is useful to describeimpulsive forces in mechanical systems.
(b) An impulsive force transmits a finite momentum in aninfinitely short time.
(c) For example: The momentum transmitted to a pendulumwhen hit by a hammer. Newton’s law of motion says,
m v ′(t) = F (t),
with F (t) = F0 δ(t − t0).
The momentum transfer is:
∆I = lim∆t→0
mv(t)∣∣∣t0+∆t
t0−∆t= lim
∆t→0
∫ t0+∆t
t0−∆tF (t) dt = F0.
That is, ∆I = F0.
Dirac’s delta in Physics.
Remarks:
(a) Dirac’s delta generalized function is useful to describeimpulsive forces in mechanical systems.
(b) An impulsive force transmits a finite momentum in aninfinitely short time.
(c) For example: The momentum transmitted to a pendulumwhen hit by a hammer. Newton’s law of motion says,
m v ′(t) = F (t), with F (t) = F0 δ(t − t0).
The momentum transfer is:
∆I = lim∆t→0
mv(t)∣∣∣t0+∆t
t0−∆t= lim
∆t→0
∫ t0+∆t
t0−∆tF (t) dt = F0.
That is, ∆I = F0.
Dirac’s delta in Physics.
Remarks:
(a) Dirac’s delta generalized function is useful to describeimpulsive forces in mechanical systems.
(b) An impulsive force transmits a finite momentum in aninfinitely short time.
(c) For example: The momentum transmitted to a pendulumwhen hit by a hammer. Newton’s law of motion says,
m v ′(t) = F (t), with F (t) = F0 δ(t − t0).
The momentum transfer is:
∆I = lim∆t→0
mv(t)∣∣∣t0+∆t
t0−∆t
= lim∆t→0
∫ t0+∆t
t0−∆tF (t) dt = F0.
That is, ∆I = F0.
Dirac’s delta in Physics.
Remarks:
(a) Dirac’s delta generalized function is useful to describeimpulsive forces in mechanical systems.
(b) An impulsive force transmits a finite momentum in aninfinitely short time.
(c) For example: The momentum transmitted to a pendulumwhen hit by a hammer. Newton’s law of motion says,
m v ′(t) = F (t), with F (t) = F0 δ(t − t0).
The momentum transfer is:
∆I = lim∆t→0
mv(t)∣∣∣t0+∆t
t0−∆t= lim
∆t→0
∫ t0+∆t
t0−∆tF (t) dt
= F0.
That is, ∆I = F0.
Dirac’s delta in Physics.
Remarks:
(a) Dirac’s delta generalized function is useful to describeimpulsive forces in mechanical systems.
(b) An impulsive force transmits a finite momentum in aninfinitely short time.
(c) For example: The momentum transmitted to a pendulumwhen hit by a hammer. Newton’s law of motion says,
m v ′(t) = F (t), with F (t) = F0 δ(t − t0).
The momentum transfer is:
∆I = lim∆t→0
mv(t)∣∣∣t0+∆t
t0−∆t= lim
∆t→0
∫ t0+∆t
t0−∆tF (t) dt = F0.
That is, ∆I = F0.
Dirac’s delta in Physics.
Remarks:
(a) Dirac’s delta generalized function is useful to describeimpulsive forces in mechanical systems.
(b) An impulsive force transmits a finite momentum in aninfinitely short time.
(c) For example: The momentum transmitted to a pendulumwhen hit by a hammer. Newton’s law of motion says,
m v ′(t) = F (t), with F (t) = F0 δ(t − t0).
The momentum transfer is:
∆I = lim∆t→0
mv(t)∣∣∣t0+∆t
t0−∆t= lim
∆t→0
∫ t0+∆t
t0−∆tF (t) dt = F0.
That is, ∆I = F0.
Generalized sources (Sect. 6.5).
I The Dirac delta generalized function.
I Properties of Dirac’s delta.
I Relation between deltas and steps.
I Dirac’s delta in Physics.
I The Laplace Transform of Dirac’s delta.
I Differential equations with Dirac’s delta sources.
