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Transforms and New Formulas An Example Double Check
Laplace Transforms and IntegralEquations
Bernd Schroder
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Laplace Transforms and Integral Equations
logo1
Transforms and New Formulas An Example Double Check
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 and Integral Equations
logo1
Transforms and New Formulas An Example Double Check
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 and Integral Equations
logo1
Transforms and New Formulas An Example Double Check
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 and Integral Equations
logo1
Transforms and New Formulas An Example Double Check
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 and Integral Equations
logo1
Transforms and New Formulas An Example Double Check
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 and Integral Equations
logo1
Transforms and New Formulas An Example Double Check
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 and Integral Equations
logo1
Transforms and New Formulas An Example Double Check
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 and Integral Equations
logo1
Transforms and New Formulas An Example Double Check
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 and Integral Equations
logo1
Transforms and New Formulas An Example Double Check
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 and Integral Equations
logo1
Transforms and New Formulas An Example Double Check
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 and Integral Equations
logo1
Transforms and New Formulas An Example Double Check
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 and Integral Equations
logo1
Transforms and New Formulas An Example Double Check
The Laplace Transform of an Integral
1. Definite integrals of the form∫ t
0f (τ)dτ arise in circuit
theory: The charge of a capacitor is the integral of thecurrent over time.(We assume the capacitor is initially uncharged.)
2. L
{∫ t
0f (τ)dτ
}=
F(s)s
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Laplace Transforms and Integral Equations
logo1
Transforms and New Formulas An Example Double Check
The Laplace Transform of an Integral1. Definite integrals of the form
∫ t
0f (τ)dτ arise in circuit
theory: The charge of a capacitor is the integral of thecurrent over time.
(We assume the capacitor is initially uncharged.)
2. L
{∫ t
0f (τ)dτ
}=
F(s)s
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Laplace Transforms and Integral Equations
logo1
Transforms and New Formulas An Example Double Check
The Laplace Transform of an Integral1. Definite integrals of the form
∫ t
0f (τ)dτ arise in circuit
theory: The charge of a capacitor is the integral of thecurrent over time.(We assume the capacitor is initially uncharged.)
2. L
{∫ t
0f (τ)dτ
}=
F(s)s
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Laplace Transforms and Integral Equations
logo1
Transforms and New Formulas An Example Double Check
The Laplace Transform of an Integral1. Definite integrals of the form
∫ t
0f (τ)dτ arise in circuit
theory: The charge of a capacitor is the integral of thecurrent over time.(We assume the capacitor is initially uncharged.)
