laplace transforms of periodic functions · 2008. 11. 6. · periodic functions 1. a function f is...
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logo1
Transforms and New Formulas An Example Double Check Visualization
Laplace Transforms of Periodic Functions
Bernd Schroder
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Laplace Transforms of Periodic Functions
logo1
Transforms and New Formulas An Example Double Check Visualization
Everything Remains As It Was
No matter what functions arise, the idea for solving differentialequations with Laplace transforms stays the same.
Time Domain (t)
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Laplace Transforms of Periodic Functions
logo1
Transforms and New Formulas An Example Double Check Visualization
Everything Remains As It WasNo matter what functions arise, the idea for solving differentialequations with Laplace transforms stays the same.
Time Domain (t)
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Laplace Transforms of Periodic Functions
logo1
Transforms and New Formulas An Example Double Check Visualization
Everything Remains As It WasNo matter what functions arise, the idea for solving differentialequations with Laplace transforms stays the same.
Time Domain (t)
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Laplace Transforms of Periodic Functions
logo1
Transforms and New Formulas An Example Double Check Visualization
Everything Remains As It WasNo matter what functions arise, the idea for solving differentialequations with Laplace transforms stays the same.
Time Domain (t)
OriginalDE & IVP
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Laplace Transforms of Periodic Functions
logo1
Transforms and New Formulas An Example Double Check Visualization
Everything Remains As It WasNo matter what functions arise, the idea for solving differentialequations with Laplace transforms stays the same.
Time Domain (t)
OriginalDE & IVP
-L
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Laplace Transforms of Periodic Functions
logo1
Transforms and New Formulas An Example Double Check Visualization
Everything Remains As It WasNo matter what functions arise, the idea for solving differentialequations with Laplace transforms stays the same.
Time Domain (t)
OriginalDE & IVP
Algebraic equation forthe Laplace transform
-L
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Laplace Transforms of Periodic Functions
logo1
Transforms and New Formulas An Example Double Check Visualization
Everything Remains As It WasNo matter what functions arise, the idea for solving differentialequations with Laplace transforms stays the same.
Time Domain (t) Transform domain (s)
OriginalDE & IVP
Algebraic equation forthe Laplace transform
-L
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Laplace Transforms of Periodic Functions
logo1
Transforms and New Formulas An Example Double Check Visualization
Everything Remains As It WasNo matter what functions arise, the idea for solving differentialequations with Laplace transforms stays the same.
Time Domain (t) Transform domain (s)
OriginalDE & IVP
Algebraic equation forthe Laplace transform
-L
Algebraic solution,partial fractions
?
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Laplace Transforms of Periodic Functions
logo1
Transforms and New Formulas An Example Double Check Visualization
Everything Remains As It WasNo matter what functions arise, the idea for solving differentialequations with Laplace transforms stays the same.
Time Domain (t) Transform domain (s)
OriginalDE & IVP
Algebraic equation forthe Laplace transform
Laplace transformof the solution
-L
Algebraic solution,partial fractions
?
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Laplace Transforms of Periodic Functions
logo1
Transforms and New Formulas An Example Double Check Visualization
Everything Remains As It WasNo matter what functions arise, the idea for solving differentialequations with Laplace transforms stays the same.
Time Domain (t) Transform domain (s)
OriginalDE & IVP
Algebraic equation forthe Laplace transform
Laplace transformof the solution
-
�
L
L −1
Algebraic solution,partial fractions
?
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Laplace Transforms of Periodic Functions
logo1
Transforms and New Formulas An Example Double Check Visualization
Everything Remains As It WasNo matter what functions arise, the idea for solving differentialequations with Laplace transforms stays the same.
Time Domain (t) Transform domain (s)
OriginalDE & IVP
Algebraic equation forthe Laplace transform
Laplace transformof the solutionSolution
-
�
L
L −1
Algebraic solution,partial fractions
?
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Laplace Transforms of Periodic Functions
logo1
Transforms and New Formulas An Example Double Check Visualization
Periodic Functions
1. A function f is periodic with period T > 0 if and only if forall t we have f (t +T) = f (t).
2. If f is bounded, piecewise continuous and periodic withperiod T , then
L{
f (t)}
=1
1− e−sT
∫ T
0e−stf (t) dt
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Laplace Transforms of Periodic Functions
logo1
Transforms and New Formulas An Example Double Check Visualization
Periodic Functions1. A function f is periodic with period T > 0 if and only if for
all t we have f (t +T) = f (t).
2. If f is bounded, piecewise continuous and periodic withperiod T , then
L{
f (t)}
=1
1− e−sT
∫ T
0e−stf (t) dt
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Laplace Transforms of Periodic Functions
logo1
Transforms and New Formulas An Example Double Check Visualization
Periodic Functions1. A function f is periodic with period T > 0 if and only if for
all t we have f (t +T) = f (t).2. If f is bounded, piecewise continuous and periodic with
period T , then
L{
f (t)}
=1
1− e−sT
∫ T
0e−stf (t) dt
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Laplace Transforms of Periodic Functions
logo1
Transforms and New Formulas An Example Double Check Visualization
How Did We Get That?
L{
f (t)}
=∫
∞
0e−stf (t) dt =
∞
∑n=0
∫ (n+1)T
nTe−stf (t) dt
=∞
∑n=0
∫ (n+1)T
nTe−s((t−nT)+nT
)f (t) dt =
∞
∑n=0
∫ T
0e−s(u+nT)f (u) du
=∞
∑n=0
e−nsT∫ T
0e−suf (u) du =
[∞
∑n=0
(e−sT)n
]∫ T
0e−stf (t) dt
=1
1− e−sT
∫ T
0e−stf (t) dt
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Laplace Transforms of Periodic Functions
logo1
Transforms and New Formulas An Example Double Check Visualization
How Did We Get That?
L{
f (t)}
=∫
∞
0e−stf (t) dt =
∞
∑n=0
∫ (n+1)T
nTe−stf (t) dt
=∞
∑n=0
∫ (n+1)T
nTe−s((t−nT)+nT
)f (t) dt =
∞
∑n=0
∫ T
0e−s(u+nT)f (u) du
=∞
∑n=0
e−nsT∫ T
0e−suf (u) du =
[∞
∑n=0
(e−sT)n
]∫ T
0e−stf (t) dt
=1
1− e−sT
∫ T
0e−stf (t) dt
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Laplace Transforms of Periodic Functions
logo1
Transforms and New Formulas An Example Double Check Visualization
How Did We Get That?
L{
f (t)}
=∫
∞
0e−stf (t) dt
=∞
∑n=0
∫ (n+1)T
nTe−stf (t) dt
=∞
∑n=0
∫ (n+1)T
nTe−s((t−nT)+nT
)f (t) dt =
∞
∑n=0
∫ T
0e−s(u+nT)f (u) du
=∞
∑n=0
e−nsT∫ T
0e−suf (u) du =
[∞
∑n=0
(e−sT)n
]∫ T
0e−stf (t) dt
=1
1− e−sT
∫ T
0e−stf (t) dt
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Laplace Transforms of Periodic Functions
logo1
Transforms and New Formulas An Example Double Check Visualization
How Did We Get That?