The Laplace Transform of Dirac’s delta.
Recall: The Laplace Transform can be generalized from functionsto δ,
as follows, L[δ(t − c)] = limn→∞
L[δn(t − c)].
TheoremL[δ(t − c)] = e−cs .
Proof:
L[δ(t − c)] = limn→∞
L[δn(t − c)], δn(t) = n[u(t)− u
(t − 1
n
)].
L[δ(t − c)] = limn→∞
n(L[u(t − c)]− L
[u(t − c − 1
n
)])L[δ(t − c)] = lim
n→∞n(e−cs
s− e−(c+ 1
n)s
s
)= e−cs lim
n→∞
(1− e−sn )(
sn
) .
This is a singular limit, 00 . Use l’Hopital rule.
The Laplace Transform of Dirac’s delta.
Recall: The Laplace Transform can be generalized from functionsto δ, as follows, L[δ(t − c)] = lim
n→∞L[δn(t − c)].
TheoremL[δ(t − c)] = e−cs .
Proof:
L[δ(t − c)] = limn→∞
L[δn(t − c)], δn(t) = n[u(t)− u
(t − 1
n
)].
L[δ(t − c)] = limn→∞
n(L[u(t − c)]− L
[u(t − c − 1
n
)])L[δ(t − c)] = lim
n→∞n(e−cs
s− e−(c+ 1
n)s
s
)= e−cs lim
n→∞
(1− e−sn )(
sn
) .
This is a singular limit, 00 . Use l’Hopital rule.
The Laplace Transform of Dirac’s delta.
Recall: The Laplace Transform can be generalized from functionsto δ, as follows, L[δ(t − c)] = lim
n→∞L[δn(t − c)].
TheoremL[δ(t − c)] = e−cs .
Proof:
L[δ(t − c)] = limn→∞
L[δn(t − c)], δn(t) = n[u(t)− u
(t − 1
n
)].
L[δ(t − c)] = limn→∞
n(L[u(t − c)]− L
[u(t − c − 1
n
)])L[δ(t − c)] = lim
n→∞n(e−cs
s− e−(c+ 1
n)s
s
)= e−cs lim
n→∞
(1− e−sn )(
sn
) .
This is a singular limit, 00 . Use l’Hopital rule.
The Laplace Transform of Dirac’s delta.
Recall: The Laplace Transform can be generalized from functionsto δ, as follows, L[δ(t − c)] = lim
n→∞L[δn(t − c)].
TheoremL[δ(t − c)] = e−cs .
Proof:
L[δ(t − c)] = limn→∞
L[δn(t − c)],
δn(t) = n[u(t)− u
(t − 1
n
)].
L[δ(t − c)] = limn→∞
n(L[u(t − c)]− L
[u(t − c − 1
n
)])L[δ(t − c)] = lim
n→∞n(e−cs
s− e−(c+ 1
n)s
s
)= e−cs lim
n→∞
(1− e−sn )(
sn
) .
This is a singular limit, 00 . Use l’Hopital rule.
The Laplace Transform of Dirac’s delta.
Recall: The Laplace Transform can be generalized from functionsto δ, as follows, L[δ(t − c)] = lim
n→∞L[δn(t − c)].
TheoremL[δ(t − c)] = e−cs .
Proof:
L[δ(t − c)] = limn→∞
L[δn(t − c)], δn(t) = n[u(t)− u
(t − 1
n
)].
L[δ(t − c)] = limn→∞
n(L[u(t − c)]− L
[u(t − c − 1
n
)])L[δ(t − c)] = lim
n→∞n(e−cs
s− e−(c+ 1
n)s
s
)= e−cs lim
n→∞
(1− e−sn )(
sn
) .
This is a singular limit, 00 . Use l’Hopital rule.
The Laplace Transform of Dirac’s delta.
Recall: The Laplace Transform can be generalized from functionsto δ, as follows, L[δ(t − c)] = lim
n→∞L[δn(t − c)].
TheoremL[δ(t − c)] = e−cs .