2. L
{∫ t
0f (τ)dτ
}=
F(s)s
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Laplace Transforms and Integral Equations
logo1
Transforms and New Formulas An Example Double Check
Solve the Initial Value Problem
f ′(t)+ f (t)−2∫ t
0f (z) dz = t, f (0) = 0
f ′(t)+ f (t)−2∫ t
0f (z) dz = t, f (0) = 0
sF +F−2Fs
=1s2
F(
s+1− 2s
)=
1s2
Fs2 + s−2
s=
1s2
F =s
s2(s−1)(s+2)
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Laplace Transforms and Integral Equations
logo1
Transforms and New Formulas An Example Double Check
Solve the Initial Value Problem
f ′(t)+ f (t)−2∫ t
0f (z) dz = t, f (0) = 0
f ′(t)+ f (t)−2∫ t
0f (z) dz = t, f (0) = 0
sF +F−2Fs
=1s2
F(
s+1− 2s
)=
1s2
Fs2 + s−2
s=
1s2
F =s
s2(s−1)(s+2)
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Laplace Transforms and Integral Equations
logo1
Transforms and New Formulas An Example Double Check
Solve the Initial Value Problem
f ′(t)+ f (t)−2∫ t
0f (z) dz = t, f (0) = 0
f ′(t)+ f (t)−2∫ t
0f (z) dz = t, f (0) = 0
sF +F−2Fs
=1s2
F(
s+1− 2s
)=
1s2
Fs2 + s−2
s=
1s2
F =s
s2(s−1)(s+2)
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Laplace Transforms and Integral Equations
logo1
Transforms and New Formulas An Example Double Check
Solve the Initial Value Problem
f ′(t)+ f (t)−2∫ t
0f (z) dz = t, f (0) = 0
f ′(t)+ f (t)−2∫ t
0f (z) dz = t, f (0) = 0
sF +F−2Fs
=1s2
F(
s+1− 2s
)=
1s2
Fs2 + s−2
s=
1s2
F =s
s2(s−1)(s+2)
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Laplace Transforms and Integral Equations
logo1
Transforms and New Formulas An Example Double Check
Solve the Initial Value Problem
f ′(t)+ f (t)−2∫ t
0f (z) dz = t, f (0) = 0
f ′(t)+ f (t)−2∫ t
0f (z) dz = t, f (0) = 0
sF +F−2Fs
=1s2
F(
s+1− 2s
)=
1s2
Fs2 + s−2
s=
1s2
F =s
s2(s−1)(s+2)
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Laplace Transforms and Integral Equations
logo1
Transforms and New Formulas An Example Double Check
Solve the Initial Value Problem
f ′(t)+ f (t)−2∫ t
0f (z) dz = t, f (0) = 0
f ′(t)+ f (t)−2∫ t
0f (z) dz = t, f (0) = 0
sF +F−2Fs
=1s2
F(
s+1− 2s
)=
1s2
Fs2 + s−2
s=
1s2
F =s
s2(s−1)(s+2)Bernd Schroder Louisiana Tech University, College of Engineering and Science
Laplace Transforms and Integral Equations
logo1
Transforms and New Formulas An Example Double Check
Solve f ′(t)+ f (t)−2∫ t
0f (z) dz = t, f (0) = 0
F =s
s2(s−1)(s+2)=
As
+Bs2 +
Cs−1
+D
s+2
s = As(s−1)(s+2)+B(s−1)(s+2)+Cs2(s+2)+Ds2(s−1)s = 0 : 0 = B · (−1) ·2, B = 0
s = 1 : 1 = C ·12 ·3, C =13
s = −2 : −2 = D · (−2)2 · (−3), D =16
s = −1 : −1=A·(−1)·(−2)·1+0+13·(−1)2·1+
16·(−1)2·(−2), A = −1
2
F = −12
1s
+0 · 1s2 +
13
1s−1
+16
1s+2
f (t) = −12
+13
et +16
e−2t
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Laplace Transforms and Integral Equations
logo1
Transforms and New Formulas An Example Double Check
Solve f ′(t)+ f (t)−2∫ t
0f (z) dz = t, f (0) = 0
F =s
s2(s−1)(s+2)=
As
+Bs2 +
Cs−1
+D
s+2
s = As(s−1)(s+2)+B(s−1)(s+2)+Cs2(s+2)+Ds2(s−1)s = 0 : 0 = B · (−1) ·2, B = 0
s = 1 : 1 = C ·12 ·3, C =13
s = −2 : −2 = D · (−2)2 · (−3), D =16
s = −1 : −1=A·(−1)·(−2)·1+0+13·(−1)2·1+
16·(−1)2·(−2), A = −1
2
F = −12
1s
+0 · 1s2 +
13
1s−1
+16
1s+2
f (t) = −12
+13
et +16
e−2t
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Laplace Transforms and Integral Equations