L{
f (t)}
=∫
∞
0e−stf (t) dt =
∞
∑n=0
∫ (n+1)T
nTe−stf (t) dt
=∞
∑n=0
∫ (n+1)T
nTe−s((t−nT)+nT
)f (t) dt =
∞
∑n=0
∫ T
0e−s(u+nT)f (u) du
=∞
∑n=0
e−nsT∫ T
0e−suf (u) du =
[∞
∑n=0
(e−sT)n
]∫ T
0e−stf (t) dt
=1
1− e−sT
∫ T
0e−stf (t) dt
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Laplace Transforms of Periodic Functions
logo1
Transforms and New Formulas An Example Double Check Visualization
How Did We Get That?
L{
f (t)}
=∫
∞
0e−stf (t) dt =
∞
∑n=0
∫ (n+1)T
nTe−stf (t) dt
=∞
∑n=0
∫ (n+1)T
nTe−s((t−nT)+nT
)f (t) dt
=∞
∑n=0
∫ T
0e−s(u+nT)f (u) du
=∞
∑n=0
e−nsT∫ T
0e−suf (u) du =
[∞
∑n=0
(e−sT)n
]∫ T
0e−stf (t) dt
=1
1− e−sT
∫ T
0e−stf (t) dt
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Laplace Transforms of Periodic Functions
logo1
Transforms and New Formulas An Example Double Check Visualization
How Did We Get That?
L{
f (t)}
=∫
∞
0e−stf (t) dt =
∞
∑n=0
∫ (n+1)T
nTe−stf (t) dt
=∞
∑n=0
∫ (n+1)T
nTe−s((t−nT)+nT
)f (t) dt =
∞
∑n=0
∫ T
0e−s(u+nT)f (u) du
=∞
∑n=0
e−nsT∫ T
0e−suf (u) du =
[∞
∑n=0
(e−sT)n
]∫ T
0e−stf (t) dt
=1
1− e−sT
∫ T
0e−stf (t) dt
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Laplace Transforms of Periodic Functions
logo1
Transforms and New Formulas An Example Double Check Visualization
How Did We Get That?
L{
f (t)}
=∫
∞
0e−stf (t) dt =
∞
∑n=0
∫ (n+1)T
nTe−stf (t) dt
=∞
∑n=0
∫ (n+1)T
nTe−s((t−nT)+nT
)f (t) dt =
∞
∑n=0
∫ T
0e−s(u+nT)f (u) du
=∞
∑n=0
e−nsT∫ T
0e−suf (u) du
=
[∞
∑n=0
(e−sT)n
]∫ T
0e−stf (t) dt
=1
1− e−sT
∫ T
0e−stf (t) dt
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Laplace Transforms of Periodic Functions
logo1
Transforms and New Formulas An Example Double Check Visualization
How Did We Get That?
L{
f (t)}
=∫
∞
0e−stf (t) dt =
∞
∑n=0
∫ (n+1)T
nTe−stf (t) dt
=∞
∑n=0
∫ (n+1)T
nTe−s((t−nT)+nT
)f (t) dt =
∞
∑n=0
∫ T
0e−s(u+nT)f (u) du
=∞
∑n=0
e−nsT∫ T
0e−suf (u) du =
[∞
∑n=0
(e−sT)n
]∫ T
0e−stf (t) dt
=1
1− e−sT
∫ T
0e−stf (t) dt
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Laplace Transforms of Periodic Functions
logo1
Transforms and New Formulas An Example Double Check Visualization
How Did We Get That?
L{
f (t)}
=∫
∞
0e−stf (t) dt =
∞
∑n=0
∫ (n+1)T
nTe−stf (t) dt
=∞
∑n=0
∫ (n+1)T
nTe−s((t−nT)+nT
)f (t) dt =
∞
∑n=0
∫ T
0e−s(u+nT)f (u) du
=∞
∑n=0
e−nsT∫ T
0e−suf (u) du =
[∞
∑n=0
(e−sT)n
]∫ T
0e−stf (t) dt
=1
1− e−sT
∫ T
0e−stf (t) dt
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Laplace Transforms of Periodic Functions
logo1
Transforms and New Formulas An Example Double Check Visualization
Solve the Initial Value Problem3y′+2y =
∣∣sin(t)∣∣, y(0) = 0
L{|sin(t)|
}=
11− e−sπ
∫π
0e−st∣∣sin(t)
∣∣ dt
=1
1− e−sπ
∫π
0e−st sin(t) dt
=1
1− e−sπ
∫∞
0e−st(1−U (t−π)
)sin(t) dt
=1
1− e−sπ
[L{
sin(t)}
+L{U (t−π)sin(t−π)
}]=
11− e−sπ
[1
s2 +1+ e−πs 1
s2 +1
]=
11− e−sπ
[1+ e−πs] 1
s2 +1
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Laplace Transforms of Periodic Functions
logo1
Transforms and New Formulas An Example Double Check Visualization
Solve the Initial Value Problem3y′+2y =
∣∣sin(t)∣∣, y(0) = 0
L{|sin(t)|
}=
11− e−sπ
∫π
0e−st∣∣sin(t)
∣∣ dt
=1
1− e−sπ
∫π
0e−st sin(t) dt
=1
1− e−sπ
∫∞
0e−st(1−U (t−π)
)sin(t) dt
=1
1− e−sπ
[L{
sin(t)}
+L{U (t−π)sin(t−π)
}]=
11− e−sπ
[1
s2 +1+ e−πs 1
s2 +1
]=
11− e−sπ
[1+ e−πs] 1
s2 +1
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Laplace Transforms of Periodic Functions
logo1
Transforms and New Formulas An Example Double Check Visualization
Solve the Initial Value Problem3y′+2y =
∣∣sin(t)∣∣, y(0) = 0
L{|sin(t)|
}=
11− e−sπ
∫π
0e−st∣∣sin(t)
∣∣ dt
=1
1− e−sπ
∫π
0e−st sin(t) dt
=1
1− e−sπ
∫∞
0e−st(1−U (t−π)
)sin(t) dt
=1
1− e−sπ
[L{
sin(t)}
+L{U (t−π)sin(t−π)
}]=
11− e−sπ
[1
s2 +1+ e−πs 1
s2 +1
]=
11− e−sπ
[1+ e−πs] 1
s2 +1
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Laplace Transforms of Periodic Functions
logo1
Transforms and New Formulas An Example Double Check Visualization
Solve the Initial Value Problem3y′+2y =
∣∣sin(t)∣∣, y(0) = 0
L{|sin(t)|
}=
11− e−sπ
∫π
0e−st∣∣sin(t)
∣∣ dt
=1
1− e−sπ
∫π
0e−st sin(t) dt
=1
1− e−sπ
∫∞
0e−st(1−U (t−π)
)sin(t) dt
=1
1− e−sπ
[L{
sin(t)}
+L{U (t−π)sin(t−π)
}]=
11− e−sπ
[1
s2 +1+ e−πs 1
s2 +1
]=
11− e−sπ
[1+ e−πs] 1
s2 +1
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Laplace Transforms of Periodic Functions