Proof:
L[δ(t − c)] = limn→∞
L[δn(t − c)], δn(t) = n[u(t)− u
(t − 1
n
)].
L[δ(t − c)] = limn→∞
n(L[u(t − c)]− L
[u(t − c − 1
n
)])
L[δ(t − c)] = limn→∞
n(e−cs
s− e−(c+ 1
n)s
s
)= e−cs lim
n→∞
(1− e−sn )(
sn
) .
This is a singular limit, 00 . Use l’Hopital rule.
The Laplace Transform of Dirac’s delta.
Recall: The Laplace Transform can be generalized from functionsto δ, as follows, L[δ(t − c)] = lim
n→∞L[δn(t − c)].
TheoremL[δ(t − c)] = e−cs .
Proof:
L[δ(t − c)] = limn→∞
L[δn(t − c)], δn(t) = n[u(t)− u
(t − 1
n
)].
L[δ(t − c)] = limn→∞
n(L[u(t − c)]− L
[u(t − c − 1
n
)])L[δ(t − c)] = lim
n→∞n(e−cs
s− e−(c+ 1
n)s
s
)
= e−cs limn→∞
(1− e−sn )(
sn
) .
This is a singular limit, 00 . Use l’Hopital rule.
The Laplace Transform of Dirac’s delta.
Recall: The Laplace Transform can be generalized from functionsto δ, as follows, L[δ(t − c)] = lim
n→∞L[δn(t − c)].
TheoremL[δ(t − c)] = e−cs .
Proof:
L[δ(t − c)] = limn→∞
L[δn(t − c)], δn(t) = n[u(t)− u
(t − 1
n
)].
L[δ(t − c)] = limn→∞
n(L[u(t − c)]− L
[u(t − c − 1
n
)])L[δ(t − c)] = lim
n→∞n(e−cs
s− e−(c+ 1
n)s
s
)= e−cs lim
n→∞
(1− e−sn )(
sn
) .
This is a singular limit, 00 . Use l’Hopital rule.
The Laplace Transform of Dirac’s delta.
Recall: The Laplace Transform can be generalized from functionsto δ, as follows, L[δ(t − c)] = lim
n→∞L[δn(t − c)].
TheoremL[δ(t − c)] = e−cs .
Proof:
L[δ(t − c)] = limn→∞
L[δn(t − c)], δn(t) = n[u(t)− u
(t − 1
n
)].
L[δ(t − c)] = limn→∞
n(L[u(t − c)]− L
[u(t − c − 1
n
)])L[δ(t − c)] = lim
n→∞n(e−cs
s− e−(c+ 1
n)s
s
)= e−cs lim
n→∞
(1− e−sn )(
sn
) .
This is a singular limit, 00 .
Use l’Hopital rule.
The Laplace Transform of Dirac’s delta.
Recall: The Laplace Transform can be generalized from functionsto δ, as follows, L[δ(t − c)] = lim
n→∞L[δn(t − c)].
TheoremL[δ(t − c)] = e−cs .
Proof:
L[δ(t − c)] = limn→∞
L[δn(t − c)], δn(t) = n[u(t)− u
(t − 1
n
)].
L[δ(t − c)] = limn→∞
n(L[u(t − c)]− L
[u(t − c − 1
n
)])L[δ(t − c)] = lim
n→∞n(e−cs
s− e−(c+ 1
n)s
s
)= e−cs lim
n→∞
(1− e−sn )(
sn
) .
This is a singular limit, 00 . Use l’Hopital rule.
The Laplace Transform of Dirac’s delta.
Proof: Recall: L[δ(t − c)] = e−cs limn→∞
(1− e−sn )(
sn
) .
limn→∞
(1− e−sn )(
sn
) = limn→∞
(− sn2 e−
sn )(
− sn2
) = limn→∞
e−sn = 1.
We therefore conclude that L[δ(t − c)] = e−cs .
Remarks:
(a) This result is consistent with a previous result:∫ t0+a
t0−aδ(t − t0) f (t) dt = f (t0).
(b) L[δ(t − c)] =
∫ ∞
0δ(t − c) e−st dt = e−cs .