logo1
Transforms and New Formulas An Example Double Check
Solve f ′(t)+ f (t)−2∫ t
0f (z) dz = t, f (0) = 0
F =s
s2(s−1)(s+2)=
As
+Bs2 +
Cs−1
+D
s+2
s = As(s−1)(s+2)+B(s−1)(s+2)+Cs2(s+2)+Ds2(s−1)
s = 0 : 0 = B · (−1) ·2, B = 0
s = 1 : 1 = C ·12 ·3, C =13
s = −2 : −2 = D · (−2)2 · (−3), D =16
s = −1 : −1=A·(−1)·(−2)·1+0+13·(−1)2·1+
16·(−1)2·(−2), A = −1
2
F = −12
1s
+0 · 1s2 +
13
1s−1
+16
1s+2
f (t) = −12
+13
et +16
e−2t
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Laplace Transforms and Integral Equations
logo1
Transforms and New Formulas An Example Double Check
Solve f ′(t)+ f (t)−2∫ t
0f (z) dz = t, f (0) = 0
F =s
s2(s−1)(s+2)=
As
+Bs2 +
Cs−1
+D
s+2
s = As(s−1)(s+2)+B(s−1)(s+2)+Cs2(s+2)+Ds2(s−1)s = 0 : 0 = B · (−1) ·2
, B = 0
s = 1 : 1 = C ·12 ·3, C =13
s = −2 : −2 = D · (−2)2 · (−3), D =16
s = −1 : −1=A·(−1)·(−2)·1+0+13·(−1)2·1+
16·(−1)2·(−2), A = −1
2
F = −12
1s
+0 · 1s2 +
13
1s−1
+16
1s+2
f (t) = −12
+13
et +16
e−2t
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Laplace Transforms and Integral Equations
logo1
Transforms and New Formulas An Example Double Check
Solve f ′(t)+ f (t)−2∫ t
0f (z) dz = t, f (0) = 0
F =s
s2(s−1)(s+2)=
As
+Bs2 +
Cs−1
+D
s+2
s = As(s−1)(s+2)+B(s−1)(s+2)+Cs2(s+2)+Ds2(s−1)s = 0 : 0 = B · (−1) ·2, B = 0
s = 1 : 1 = C ·12 ·3, C =13
s = −2 : −2 = D · (−2)2 · (−3), D =16
s = −1 : −1=A·(−1)·(−2)·1+0+13·(−1)2·1+
16·(−1)2·(−2), A = −1
2
F = −12
1s
+0 · 1s2 +
13
1s−1
+16
1s+2
f (t) = −12
+13
et +16
e−2t
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Laplace Transforms and Integral Equations
logo1
Transforms and New Formulas An Example Double Check
Solve f ′(t)+ f (t)−2∫ t
0f (z) dz = t, f (0) = 0
F =s
s2(s−1)(s+2)=
As
+Bs2 +
Cs−1
+D
s+2
s = As(s−1)(s+2)+B(s−1)(s+2)+Cs2(s+2)+Ds2(s−1)s = 0 : 0 = B · (−1) ·2, B = 0
s = 1 : 1 = C ·12 ·3
, C =13
s = −2 : −2 = D · (−2)2 · (−3), D =16
s = −1 : −1=A·(−1)·(−2)·1+0+13·(−1)2·1+
16·(−1)2·(−2), A = −1
2
F = −12
1s
+0 · 1s2 +
13
1s−1
+16
1s+2
f (t) = −12
+13
et +16
e−2t
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Laplace Transforms and Integral Equations
logo1
Transforms and New Formulas An Example Double Check
Solve f ′(t)+ f (t)−2∫ t
0f (z) dz = t, f (0) = 0
F =s
s2(s−1)(s+2)=
As
+Bs2 +
Cs−1
+D
s+2
s = As(s−1)(s+2)+B(s−1)(s+2)+Cs2(s+2)+Ds2(s−1)s = 0 : 0 = B · (−1) ·2, B = 0
s = 1 : 1 = C ·12 ·3, C =13
s = −2 : −2 = D · (−2)2 · (−3), D =16
s = −1 : −1=A·(−1)·(−2)·1+0+13·(−1)2·1+
16·(−1)2·(−2), A = −1
2
F = −12
1s
+0 · 1s2 +
13
1s−1
+16
1s+2
f (t) = −12
+13
et +16
e−2t
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Laplace Transforms and Integral Equations
logo1
Transforms and New Formulas An Example Double Check
Solve f ′(t)+ f (t)−2∫ t
0f (z) dz = t, f (0) = 0
F =s
s2(s−1)(s+2)=
As
+Bs2 +
Cs−1
+D
s+2
s = As(s−1)(s+2)+B(s−1)(s+2)+Cs2(s+2)+Ds2(s−1)s = 0 : 0 = B · (−1) ·2, B = 0
s = 1 : 1 = C ·12 ·3, C =13
s = −2 : −2 = D · (−2)2 · (−3)
, D =16
s = −1 : −1=A·(−1)·(−2)·1+0+13·(−1)2·1+
16·(−1)2·(−2), A = −1
2
F = −12
1s
+0 · 1s2 +
13
1s−1
+16
1s+2
f (t) = −12
+13
et +16
e−2t
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Laplace Transforms and Integral Equations
logo1
Transforms and New Formulas An Example Double Check
Solve f ′(t)+ f (t)−2∫ t
0f (z) dz = t, f (0) = 0
F =s