logo1
Transforms and New Formulas An Example Double Check Visualization
Solve the Initial Value Problem3y′+2y =
∣∣sin(t)∣∣, y(0) = 0
L{|sin(t)|
}=
11− e−sπ
∫π
0e−st∣∣sin(t)
∣∣ dt
=1
1− e−sπ
∫π
0e−st sin(t) dt
=1
1− e−sπ
∫∞
0e−st(1−U (t−π)
)sin(t) dt
=1
1− e−sπ
[L{
sin(t)}
+L{U (t−π)sin(t−π)
}]=
11− e−sπ
[1
s2 +1+ e−πs 1
s2 +1
]=
11− e−sπ
[1+ e−πs] 1
s2 +1
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Laplace Transforms of Periodic Functions
logo1
Transforms and New Formulas An Example Double Check Visualization
Solve the Initial Value Problem3y′+2y =
∣∣sin(t)∣∣, y(0) = 0
L{|sin(t)|
}=
11− e−sπ
∫π
0e−st∣∣sin(t)
∣∣ dt
=1
1− e−sπ
∫π
0e−st sin(t) dt
=1
1− e−sπ
∫∞
0e−st(1−U (t−π)
)sin(t) dt
=1
1− e−sπ
[L{
sin(t)}
+L{U (t−π)sin(t−π)
}]
=1
1− e−sπ
[1
s2 +1+ e−πs 1
s2 +1
]=
11− e−sπ
[1+ e−πs] 1
s2 +1
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Laplace Transforms of Periodic Functions
logo1
Transforms and New Formulas An Example Double Check Visualization
Solve the Initial Value Problem3y′+2y =
∣∣sin(t)∣∣, y(0) = 0
L{|sin(t)|
}=
11− e−sπ
∫π
0e−st∣∣sin(t)
∣∣ dt
=1
1− e−sπ
∫π
0e−st sin(t) dt
=1
1− e−sπ
∫∞
0e−st(1−U (t−π)
)sin(t) dt
=1
1− e−sπ
[L{
sin(t)}
+L{U (t−π)sin(t−π)
}]=
11− e−sπ
[1
s2 +1
+ e−πs 1s2 +1
]=
11− e−sπ
[1+ e−πs] 1
s2 +1
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Laplace Transforms of Periodic Functions
logo1
Transforms and New Formulas An Example Double Check Visualization
Solve the Initial Value Problem3y′+2y =
∣∣sin(t)∣∣, y(0) = 0
L{|sin(t)|
}=
11− e−sπ
∫π
0e−st∣∣sin(t)
∣∣ dt
=1
1− e−sπ
∫π
0e−st sin(t) dt
=1
1− e−sπ
∫∞
0e−st(1−U (t−π)
)sin(t) dt
=1
1− e−sπ
[L{
sin(t)}
+L{U (t−π)sin(t−π)
}]=
11− e−sπ
[1
s2 +1+ e−πs 1
s2 +1
]
=1
1− e−sπ
[1+ e−πs] 1
s2 +1
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Laplace Transforms of Periodic Functions
logo1
Transforms and New Formulas An Example Double Check Visualization
Solve the Initial Value Problem3y′+2y =
∣∣sin(t)∣∣, y(0) = 0
L{|sin(t)|
}=
11− e−sπ
∫π
0e−st∣∣sin(t)
∣∣ dt
=1
1− e−sπ
∫π
0e−st sin(t) dt
=1
1− e−sπ
∫∞
0e−st(1−U (t−π)
)sin(t) dt
=1
1− e−sπ
[L{
sin(t)}
+L{U (t−π)sin(t−π)
}]=
11− e−sπ
[1
s2 +1+ e−πs 1
s2 +1
]=
11− e−sπ
[1+ e−πs] 1
s2 +1Bernd Schroder Louisiana Tech University, College of Engineering and Science
Laplace Transforms of Periodic Functions
logo1
Transforms and New Formulas An Example Double Check Visualization
Solve the Initial Value Problem3y′+2y =
∣∣sin(t)∣∣, y(0) = 0
3sY +2Y =1
1− e−sπ
[1+ e−πs] 1
s2 +1
Y =1
1− e−sπ
[1+ e−πs] 1
(s2 +1)(3s+2)
=∞
∑n=0
(e−πs)n [1+ e−πs] 1
(s2 +1)(3s+2)
=
[∞
∑n=0
(e−πs)n +
∞
∑n=0
(e−πs)n+1
]1
(s2 +1)(3s+2)
=
[∞
∑n=0
e−nπs +∞
∑n=0
e−(n+1)πs
]1
(s2 +1)(3s+2)
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Laplace Transforms of Periodic Functions
logo1
Transforms and New Formulas An Example Double Check Visualization
Solve the Initial Value Problem3y′+2y =
∣∣sin(t)∣∣, y(0) = 0
3sY +2Y
=1
1− e−sπ
[1+ e−πs] 1
s2 +1
Y =1
1− e−sπ
[1+ e−πs] 1
(s2 +1)(3s+2)
=∞
∑n=0
(e−πs)n [1+ e−πs] 1
(s2 +1)(3s+2)
=
[∞
∑n=0
(e−πs)n +
∞
∑n=0
(e−πs)n+1
]1
(s2 +1)(3s+2)
=
[∞
∑n=0
e−nπs +∞
∑n=0
e−(n+1)πs
]1
(s2 +1)(3s+2)
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Laplace Transforms of Periodic Functions
logo1
Transforms and New Formulas An Example Double Check Visualization
Solve the Initial Value Problem3y′+2y =
∣∣sin(t)∣∣, y(0) = 0
3sY +2Y =1
1− e−sπ
[1+ e−πs] 1
s2 +1
Y =1
1− e−sπ
[1+ e−πs] 1
(s2 +1)(3s+2)
=∞
∑n=0
(e−πs)n [1+ e−πs] 1
(s2 +1)(3s+2)
=
[∞
∑n=0
(e−πs)n +
∞
∑n=0
(e−πs)n+1
]1
(s2 +1)(3s+2)
=
[∞
∑n=0
e−nπs +∞
∑n=0
e−(n+1)πs
]1
(s2 +1)(3s+2)
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Laplace Transforms of Periodic Functions
logo1
Transforms and New Formulas An Example Double Check Visualization
Solve the Initial Value Problem3y′+2y =
∣∣sin(t)∣∣, y(0) = 0
3sY +2Y =1
1− e−sπ
[1+ e−πs] 1
s2 +1
Y =1
1− e−sπ
[1+ e−πs] 1
(s2 +1)(3s+2)
=∞
∑n=0
(e−πs)n [1+ e−πs] 1
(s2 +1)(3s+2)
=
[∞
∑n=0
(e−πs)n +
∞
∑n=0
(e−πs)n+1
]1
(s2 +1)(3s+2)
=
[∞
∑n=0
e−nπs +∞
∑n=0
e−(n+1)πs
]1
(s2 +1)(3s+2)
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Laplace Transforms of Periodic Functions
logo1
Transforms and New Formulas An Example Double Check Visualization
Solve the Initial Value Problem3y′+2y =
∣∣sin(t)∣∣, y(0) = 0