(c) L[δ(t − c) f (t)] =
∫ ∞
0δ(t − c) e−st f (t) dt = e−cs f (c).
The Laplace Transform of Dirac’s delta.
Proof: Recall: L[δ(t − c)] = e−cs limn→∞
(1− e−sn )(
sn
) .
limn→∞
(1− e−sn )(
sn
)
= limn→∞
(− sn2 e−
sn )(
− sn2
) = limn→∞
e−sn = 1.
We therefore conclude that L[δ(t − c)] = e−cs .
Remarks:
(a) This result is consistent with a previous result:∫ t0+a
t0−aδ(t − t0) f (t) dt = f (t0).
(b) L[δ(t − c)] =
∫ ∞
0δ(t − c) e−st dt = e−cs .
(c) L[δ(t − c) f (t)] =
∫ ∞
0δ(t − c) e−st f (t) dt = e−cs f (c).
The Laplace Transform of Dirac’s delta.
Proof: Recall: L[δ(t − c)] = e−cs limn→∞
(1− e−sn )(
sn
) .
limn→∞
(1− e−sn )(
sn
) = limn→∞
(− sn2 e−
sn )(
− sn2
)
= limn→∞
e−sn = 1.
We therefore conclude that L[δ(t − c)] = e−cs .
Remarks:
(a) This result is consistent with a previous result:∫ t0+a
t0−aδ(t − t0) f (t) dt = f (t0).
(b) L[δ(t − c)] =
∫ ∞
0δ(t − c) e−st dt = e−cs .
(c) L[δ(t − c) f (t)] =
∫ ∞
0δ(t − c) e−st f (t) dt = e−cs f (c).
The Laplace Transform of Dirac’s delta.
Proof: Recall: L[δ(t − c)] = e−cs limn→∞
(1− e−sn )(
sn
) .
limn→∞
(1− e−sn )(
sn
) = limn→∞
(− sn2 e−
sn )(
− sn2
) = limn→∞
e−sn
= 1.
We therefore conclude that L[δ(t − c)] = e−cs .
Remarks:
(a) This result is consistent with a previous result:∫ t0+a
t0−aδ(t − t0) f (t) dt = f (t0).
(b) L[δ(t − c)] =
∫ ∞
0δ(t − c) e−st dt = e−cs .
(c) L[δ(t − c) f (t)] =
∫ ∞
0δ(t − c) e−st f (t) dt = e−cs f (c).
The Laplace Transform of Dirac’s delta.
Proof: Recall: L[δ(t − c)] = e−cs limn→∞
(1− e−sn )(
sn
) .
limn→∞
(1− e−sn )(
sn
) = limn→∞
(− sn2 e−
sn )(
− sn2
) = limn→∞
e−sn = 1.
We therefore conclude that L[δ(t − c)] = e−cs .
Remarks:
(a) This result is consistent with a previous result:∫ t0+a
t0−aδ(t − t0) f (t) dt = f (t0).
(b) L[δ(t − c)] =
∫ ∞
0δ(t − c) e−st dt = e−cs .
(c) L[δ(t − c) f (t)] =
∫ ∞
0δ(t − c) e−st f (t) dt = e−cs f (c).
The Laplace Transform of Dirac’s delta.
Proof: Recall: L[δ(t − c)] = e−cs limn→∞
(1− e−sn )(
sn
) .
limn→∞
(1− e−sn )(
sn
) = limn→∞
(− sn2 e−
sn )(
− sn2
) = limn→∞
e−sn = 1.
We therefore conclude that L[δ(t − c)] = e−cs .
Remarks:
(a) This result is consistent with a previous result:∫ t0+a
t0−aδ(t − t0) f (t) dt = f (t0).
(b) L[δ(t − c)] =
∫ ∞
0δ(t − c) e−st dt = e−cs .
(c) L[δ(t − c) f (t)] =
∫ ∞
0δ(t − c) e−st f (t) dt = e−cs f (c).
The Laplace Transform of Dirac’s delta.