s2(s−1)(s+2)=
As
+Bs2 +
Cs−1
+D
s+2
s = As(s−1)(s+2)+B(s−1)(s+2)+Cs2(s+2)+Ds2(s−1)s = 0 : 0 = B · (−1) ·2, B = 0
s = 1 : 1 = C ·12 ·3, C =13
s = −2 : −2 = D · (−2)2 · (−3), D =16
s = −1 : −1=A·(−1)·(−2)·1+0+13·(−1)2·1+
16·(−1)2·(−2), A = −1
2
F = −12
1s
+0 · 1s2 +
13
1s−1
+16
1s+2
f (t) = −12
+13
et +16
e−2t
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Laplace Transforms and Integral Equations
logo1
Transforms and New Formulas An Example Double Check
Solve f ′(t)+ f (t)−2∫ t
0f (z) dz = t, f (0) = 0
F =s
s2(s−1)(s+2)=
As
+Bs2 +
Cs−1
+D
s+2
s = As(s−1)(s+2)+B(s−1)(s+2)+Cs2(s+2)+Ds2(s−1)s = 0 : 0 = B · (−1) ·2, B = 0
s = 1 : 1 = C ·12 ·3, C =13
s = −2 : −2 = D · (−2)2 · (−3), D =16
s = −1 : −1=A·(−1)·(−2)·1+0+13·(−1)2·1+
16·(−1)2·(−2)
, A = −12
F = −12
1s
+0 · 1s2 +
13
1s−1
+16
1s+2
f (t) = −12
+13
et +16
e−2t
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Laplace Transforms and Integral Equations
logo1
Transforms and New Formulas An Example Double Check
Solve f ′(t)+ f (t)−2∫ t
0f (z) dz = t, f (0) = 0
F =s
s2(s−1)(s+2)=
As
+Bs2 +
Cs−1
+D
s+2
s = As(s−1)(s+2)+B(s−1)(s+2)+Cs2(s+2)+Ds2(s−1)s = 0 : 0 = B · (−1) ·2, B = 0
s = 1 : 1 = C ·12 ·3, C =13
s = −2 : −2 = D · (−2)2 · (−3), D =16
s = −1 : −1=A·(−1)·(−2)·1+0+13·(−1)2·1+
16·(−1)2·(−2), A = −1
2
F = −12
1s
+0 · 1s2 +
13
1s−1
+16
1s+2
f (t) = −12
+13
et +16
e−2t
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Laplace Transforms and Integral Equations
logo1
Transforms and New Formulas An Example Double Check
Solve f ′(t)+ f (t)−2∫ t
0f (z) dz = t, f (0) = 0
F =s
s2(s−1)(s+2)=
As
+Bs2 +
Cs−1
+D
s+2
s = As(s−1)(s+2)+B(s−1)(s+2)+Cs2(s+2)+Ds2(s−1)s = 0 : 0 = B · (−1) ·2, B = 0
s = 1 : 1 = C ·12 ·3, C =13
s = −2 : −2 = D · (−2)2 · (−3), D =16
s = −1 : −1=A·(−1)·(−2)·1+0+13·(−1)2·1+
16·(−1)2·(−2), A = −1
2
F = −12
1s
+0 · 1s2 +
13
1s−1
+16
1s+2
f (t) = −12
+13
et +16
e−2t
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Laplace Transforms and Integral Equations
logo1
Transforms and New Formulas An Example Double Check
Solve f ′(t)+ f (t)−2∫ t
0f (z) dz = t, f (0) = 0
F =s
s2(s−1)(s+2)=
As
+Bs2 +
Cs−1
+D
s+2
s = As(s−1)(s+2)+B(s−1)(s+2)+Cs2(s+2)+Ds2(s−1)s = 0 : 0 = B · (−1) ·2, B = 0
s = 1 : 1 = C ·12 ·3, C =13
s = −2 : −2 = D · (−2)2 · (−3), D =16
s = −1 : −1=A·(−1)·(−2)·1+0+13·(−1)2·1+
16·(−1)2·(−2), A = −1
2
F = −12
1s
+0 · 1s2 +
13
1s−1
+16
1s+2
f (t) = −12
+13
et +16
e−2t
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Laplace Transforms and Integral Equations
logo1
Transforms and New Formulas An Example Double Check
Does f (t) = −12
+13
et +16
e−2t Really Solve the Initial Value Problem
f ′(t)+ f (t)−2∫ t
0f (z) dz = t, f (0) = 0?
Initial value: Look at f ![−1
2+
13
et +16
e−2t]′
+[−1
2+
13
et +16
e−2t]−2
∫ t
0−1
2+
13
ez +16
e−2z dz
=13
et − 13
e−2t − 12
+13
et +16
e−2t −2[−1
2z+
13
ez− 112
e−2z]t
0
=13
et − 13
e−2t − 12
+13
et +16
e−2t + t−0− 23
et +23
+16
e−2t − 16
= et(
13
+13− 2
3
)+ e−2t
(−1
3+
16
+16
)+ t +
(−1
2+
23− 1
6
)= t
√
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Laplace Transforms and Integral Equations
logo1
Transforms and New Formulas An Example Double Check
Does f (t) = −12
+13
et +16
e−2t Really Solve the Initial Value Problem
f ′(t)+ f (t)−2∫ t
0f (z) dz = t, f (0) = 0?