3sY +2Y =1
1− e−sπ
[1+ e−πs] 1
s2 +1
Y =1
1− e−sπ
[1+ e−πs] 1
(s2 +1)(3s+2)
=∞
∑n=0
(e−πs)n [1+ e−πs] 1
(s2 +1)(3s+2)
=
[∞
∑n=0
(e−πs)n +
∞
∑n=0
(e−πs)n+1
]1
(s2 +1)(3s+2)
=
[∞
∑n=0
e−nπs +∞
∑n=0
e−(n+1)πs
]1
(s2 +1)(3s+2)
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Laplace Transforms of Periodic Functions
logo1
Transforms and New Formulas An Example Double Check Visualization
Solve the Initial Value Problem3y′+2y =
∣∣sin(t)∣∣, y(0) = 0
3sY +2Y =1
1− e−sπ
[1+ e−πs] 1
s2 +1
Y =1
1− e−sπ
[1+ e−πs] 1
(s2 +1)(3s+2)
=∞
∑n=0
(e−πs)n [1+ e−πs] 1
(s2 +1)(3s+2)
=
[∞
∑n=0
(e−πs)n +
∞
∑n=0
(e−πs)n+1
]1
(s2 +1)(3s+2)
=
[∞
∑n=0
e−nπs +∞
∑n=0
e−(n+1)πs
]1
(s2 +1)(3s+2)
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Laplace Transforms of Periodic Functions
logo1
Transforms and New Formulas An Example Double Check Visualization
Solve the Initial Value Problem3y′+2y =
∣∣sin(t)∣∣, y(0) = 0
3sY +2Y =1
1− e−sπ
[1+ e−πs] 1
s2 +1
Y =1
1− e−sπ
[1+ e−πs] 1
(s2 +1)(3s+2)
=∞
∑n=0
(e−πs)n [1+ e−πs] 1
(s2 +1)(3s+2)
=
[∞
∑n=0
(e−πs)n +
∞
∑n=0
(e−πs)n+1
]1
(s2 +1)(3s+2)
=
[∞
∑n=0
e−nπs +∞
∑n=0
e−(n+1)πs
]1
(s2 +1)(3s+2)
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Laplace Transforms of Periodic Functions
logo1
Transforms and New Formulas An Example Double Check Visualization
Solve the Initial Value Problem 3y′+2y =∣∣sin(t)
∣∣, y(0) = 0
1(s2 +1)(3s+2)
=As+Bs2 +1
+C
3s+2
1 = (As+B)(3s+2)+C(s2 +1
)s =−2
3: 1 = C
(49
+1)
=139
C, C =913
s = 0 : 1 = B ·2+C ·1 = 2B+9
13, B =
213
s = 1 : 1 = (A+B) ·5+C ·2 = 5A+1013
+1813
, A =− 313
1(s2 +1)(3s+2)
=1
13
(−3s+2s2 +1
+9
3s+2
)=
113
(−3
ss2 +1
+21
s2 +1+3
1s+ 2
3
)
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Laplace Transforms of Periodic Functions
logo1
Transforms and New Formulas An Example Double Check Visualization
Solve the Initial Value Problem 3y′+2y =∣∣sin(t)
∣∣, y(0) = 01
(s2 +1)(3s+2)=
As+Bs2 +1
+C
3s+2
1 = (As+B)(3s+2)+C(s2 +1
)s =−2
3: 1 = C
(49
+1)
=139
C, C =913
s = 0 : 1 = B ·2+C ·1 = 2B+9
13, B =
213
s = 1 : 1 = (A+B) ·5+C ·2 = 5A+1013
+1813
, A =− 313
1(s2 +1)(3s+2)
=1
13
(−3s+2s2 +1
+9
3s+2
)=
113
(−3
ss2 +1
+21
s2 +1+3
1s+ 2
3
)
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Laplace Transforms of Periodic Functions
logo1
Transforms and New Formulas An Example Double Check Visualization
Solve the Initial Value Problem 3y′+2y =∣∣sin(t)
∣∣, y(0) = 01
(s2 +1)(3s+2)=
As+Bs2 +1
+C
3s+2
1 = (As+B)(3s+2)+C(s2 +1
)
s =−23
: 1 = C(
49
+1)
=139
C, C =913
s = 0 : 1 = B ·2+C ·1 = 2B+9
13, B =
213
s = 1 : 1 = (A+B) ·5+C ·2 = 5A+1013
+1813
, A =− 313
1(s2 +1)(3s+2)
=1
13
(−3s+2s2 +1
+9
3s+2
)=
113
(−3
ss2 +1
+21
s2 +1+3
1s+ 2
3
)
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Laplace Transforms of Periodic Functions
logo1
Transforms and New Formulas An Example Double Check Visualization
Solve the Initial Value Problem 3y′+2y =∣∣sin(t)
∣∣, y(0) = 01
(s2 +1)(3s+2)=
As+Bs2 +1
+C
3s+2
1 = (As+B)(3s+2)+C(s2 +1
)s =−2
3:
1 = C(
49
+1)
=139
C, C =913
s = 0 : 1 = B ·2+C ·1 = 2B+9
13, B =
213
s = 1 : 1 = (A+B) ·5+C ·2 = 5A+1013
+1813
, A =− 313
1(s2 +1)(3s+2)
=1
13
(−3s+2s2 +1
+9
3s+2
)=
113
(−3
ss2 +1
+21
s2 +1+3
1s+ 2
3
)
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Laplace Transforms of Periodic Functions
logo1
Transforms and New Formulas An Example Double Check Visualization
Solve the Initial Value Problem 3y′+2y =∣∣sin(t)
∣∣, y(0) = 01
(s2 +1)(3s+2)=
As+Bs2 +1
+C
3s+2
1 = (As+B)(3s+2)+C(s2 +1
)s =−2
3: 1 = C
(49
+1)
=139
C, C =913
s = 0 : 1 = B ·2+C ·1 = 2B+9
13, B =
213
s = 1 : 1 = (A+B) ·5+C ·2 = 5A+1013
+1813
, A =− 313
1(s2 +1)(3s+2)
=1
13
(−3s+2s2 +1
+9
3s+2
)=
113
(−3
ss2 +1
+21
s2 +1+3
1s+ 2
3
)
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Laplace Transforms of Periodic Functions
logo1
Transforms and New Formulas An Example Double Check Visualization
Solve the Initial Value Problem 3y′+2y =∣∣sin(t)
∣∣, y(0) = 01
(s2 +1)(3s+2)=
As+Bs2 +1
+C
3s+2
1 = (As+B)(3s+2)+C(s2 +1
)s =−2
3: 1 = C
(49
+1)
=139
C
, C =913
s = 0 : 1 = B ·2+C ·1 = 2B+9
13, B =
213
s = 1 : 1 = (A+B) ·5+C ·2 = 5A+1013
+1813
, A =− 313
1(s2 +1)(3s+2)
=1
13
(−3s+2s2 +1
+9
3s+2
)=
113
(−3
ss2 +1
+21
s2 +1+3
1s+ 2
3
)
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Laplace Transforms of Periodic Functions