Proof: Recall: L[δ(t − c)] = e−cs limn→∞
(1− e−sn )(
sn
) .
limn→∞
(1− e−sn )(
sn
) = limn→∞
(− sn2 e−
sn )(
− sn2
) = limn→∞
e−sn = 1.
We therefore conclude that L[δ(t − c)] = e−cs .
Remarks:
(a) This result is consistent with a previous result:
∫ t0+a
t0−aδ(t − t0) f (t) dt = f (t0).
(b) L[δ(t − c)] =
∫ ∞
0δ(t − c) e−st dt = e−cs .
(c) L[δ(t − c) f (t)] =
∫ ∞
0δ(t − c) e−st f (t) dt = e−cs f (c).
The Laplace Transform of Dirac’s delta.
Proof: Recall: L[δ(t − c)] = e−cs limn→∞
(1− e−sn )(
sn
) .
limn→∞
(1− e−sn )(
sn
) = limn→∞
(− sn2 e−
sn )(
− sn2
) = limn→∞
e−sn = 1.
We therefore conclude that L[δ(t − c)] = e−cs .
Remarks:
(a) This result is consistent with a previous result:∫ t0+a
t0−aδ(t − t0) f (t) dt = f (t0).
(b) L[δ(t − c)] =
∫ ∞
0δ(t − c) e−st dt = e−cs .
(c) L[δ(t − c) f (t)] =
∫ ∞
0δ(t − c) e−st f (t) dt = e−cs f (c).
The Laplace Transform of Dirac’s delta.
Proof: Recall: L[δ(t − c)] = e−cs limn→∞
(1− e−sn )(
sn
) .
limn→∞
(1− e−sn )(
sn
) = limn→∞
(− sn2 e−
sn )(
− sn2
) = limn→∞
e−sn = 1.
We therefore conclude that L[δ(t − c)] = e−cs .
Remarks:
(a) This result is consistent with a previous result:∫ t0+a
t0−aδ(t − t0) f (t) dt = f (t0).
(b) L[δ(t − c)] =
∫ ∞
0δ(t − c) e−st dt = e−cs .
(c) L[δ(t − c) f (t)] =
∫ ∞
0δ(t − c) e−st f (t) dt = e−cs f (c).
The Laplace Transform of Dirac’s delta.
Proof: Recall: L[δ(t − c)] = e−cs limn→∞
(1− e−sn )(
sn
) .
limn→∞
(1− e−sn )(
sn
) = limn→∞
(− sn2 e−
sn )(
− sn2
) = limn→∞
e−sn = 1.
We therefore conclude that L[δ(t − c)] = e−cs .
Remarks:
(a) This result is consistent with a previous result:∫ t0+a
t0−aδ(t − t0) f (t) dt = f (t0).
(b) L[δ(t − c)] =
∫ ∞
0δ(t − c) e−st dt = e−cs .
(c) L[δ(t − c) f (t)] =
∫ ∞
0δ(t − c) e−st f (t) dt = e−cs f (c).
Generalized sources (Sect. 6.5).
I The Dirac delta generalized function.
I Properties of Dirac’s delta.
I Relation between deltas and steps.
I Dirac’s delta in Physics.
I The Laplace Transform of Dirac’s delta.
I Differential equations with Dirac’s delta sources.
Differential equations with Dirac’s delta sources.
Example
Find the solution y to the initial value problem
y ′′ − y = −20 δ(t − 3), y(0) = 1, y ′(0) = 0.
Solution: Compute: L[y ′′]− L[y ] = −20L[δ(t − 3)].
L[y ′′] = s2 L[y ]−s y(0)−y ′(0) ⇒ (s2−1)L[y ]−s = −20 e−3s ,
We arrive to the equation L[y ] =s
(s2 − 1)− 20 e−3s 1
(s2 − 1),
L[y ] = L[cosh(t)]− 20L[u(t − 3) sinh(t − 3)],
We conclude: y(t) = cosh(t)− 20 u(t − 3) sinh(t − 3). C
Differential equations with Dirac’s delta sources.
Example
Find the solution y to the initial value problem
y ′′ − y = −20 δ(t − 3), y(0) = 1, y ′(0) = 0.
Solution: Compute: L[y ′′]− L[y ] = −20L[δ(t − 3)].
L[y ′′] = s2 L[y ]−s y(0)−y ′(0) ⇒ (s2−1)L[y ]−s = −20 e−3s ,
We arrive to the equation L[y ] =s
(s2 − 1)− 20 e−3s 1
(s2 − 1),
L[y ] = L[cosh(t)]− 20L[u(t − 3) sinh(t − 3)],
We conclude: y(t) = cosh(t)− 20 u(t − 3) sinh(t − 3). C
Differential equations with Dirac’s delta sources.
Example
Find the solution y to the initial value problem
y ′′ − y = −20 δ(t − 3), y(0) = 1, y ′(0) = 0.
Solution: Compute: L[y ′′]− L[y ] = −20L[δ(t − 3)].
L[y ′′] = s2 L[y ]−s y(0)−y ′(0)
⇒ (s2−1)L[y ]−s = −20 e−3s ,
We arrive to the equation L[y ] =s
(s2 − 1)− 20 e−3s 1
(s2 − 1),
L[y ] = L[cosh(t)]− 20L[u(t − 3) sinh(t − 3)],
We conclude: y(t) = cosh(t)− 20 u(t − 3) sinh(t − 3). C
Differential equations with Dirac’s delta sources.
Example
Find the solution y to the initial value problem
y ′′ − y = −20 δ(t − 3), y(0) = 1, y ′(0) = 0.
Solution: Compute: L[y ′′]− L[y ] = −20L[δ(t − 3)].
L[y ′′] = s2 L[y ]−s y(0)−y ′(0) ⇒ (s2−1)L[y ]−s = −20 e−3s ,
We arrive to the equation L[y ] =s
(s2 − 1)− 20 e−3s 1
(s2 − 1),
L[y ] = L[cosh(t)]− 20L[u(t − 3) sinh(t − 3)],
We conclude: y(t) = cosh(t)− 20 u(t − 3) sinh(t − 3). C
Differential equations with Dirac’s delta sources.
Example
Find the solution y to the initial value problem
y ′′ − y = −20 δ(t − 3), y(0) = 1, y ′(0) = 0.
Solution: Compute: L[y ′′]− L[y ] = −20L[δ(t − 3)].
L[y ′′] = s2 L[y ]−s y(0)−y ′(0) ⇒ (s2−1)L[y ]−s = −20 e−3s ,
We arrive to the equation L[y ] =s
(s2 − 1)− 20 e−3s 1
(s2 − 1),
L[y ] = L[cosh(t)]− 20L[u(t − 3) sinh(t − 3)],
We conclude: y(t) = cosh(t)− 20 u(t − 3) sinh(t − 3). C
Differential equations with Dirac’s delta sources.
Example
Find the solution y to the initial value problem
y ′′ − y = −20 δ(t − 3), y(0) = 1, y ′(0) = 0.
Solution: Compute: L[y ′′]− L[y ] = −20L[δ(t − 3)].
L[y ′′] = s2 L[y ]−s y(0)−y ′(0) ⇒ (s2−1)L[y ]−s = −20 e−3s ,
We arrive to the equation L[y ] =s
(s2 − 1)− 20 e−3s 1
(s2 − 1),
L[y ] = L[cosh(t)]− 20L[u(t − 3) sinh(t − 3)],
We conclude: y(t) = cosh(t)− 20 u(t − 3) sinh(t − 3). C
Differential equations with Dirac’s delta sources.
Example
Find the solution y to the initial value problem
y ′′ − y = −20 δ(t − 3), y(0) = 1, y ′(0) = 0.
Solution: Compute: L[y ′′]− L[y ] = −20L[δ(t − 3)].