logo1
Transforms and New Formulas An Example Double Check Visualization
Solve the Initial Value Problem 3y′+2y =∣∣sin(t)
∣∣, y(0) = 01
(s2 +1)(3s+2)=
As+Bs2 +1
+C
3s+2
1 = (As+B)(3s+2)+C(s2 +1
)s =−2
3: 1 = C
(49
+1)
=139
C, C =913
s = 0 : 1 = B ·2+C ·1 = 2B+9
13, B =
213
s = 1 : 1 = (A+B) ·5+C ·2 = 5A+1013
+1813
, A =− 313
1(s2 +1)(3s+2)
=1
13
(−3s+2s2 +1
+9
3s+2
)=
113
(−3
ss2 +1
+21
s2 +1+3
1s+ 2
3
)
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Laplace Transforms of Periodic Functions
logo1
Transforms and New Formulas An Example Double Check Visualization
Solve the Initial Value Problem 3y′+2y =∣∣sin(t)
∣∣, y(0) = 01
(s2 +1)(3s+2)=
As+Bs2 +1
+C
3s+2
1 = (As+B)(3s+2)+C(s2 +1
)s =−2
3: 1 = C
(49
+1)
=139
C, C =913
s = 0 :
1 = B ·2+C ·1 = 2B+9
13, B =
213
s = 1 : 1 = (A+B) ·5+C ·2 = 5A+1013
+1813
, A =− 313
1(s2 +1)(3s+2)
=1
13
(−3s+2s2 +1
+9
3s+2
)=
113
(−3
ss2 +1
+21
s2 +1+3
1s+ 2
3
)
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Laplace Transforms of Periodic Functions
logo1
Transforms and New Formulas An Example Double Check Visualization
Solve the Initial Value Problem 3y′+2y =∣∣sin(t)
∣∣, y(0) = 01
(s2 +1)(3s+2)=
As+Bs2 +1
+C
3s+2
1 = (As+B)(3s+2)+C(s2 +1
)s =−2
3: 1 = C
(49
+1)
=139
C, C =913
s = 0 : 1 = B ·2+C ·1
= 2B+9
13, B =
213
s = 1 : 1 = (A+B) ·5+C ·2 = 5A+1013
+1813
, A =− 313
1(s2 +1)(3s+2)
=1
13
(−3s+2s2 +1
+9
3s+2
)=
113
(−3
ss2 +1
+21
s2 +1+3
1s+ 2
3
)
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Laplace Transforms of Periodic Functions
logo1
Transforms and New Formulas An Example Double Check Visualization
Solve the Initial Value Problem 3y′+2y =∣∣sin(t)
∣∣, y(0) = 01
(s2 +1)(3s+2)=
As+Bs2 +1
+C
3s+2
1 = (As+B)(3s+2)+C(s2 +1
)s =−2
3: 1 = C
(49
+1)
=139
C, C =913
s = 0 : 1 = B ·2+C ·1 = 2B+9
13
, B =213
s = 1 : 1 = (A+B) ·5+C ·2 = 5A+1013
+1813
, A =− 313
1(s2 +1)(3s+2)
=1
13
(−3s+2s2 +1
+9
3s+2
)=
113
(−3
ss2 +1
+21
s2 +1+3
1s+ 2
3
)
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Laplace Transforms of Periodic Functions
logo1
Transforms and New Formulas An Example Double Check Visualization
Solve the Initial Value Problem 3y′+2y =∣∣sin(t)
∣∣, y(0) = 01
(s2 +1)(3s+2)=
As+Bs2 +1
+C
3s+2
1 = (As+B)(3s+2)+C(s2 +1
)s =−2
3: 1 = C
(49
+1)
=139
C, C =913
s = 0 : 1 = B ·2+C ·1 = 2B+9
13, B =
213
s = 1 : 1 = (A+B) ·5+C ·2 = 5A+1013
+1813
, A =− 313
1(s2 +1)(3s+2)
=1
13
(−3s+2s2 +1
+9
3s+2
)=
113
(−3
ss2 +1
+21
s2 +1+3
1s+ 2
3
)
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Laplace Transforms of Periodic Functions
logo1
Transforms and New Formulas An Example Double Check Visualization
Solve the Initial Value Problem 3y′+2y =∣∣sin(t)
∣∣, y(0) = 01
(s2 +1)(3s+2)=
As+Bs2 +1
+C
3s+2
1 = (As+B)(3s+2)+C(s2 +1
)s =−2
3: 1 = C
(49
+1)
=139
C, C =913
s = 0 : 1 = B ·2+C ·1 = 2B+9
13, B =
213
s = 1 :
1 = (A+B) ·5+C ·2 = 5A+1013
+1813
, A =− 313
1(s2 +1)(3s+2)
=1
13
(−3s+2s2 +1
+9
3s+2
)=
113
(−3
ss2 +1
+21
s2 +1+3
1s+ 2
3
)
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Laplace Transforms of Periodic Functions
logo1
Transforms and New Formulas An Example Double Check Visualization
Solve the Initial Value Problem 3y′+2y =∣∣sin(t)
∣∣, y(0) = 01
(s2 +1)(3s+2)=
As+Bs2 +1
+C
3s+2
1 = (As+B)(3s+2)+C(s2 +1
)s =−2
3: 1 = C
(49
+1)
=139
C, C =913
s = 0 : 1 = B ·2+C ·1 = 2B+9
13, B =
213
s = 1 : 1 = (A+B) ·5+C ·2
= 5A+1013
+1813
, A =− 313
1(s2 +1)(3s+2)
=1
13
(−3s+2s2 +1
+9
3s+2
)=
113
(−3
ss2 +1
+21
s2 +1+3
1s+ 2
3
)
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Laplace Transforms of Periodic Functions
logo1
Transforms and New Formulas An Example Double Check Visualization
Solve the Initial Value Problem 3y′+2y =∣∣sin(t)
∣∣, y(0) = 01
(s2 +1)(3s+2)=
As+Bs2 +1
+C
3s+2
1 = (As+B)(3s+2)+C(s2 +1
)s =−2
3: 1 = C
(49
+1)
=139
C, C =913
s = 0 : 1 = B ·2+C ·1 = 2B+9
13, B =
213
s = 1 : 1 = (A+B) ·5+C ·2 = 5A+1013
+1813
, A =− 313
1(s2 +1)(3s+2)
=1
13
(−3s+2s2 +1
+9
3s+2
)=
113
(−3
ss2 +1
+21
s2 +1+3
1s+ 2
3
)
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Laplace Transforms of Periodic Functions
logo1
Transforms and New Formulas An Example Double Check Visualization
Solve the Initial Value Problem 3y′+2y =∣∣sin(t)
∣∣, y(0) = 01
(s2 +1)(3s+2)=
As+Bs2 +1
+C
3s+2
1 = (As+B)(3s+2)+C(s2 +1
)s =−2
3: 1 = C