L[y ′′] = s2 L[y ]−s y(0)−y ′(0) ⇒ (s2−1)L[y ]−s = −20 e−3s ,
We arrive to the equation L[y ] =s
(s2 − 1)− 20 e−3s 1
(s2 − 1),
L[y ] = L[cosh(t)]− 20L[u(t − 3) sinh(t − 3)],
We conclude: y(t) = cosh(t)− 20 u(t − 3) sinh(t − 3). C
Differential equations with Dirac’s delta sources.
Example
Find the solution to the initial value problem
y ′′ + 4y = δ(t − π)− δ(t − 2π), y(0) = 0, y ′(0) = 0.
Solution: Compute: L[y ′′] + 4L[y ] = L[δ(t − π)]− L[δ(t − 2π)],
(s2 + 4)L[y ] = e−πs − e−2πs ⇒ L[y ] =e−πs
(s2 + 4)− e−2πs
(s2 + 4),
that is, L[y ] =e−πs
2
2
(s2 + 4)− e−2πs
2
2
(s2 + 4).
Recall: e−cs L[f (t)] = L[u(t − c) f (t − c)]. Therefore,
L[y ] =1
2L
[u(t−π) sin
[2(t−π)
]]− 1
2L
[u(t−2π) sin
[2(t−2π)
]].
Differential equations with Dirac’s delta sources.
Example
Find the solution to the initial value problem
y ′′ + 4y = δ(t − π)− δ(t − 2π), y(0) = 0, y ′(0) = 0.
Solution: Compute: L[y ′′] + 4L[y ] = L[δ(t − π)]− L[δ(t − 2π)],
(s2 + 4)L[y ] = e−πs − e−2πs ⇒ L[y ] =e−πs
(s2 + 4)− e−2πs
(s2 + 4),
that is, L[y ] =e−πs
2
2
(s2 + 4)− e−2πs
2
2
(s2 + 4).
Recall: e−cs L[f (t)] = L[u(t − c) f (t − c)]. Therefore,
L[y ] =1
2L
[u(t−π) sin
[2(t−π)
]]− 1
2L
[u(t−2π) sin
[2(t−2π)
]].
Differential equations with Dirac’s delta sources.
Example
Find the solution to the initial value problem
y ′′ + 4y = δ(t − π)− δ(t − 2π), y(0) = 0, y ′(0) = 0.
Solution: Compute: L[y ′′] + 4L[y ] = L[δ(t − π)]− L[δ(t − 2π)],
(s2 + 4)L[y ] = e−πs − e−2πs
⇒ L[y ] =e−πs
(s2 + 4)− e−2πs
(s2 + 4),
that is, L[y ] =e−πs
2
2
(s2 + 4)− e−2πs
2
2
(s2 + 4).
Recall: e−cs L[f (t)] = L[u(t − c) f (t − c)]. Therefore,
L[y ] =1
2L
[u(t−π) sin
[2(t−π)
]]− 1
2L
[u(t−2π) sin
[2(t−2π)
]].
Differential equations with Dirac’s delta sources.
Example
Find the solution to the initial value problem
y ′′ + 4y = δ(t − π)− δ(t − 2π), y(0) = 0, y ′(0) = 0.
Solution: Compute: L[y ′′] + 4L[y ] = L[δ(t − π)]− L[δ(t − 2π)],
(s2 + 4)L[y ] = e−πs − e−2πs ⇒ L[y ] =e−πs
(s2 + 4)− e−2πs
(s2 + 4),
that is, L[y ] =e−πs
2
2
(s2 + 4)− e−2πs
2
2
(s2 + 4).
Recall: e−cs L[f (t)] = L[u(t − c) f (t − c)]. Therefore,
L[y ] =1
2L
[u(t−π) sin
[2(t−π)
]]− 1
2L
[u(t−2π) sin
[2(t−2π)
]].
Differential equations with Dirac’s delta sources.
Example
Find the solution to the initial value problem
y ′′ + 4y = δ(t − π)− δ(t − 2π), y(0) = 0, y ′(0) = 0.
Solution: Compute: L[y ′′] + 4L[y ] = L[δ(t − π)]− L[δ(t − 2π)],
(s2 + 4)L[y ] = e−πs − e−2πs ⇒ L[y ] =e−πs
(s2 + 4)− e−2πs
(s2 + 4),
that is, L[y ] =e−πs
2
2
(s2 + 4)− e−2πs
2
2
(s2 + 4).
Recall: e−cs L[f (t)] = L[u(t − c) f (t − c)]. Therefore,
L[y ] =1
2L
[u(t−π) sin
[2(t−π)
]]− 1
2L
[u(t−2π) sin
[2(t−2π)
]].
Differential equations with Dirac’s delta sources.
Example
Find the solution to the initial value problem
y ′′ + 4y = δ(t − π)− δ(t − 2π), y(0) = 0, y ′(0) = 0.
Solution: Compute: L[y ′′] + 4L[y ] = L[δ(t − π)]− L[δ(t − 2π)],
(s2 + 4)L[y ] = e−πs − e−2πs ⇒ L[y ] =e−πs
(s2 + 4)− e−2πs
(s2 + 4),
that is, L[y ] =e−πs
2
2
(s2 + 4)− e−2πs
2
2
(s2 + 4).
Recall: e−cs L[f (t)] = L[u(t − c) f (t − c)].
Therefore,
L[y ] =1
2L
[u(t−π) sin
[2(t−π)
]]− 1
2L
[u(t−2π) sin
[2(t−2π)
]].
Differential equations with Dirac’s delta sources.
Example
Find the solution to the initial value problem
y ′′ + 4y = δ(t − π)− δ(t − 2π), y(0) = 0, y ′(0) = 0.
Solution: Compute: L[y ′′] + 4L[y ] = L[δ(t − π)]− L[δ(t − 2π)],
(s2 + 4)L[y ] = e−πs − e−2πs ⇒ L[y ] =e−πs
(s2 + 4)− e−2πs
(s2 + 4),
that is, L[y ] =e−πs
2
2
(s2 + 4)− e−2πs
2
2
(s2 + 4).
Recall: e−cs L[f (t)] = L[u(t − c) f (t − c)]. Therefore,
L[y ] =1
2L
[u(t−π) sin
[2(t−π)
]]− 1
2L
[u(t−2π) sin
[2(t−2π)
]].
Differential equations with Dirac’s delta sources.
Example
Find the solution to the initial value problem
y ′′ + 4y = δ(t − π)− δ(t − 2π), y(0) = 0, y ′(0) = 0.
Solution: Recall:
L[y ] =1
2L
[u(t−π) sin
[2(t−π)
]]− 1
2L
[u(t−2π) sin
[2(t−2π)
]].
This implies that,
y(t) =1
2u(t − π) sin
[2(t − π)
]− 1
2u(t − 2π) sin
[2(t − 2π)
],
We conclude: y(t) =1
2
[u(t − π)− u(t − 2π)
]sin(2t). C
Differential equations with Dirac’s delta sources.
Example
Find the solution to the initial value problem
y ′′ + 4y = δ(t − π)− δ(t − 2π), y(0) = 0, y ′(0) = 0.
Solution: Recall:
L[y ] =1
2L
[u(t−π) sin
[2(t−π)
]]− 1
2L
[u(t−2π) sin
[2(t−2π)
]].
This implies that,
y(t) =1
2u(t − π) sin
[2(t − π)
]− 1
2u(t − 2π) sin
[2(t − 2π)
],
We conclude: y(t) =1
2
[u(t − π)− u(t − 2π)
]sin(2t). C
Differential equations with Dirac’s delta sources.
Example
Find the solution to the initial value problem
y ′′ + 4y = δ(t − π)− δ(t − 2π), y(0) = 0, y ′(0) = 0.
Solution: Recall:
L[y ] =1
2L
[u(t−π) sin
[2(t−π)
]]− 1
2L
[u(t−2π) sin
[2(t−2π)
]].
This implies that,
y(t) =1
2u(t − π) sin
[2(t − π)
]− 1
2u(t − 2π) sin
[2(t − 2π)
],
We conclude: y(t) =1
2
[u(t − π)− u(t − 2π)
]sin(2t). C