(49
+1)
=139
C, C =913
s = 0 : 1 = B ·2+C ·1 = 2B+9
13, B =
213
s = 1 : 1 = (A+B) ·5+C ·2 = 5A+1013
+1813
, A =− 313
1(s2 +1)(3s+2)
=1
13
(−3s+2s2 +1
+9
3s+2
)=
113
(−3
ss2 +1
+21
s2 +1+3
1s+ 2
3
)
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Laplace Transforms of Periodic Functions
logo1
Transforms and New Formulas An Example Double Check Visualization
Solve the Initial Value Problem 3y′+2y =∣∣sin(t)
∣∣, y(0) = 01
(s2 +1)(3s+2)=
As+Bs2 +1
+C
3s+2
1 = (As+B)(3s+2)+C(s2 +1
)s =−2
3: 1 = C
(49
+1)
=139
C, C =913
s = 0 : 1 = B ·2+C ·1 = 2B+9
13, B =
213
s = 1 : 1 = (A+B) ·5+C ·2 = 5A+1013
+1813
, A =− 313
1(s2 +1)(3s+2)
=1
13
(−3s+2s2 +1
+9
3s+2
)=
113
(−3
ss2 +1
+21
s2 +1+3
1s+ 2
3
)
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Laplace Transforms of Periodic Functions
logo1
Transforms and New Formulas An Example Double Check Visualization
Solve the Initial Value Problem 3y′+2y =∣∣sin(t)
∣∣, y(0) = 01
(s2 +1)(3s+2)=
As+Bs2 +1
+C
3s+2
1 = (As+B)(3s+2)+C(s2 +1
)s =−2
3: 1 = C
(49
+1)
=139
C, C =913
s = 0 : 1 = B ·2+C ·1 = 2B+9
13, B =
213
s = 1 : 1 = (A+B) ·5+C ·2 = 5A+1013
+1813
, A =− 313
1(s2 +1)(3s+2)
=1
13
(−3s+2s2 +1
+9
3s+2
)=
113
(−3
ss2 +1
+21
s2 +1+3
1s+ 2
3
)
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Laplace Transforms of Periodic Functions
logo1
Transforms and New Formulas An Example Double Check Visualization
Solve the Initial Value Problem 3y′+2y =∣∣sin(t)
∣∣, y(0) = 01
(s2 +1)(3s+2)=
As+Bs2 +1
+C
3s+2
1 = (As+B)(3s+2)+C(s2 +1
)s =−2
3: 1 = C
(49
+1)
=139
C, C =913
s = 0 : 1 = B ·2+C ·1 = 2B+9
13, B =
213
s = 1 : 1 = (A+B) ·5+C ·2 = 5A+1013
+1813
, A =− 313
1(s2 +1)(3s+2)
=1
13
(−3s+2s2 +1
+9
3s+2
)=
113
(−3
ss2 +1
+21
s2 +1+3
1s+ 2
3
)
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Laplace Transforms of Periodic Functions
logo1
Transforms and New Formulas An Example Double Check Visualization
Solve the Initial Value Problem 3y′+2y =∣∣sin(t)
∣∣, y(0) = 01
(s2 +1)(3s+2)=
As+Bs2 +1
+C
3s+2
1 = (As+B)(3s+2)+C(s2 +1
)s =−2
3: 1 = C
(49
+1)
=139
C, C =913
s = 0 : 1 = B ·2+C ·1 = 2B+9
13, B =
213
s = 1 : 1 = (A+B) ·5+C ·2 = 5A+1013
+1813
, A =− 313
1(s2 +1)(3s+2)
=1
13
(−3s+2s2 +1
+9
3s+2
)
=1
13
(−3
ss2 +1
+21
s2 +1+3
1s+ 2
3
)
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Laplace Transforms of Periodic Functions
logo1
Transforms and New Formulas An Example Double Check Visualization
Solve the Initial Value Problem 3y′+2y =∣∣sin(t)
∣∣, y(0) = 01
(s2 +1)(3s+2)=
As+Bs2 +1
+C
3s+2
1 = (As+B)(3s+2)+C(s2 +1
)s =−2
3: 1 = C
(49
+1)
=139
C, C =913
s = 0 : 1 = B ·2+C ·1 = 2B+9
13, B =
213
s = 1 : 1 = (A+B) ·5+C ·2 = 5A+1013
+1813
, A =− 313
1(s2 +1)(3s+2)
=1
13
(−3s+2s2 +1
+9
3s+2
)=
113
(−3
ss2 +1
+21
s2 +1+3
1s+ 2
3
)Bernd Schroder Louisiana Tech University, College of Engineering and Science
Laplace Transforms of Periodic Functions
logo1
Transforms and New Formulas An Example Double Check Visualization
Solve the Initial Value Problem3y′+2y =
∣∣sin(t)∣∣, y(0) = 0
Y =
[∞
∑n=0
e−nπs +∞
∑n=0
e−(n+1)πs
]1
(s2 +1)(3s+2)
=
[∞
∑n=0
e−nπs +∞
∑n=0
e−(n+1)πs
]1
13
(−3
ss2 +1
+21
s2 +1+3
1s+ 2
3
)
=
[∞
∑n=0
e−nπs +∞
∑n=1
e−nπs
]113
(−3
ss2 +1
+21
s2 +1+3
1s+ 2
3
)
=
[1+2
∞
∑n=1
e−nπs
]1
13
(−3
ss2 +1
+21
s2 +1+3
1s+ 2
3
)
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Laplace Transforms of Periodic Functions
logo1
Transforms and New Formulas An Example Double Check Visualization
Solve the Initial Value Problem3y′+2y =
∣∣sin(t)∣∣, y(0) = 0
Y =
[∞
∑n=0
e−nπs +∞
∑n=0
e−(n+1)πs
]1
(s2 +1)(3s+2)
=
[∞
∑n=0
e−nπs +∞
∑n=0
e−(n+1)πs
]1
13
(−3
ss2 +1
+21
s2 +1+3
1s+ 2
3
)
=
[∞
∑n=0
e−nπs +∞
∑n=1
e−nπs
]113
(−3
ss2 +1
+21
s2 +1+3
1s+ 2
3
)
=
[1+2
∞
∑n=1
e−nπs
]1
13
(−3
ss2 +1
+21
s2 +1+3
1s+ 2
3
)
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Laplace Transforms of Periodic Functions
logo1
Transforms and New Formulas An Example Double Check Visualization
Solve the Initial Value Problem3y′+2y =
∣∣sin(t)∣∣, y(0) = 0
Y =
[∞
∑n=0
e−nπs +∞
∑n=0
e−(n+1)πs
]1
(s2 +1)(3s+2)
=
[∞
∑n=0
e−nπs +∞
∑n=0
e−(n+1)πs
]1
13
(−3
ss2 +1
+21
s2 +1+3
1s+ 2
3
)
=
[∞
∑n=0
e−nπs +∞
∑n=1
e−nπs
]113
(−3
ss2 +1
+21
s2 +1+3
1s+ 2
3
)
=
[1+2
∞
∑n=1
e−nπs
]1
13
(−3
ss2 +1
+21
s2 +1+3
1s+ 2
3
)
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Laplace Transforms of Periodic Functions
logo1
Transforms and New Formulas An Example Double Check Visualization
Solve the Initial Value Problem3y′+2y =
∣∣sin(t)∣∣, y(0) = 0
Y =
[∞
∑n=0
e−nπs +∞
∑n=0
e−(n+1)πs
]1
(s2 +1)(3s+2)
=
[∞
∑n=0
e−nπs +∞
∑n=0
e−(n+1)πs
]1
13
(−3
ss2 +1
+21
s2 +1+3
1s+ 2
3
)
=
[∞
∑n=0
e−nπs +∞
∑n=1
e−nπs
]113
(−3
ss2 +1
+21
s2 +1+3
1s+ 2
3
)
=
[1+2
∞
∑n=1
e−nπs
]1
13
(−3
ss2 +1
+21
s2 +1+3
1s+ 2
3
)
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Laplace Transforms of Periodic Functions
logo1
Transforms and New Formulas An Example Double Check Visualization
Solve the Initial Value Problem3y′+2y =
∣∣sin(t)∣∣, y(0) = 0
Y =
[∞
∑n=0
e−nπs +∞
∑n=0
e−(n+1)πs
]1
(s2 +1)(3s+2)
=
[∞
∑n=0
e−nπs +∞
∑n=0
e−(n+1)πs
]1
13
(−3
ss2 +1
+21
s2 +1+3
1s+ 2
3
)
=
[∞
∑n=0
e−nπs +∞
∑n=1
e−nπs
]113
(−3
ss2 +1
+21
s2 +1+3
1s+ 2
3
)
=
[1+2
∞
∑n=1
e−nπs
]1
13
(−3
ss2 +1
+21
s2 +1+3
1s+ 2
3
)
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Laplace Transforms of Periodic Functions
logo1
Transforms and New Formulas An Example Double Check Visualization
Solve the Initial Value Problem 3y′+2y =∣∣sin(t)
∣∣, y(0) = 0
y =113
[−3cos(t)+2sin(t)+3e−
23 t]
+213
∞
∑n=1
U(t−nπ
)[−3cos(t)+2sin(t)+3e−
23 t]
t→t−nπ
=113
[−3cos(t)+2sin(t)+3e−
23 t]
+213
∞
∑n=1
U (t−nπ)[−3cos(t−nπ)+2sin(t−nπ)+3e−
23 (t−nπ)
]
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Laplace Transforms of Periodic Functions
logo1
Transforms and New Formulas An Example Double Check Visualization
Solve the Initial Value Problem 3y′+2y =∣∣sin(t)
∣∣, y(0) = 0
y =113
[−3cos(t)+2sin(t)+3e−
23 t]
+213
∞
∑n=1
U(t−nπ
)[−3cos(t)+2sin(t)+3e−
23 t]
t→t−nπ
=113
[−3cos(t)+2sin(t)+3e−
23 t]
+213
∞
∑n=1
U (t−nπ)[−3cos(t−nπ)+2sin(t−nπ)+3e−
23 (t−nπ)
]
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Laplace Transforms of Periodic Functions
logo1
Transforms and New Formulas An Example Double Check Visualization
Solve the Initial Value Problem 3y′+2y =∣∣sin(t)
∣∣, y(0) = 0
y =113
[−3cos(t)+2sin(t)+3e−
23 t]
+213
∞
∑n=1
U(t−nπ
)[−3cos(t)+2sin(t)+3e−
23 t]
t→t−nπ
=113
[−3cos(t)+2sin(t)+3e−
23 t]
+213
∞
∑n=1
U (t−nπ)[−3cos(t−nπ)+2sin(t−nπ)+3e−
23 (t−nπ)
]
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Laplace Transforms of Periodic Functions
logo1
Transforms and New Formulas An Example Double Check Visualization
Does y =113
[−3cos(t)+2sin(t)+3e−
23 t]
+213
∞
∑n=1
U (t−nπ)[−3cos(t−nπ)+2sin(t−nπ)+3e−
23 (t−nπ)
]Really Solve
the Initial Value Problem 3y′+2y =∣∣sin(t)
∣∣, y(0) = 0?
y(0) =113
[−3cos(0)+2sin(0)+3e−
23 0]
+213
∞
∑n=1
U (0−nπ)[−3cos(0−nπ)+2sin(0−nπ)+3e−
23 (0−nπ)
]= 0
√
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Laplace Transforms of Periodic Functions
logo1
Transforms and New Formulas An Example Double Check Visualization
Does y =113
[−3cos(t)+2sin(t)+3e−
23 t]
+213
∞
∑n=1
U (t−nπ)[−3cos(t−nπ)+2sin(t−nπ)+3e−
23 (t−nπ)
]Really Solve
the Initial Value Problem 3y′+2y =∣∣sin(t)
∣∣, y(0) = 0?
y(0)
=113
[−3cos(0)+2sin(0)+3e−
23 0]
+213
∞
∑n=1
U (0−nπ)[−3cos(0−nπ)+2sin(0−nπ)+3e−
23 (0−nπ)
]= 0
√
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Laplace Transforms of Periodic Functions
logo1
Transforms and New Formulas An Example Double Check Visualization
Does y =113
[−3cos(t)+2sin(t)+3e−
23 t]
+213
∞
∑n=1
U (t−nπ)[−3cos(t−nπ)+2sin(t−nπ)+3e−
23 (t−nπ)
]Really Solve
the Initial Value Problem 3y′+2y =∣∣sin(t)
∣∣, y(0) = 0?
y(0) =113
[−3cos(0)+2sin(0)+3e−
23 0]
+213
∞
∑n=1
U (0−nπ)[−3cos(0−nπ)+2sin(0−nπ)+3e−
23 (0−nπ)
]
= 0√
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Laplace Transforms of Periodic Functions
logo1
Transforms and New Formulas An Example Double Check Visualization
Does y =113
[−3cos(t)+2sin(t)+3e−
23 t]
+213
∞
∑n=1
U (t−nπ)[−3cos(t−nπ)+2sin(t−nπ)+3e−
23 (t−nπ)
]Really Solve
the Initial Value Problem 3y′+2y =∣∣sin(t)
∣∣, y(0) = 0?
y(0) =113
[−3cos(0)+2sin(0)+3e−
23 0]
+213
∞
∑n=1
U (0−nπ)[−3cos(0−nπ)+2sin(0−nπ)+3e−
23 (0−nπ)
]= 0
√
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Laplace Transforms of Periodic Functions
logo1
Transforms and New Formulas An Example Double Check Visualization
Does y =113
[−3cos(t)+2sin(t)+3e−
23 t]
+213
∞
∑n=1
U (t−nπ)[−3cos(t−nπ)+2sin(t−nπ)+3e−
23 (t−nπ)
]Really Solve
the Initial Value Problem 3y′+2y =∣∣sin(t)
∣∣, y(0) = 0?
y(0) =113
[−3cos(0)+2sin(0)+3e−
23 0]
+213
∞
∑n=1
U (0−nπ)[−3cos(0−nπ)+2sin(0−nπ)+3e−
23 (0−nπ)
]= 0
√
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Laplace Transforms of Periodic Functions
logo1
Transforms and New Formulas An Example Double Check Visualization
y′ =1
13
[3sin(t)+2cos(t)−2e−
23 t]
+213
∞
∑n=1
U (t−nπ)[3sin(t−nπ)+2cos(t−nπ)−2e−
23 (t−nπ)
]3y′+2y =
113
[9sin(t)+6cos(t)−6e−
23 t]
+213
∞
∑n=1
U (t−nπ)[9sin(t−nπ)+6cos(t−nπ)−6e−
23 (t−nπ)
]+
113
[−6cos(t)+4sin(t)+6e−
23 t]
+213
∞
∑n=1
U (t−nπ)[−6cos(t−nπ)+4sin(t−nπ)+6e−
23 (t−nπ)
]=
113
[9sin(t)+4sin(t)]+2
13
∞
∑n=1
U (t−nπ) [9sin(t−nπ)+4sin(t−nπ)]
= sin(t)+2∞
∑n=1
U (t−nπ)sin(t−nπ) =∣∣sin(t)
∣∣ √
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Laplace Transforms of Periodic Functions
logo1
Transforms and New Formulas An Example Double Check Visualization
y′ =1
13
[3sin(t)+2cos(t)−2e−
23 t]
+213
∞
∑n=1
U (t−nπ)[3sin(t−nπ)+2cos(t−nπ)−2e−
23 (t−nπ)
]
3y′+2y =1
13
[9sin(t)+6cos(t)−6e−
23 t]
+213
∞
∑n=1
U (t−nπ)[9sin(t−nπ)+6cos(t−nπ)−6e−
23 (t−nπ)
]+
113
[−6cos(t)+4sin(t)+6e−
23 t]
+213
∞
∑n=1
U (t−nπ)[−6cos(t−nπ)+4sin(t−nπ)+6e−
23 (t−nπ)
]=
113
[9sin(t)+4sin(t)]+2
13
∞
∑n=1
U (t−nπ) [9sin(t−nπ)+4sin(t−nπ)]
= sin(t)+2∞
∑n=1
U (t−nπ)sin(t−nπ) =∣∣sin(t)
∣∣ √
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Laplace Transforms of Periodic Functions
logo1
Transforms and New Formulas An Example Double Check Visualization
y′ =1
13
[3sin(t)+2cos(t)−2e−
23 t]
+213
∞
∑n=1
U (t−nπ)[3sin(t−nπ)+2cos(t−nπ)−2e−
23 (t−nπ)
]3y′+2y =
113
[9sin(t)+6cos(t)−6e−
23 t]
+213
∞
∑n=1
U (t−nπ)[9sin(t−nπ)+6cos(t−nπ)−6e−
23 (t−nπ)
]
+113
[−6cos(t)+4sin(t)+6e−
23 t]
+213
∞
∑n=1
U (t−nπ)[−6cos(t−nπ)+4sin(t−nπ)+6e−
23 (t−nπ)
]=
113
[9sin(t)+4sin(t)]+2
13
∞
∑n=1
U (t−nπ) [9sin(t−nπ)+4sin(t−nπ)]
= sin(t)+2∞
∑n=1
U (t−nπ)sin(t−nπ) =∣∣sin(t)
∣∣ √
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Laplace Transforms of Periodic Functions
logo1
Transforms and New Formulas An Example Double Check Visualization
y′ =1
13
[3sin(t)+2cos(t)−2e−
23 t]
+213
∞
∑n=1
U (t−nπ)[3sin(t−nπ)+2cos(t−nπ)−2e−
23 (t−nπ)
]3y′+2y =
113
[9sin(t)+6cos(t)−6e−
23 t]
+213
∞
∑n=1
U (t−nπ)[9sin(t−nπ)+6cos(t−nπ)−6e−
23 (t−nπ)
]+
113
[−6cos(t)+4sin(t)+6e−
23 t]
+213
∞
∑n=1
U (t−nπ)[−6cos(t−nπ)+4sin(t−nπ)+6e−
23 (t−nπ)
]
=1
13[9sin(t)+4sin(t)]+
213
∞
∑n=1
U (t−nπ) [9sin(t−nπ)+4sin(t−nπ)]
= sin(t)+2∞
∑n=1
U (t−nπ)sin(t−nπ) =∣∣sin(t)
∣∣ √
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Laplace Transforms of Periodic Functions
logo1
Transforms and New Formulas An Example Double Check Visualization
y′ =1
13
[3sin(t)+2cos(t)−2e−
23 t]
+213
∞
∑n=1
U (t−nπ)[3sin(t−nπ)+2cos(t−nπ)−2e−
23 (t−nπ)
]3y′+2y =
113
[9sin(t)+6cos(t)−6e−
23 t]
+213
∞
∑n=1
U (t−nπ)[9sin(t−nπ)+6cos(t−nπ)−6e−
23 (t−nπ)
]+
113
[−6cos(t)+4sin(t)+6e−
23 t]
+213
∞
∑n=1
U (t−nπ)[−6cos(t−nπ)+4sin(t−nπ)+6e−
23 (t−nπ)
]=
113
[9sin(t)+4sin(t)]+2
13
∞
∑n=1
U (t−nπ) [9sin(t−nπ)+4sin(t−nπ)]
= sin(t)+2∞
∑n=1
U (t−nπ)sin(t−nπ) =∣∣sin(t)
∣∣ √
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Laplace Transforms of Periodic Functions
logo1
Transforms and New Formulas An Example Double Check Visualization
y′ =1
13
[3sin(t)+2cos(t)−2e−
23 t]
+213
∞
∑n=1
U (t−nπ)[3sin(t−nπ)+2cos(t−nπ)−2e−
23 (t−nπ)
]3y′+2y =
113
[9sin(t)+6cos(t)−6e−
23 t]
+213
∞
∑n=1
U (t−nπ)[9sin(t−nπ)+6cos(t−nπ)−6e−
23 (t−nπ)
]+
113
[−6cos(t)+4sin(t)+6e−
23 t]
+213
∞
∑n=1
U (t−nπ)[−6cos(t−nπ)+4sin(t−nπ)+6e−
23 (t−nπ)
]=
113
[9sin(t)+4sin(t)]+2
13
∞
∑n=1
U (t−nπ) [9sin(t−nπ)+4sin(t−nπ)]
= sin(t)+2∞
∑n=1
U (t−nπ)sin(t−nπ)
=∣∣sin(t)
∣∣ √
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Laplace Transforms of Periodic Functions
logo1
Transforms and New Formulas An Example Double Check Visualization
y′ =1
13
[3sin(t)+2cos(t)−2e−
23 t]
+213
∞
∑n=1
U (t−nπ)[3sin(t−nπ)+2cos(t−nπ)−2e−
23 (t−nπ)
]3y′+2y =
113
[9sin(t)+6cos(t)−6e−
23 t]
+213
∞
∑n=1
U (t−nπ)[9sin(t−nπ)+6cos(t−nπ)−6e−
23 (t−nπ)
]+
113
[−6cos(t)+4sin(t)+6e−
23 t]
+213
∞
∑n=1
U (t−nπ)[−6cos(t−nπ)+4sin(t−nπ)+6e−
23 (t−nπ)
]=
113
[9sin(t)+4sin(t)]+2
13
∞
∑n=1
U (t−nπ) [9sin(t−nπ)+4sin(t−nπ)]
= sin(t)+2∞
∑n=1
U (t−nπ)sin(t−nπ) =∣∣sin(t)
∣∣
√
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Laplace Transforms of Periodic Functions
logo1
Transforms and New Formulas An Example Double Check Visualization
y′ =1
13
[3sin(t)+2cos(t)−2e−
23 t]
+213
∞
∑n=1
U (t−nπ)[3sin(t−nπ)+2cos(t−nπ)−2e−
23 (t−nπ)
]3y′+2y =
113
[9sin(t)+6cos(t)−6e−
23 t]
+213
∞
∑n=1
U (t−nπ)[9sin(t−nπ)+6cos(t−nπ)−6e−
23 (t−nπ)
]+
113
[−6cos(t)+4sin(t)+6e−
23 t]
+213
∞
∑n=1
U (t−nπ)[−6cos(t−nπ)+4sin(t−nπ)+6e−
23 (t−nπ)
]=
113
[9sin(t)+4sin(t)]+2
13
∞
∑n=1
U (t−nπ) [9sin(t−nπ)+4sin(t−nπ)]
= sin(t)+2∞
∑n=1
U (t−nπ)sin(t−nπ) =∣∣sin(t)
∣∣ √
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Laplace Transforms of Periodic Functions
logo1
Transforms and New Formulas An Example Double Check Visualization
Comparing Output to Input
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Laplace Transforms of Periodic Functions
logo1
Transforms and New Formulas An Example Double Check Visualization
Comparing Output to Input
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Laplace Transforms of Periodic Functions
logo1
Transforms and New Formulas An Example Double Check Visualization
Comparing Output to Input
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Laplace Transforms of Periodic Functions
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