Initial value:
Look at f ![−1
2+
13
et +16
e−2t]′
+[−1
2+
13
et +16
e−2t]−2
∫ t
0−1
2+
13
ez +16
e−2z dz
=13
et − 13
e−2t − 12
+13
et +16
e−2t −2[−1
2z+
13
ez− 112
e−2z]t
0
=13
et − 13
e−2t − 12
+13
et +16
e−2t + t−0− 23
et +23
+16
e−2t − 16
= et(
13
+13− 2
3
)+ e−2t
(−1
3+
16
+16
)+ t +
(−1
2+
23− 1
6
)= t
√
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Laplace Transforms and Integral Equations
logo1
Transforms and New Formulas An Example Double Check
Does f (t) = −12
+13
et +16
e−2t Really Solve the Initial Value Problem
f ′(t)+ f (t)−2∫ t
0f (z) dz = t, f (0) = 0?
Initial value: Look at f !
[−1
2+
13
et +16
e−2t]′
+[−1
2+
13
et +16
e−2t]−2
∫ t
0−1
2+
13
ez +16
e−2z dz
=13
et − 13
e−2t − 12
+13
et +16
e−2t −2[−1
2z+
13
ez− 112
e−2z]t
0
=13
et − 13
e−2t − 12
+13
et +16
e−2t + t−0− 23
et +23
+16
e−2t − 16
= et(
13
+13− 2
3
)+ e−2t
(−1
3+
16
+16
)+ t +
(−1
2+
23− 1
6
)= t
√
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Laplace Transforms and Integral Equations
logo1
Transforms and New Formulas An Example Double Check
Does f (t) = −12
+13
et +16
e−2t Really Solve the Initial Value Problem
f ′(t)+ f (t)−2∫ t
0f (z) dz = t, f (0) = 0?
Initial value: Look at f ![−1
2+
13
et +16
e−2t]′
+[−1
2+
13
et +16
e−2t]−2
∫ t
0−1
2+
13
ez +16
e−2z dz
=13
et − 13
e−2t − 12
+13
et +16
e−2t −2[−1
2z+
13
ez− 112
e−2z]t
0
=13
et − 13
e−2t − 12
+13
et +16
e−2t + t−0− 23
et +23
+16
e−2t − 16
= et(
13
+13− 2
3
)+ e−2t
(−1
3+
16
+16
)+ t +
(−1
2+
23− 1
6
)= t
√
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Laplace Transforms and Integral Equations
logo1
Transforms and New Formulas An Example Double Check
Does f (t) = −12
+13
et +16
e−2t Really Solve the Initial Value Problem
f ′(t)+ f (t)−2∫ t
0f (z) dz = t, f (0) = 0?
Initial value: Look at f ![−1
2+
13
et +16
e−2t]′
+[−1
2+
13
et +16
e−2t]−2
∫ t
0−1
2+
13
ez +16
e−2z dz
=13
et − 13
e−2t − 12
+13
et +16
e−2t −2[−1
2z+
13
ez− 112
e−2z]t
0
=13
et − 13
e−2t − 12
+13
et +16
e−2t + t−0− 23
et +23
+16
e−2t − 16
= et(
13
+13− 2
3
)+ e−2t
(−1
3+
16
+16
)+ t +
(−1
2+
23− 1
6
)= t
√
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Laplace Transforms and Integral Equations
logo1
Transforms and New Formulas An Example Double Check
Does f (t) = −12
+13
et +16
e−2t Really Solve the Initial Value Problem
f ′(t)+ f (t)−2∫ t
0f (z) dz = t, f (0) = 0?
Initial value: Look at f ![−1
2+
13
et +16
e−2t]′
+[−1
2+
13
et +16
e−2t]−2
∫ t
0−1
2+
13
ez +16
e−2z dz
=13
et − 13
e−2t − 12
+13
et +16
e−2t −2[−1
2z+
13
ez− 112
e−2z]t
0
=13
et − 13
e−2t − 12
+13
et +16
e−2t + t−0− 23
et +23
+16
e−2t − 16
= et(
13
+13− 2
3
)+ e−2t
(−1
3+
16
+16
)+ t +
(−1
2+
23− 1
6
)= t
√
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Laplace Transforms and Integral Equations
logo1
Transforms and New Formulas An Example Double Check
Does f (t) = −12
+13
et +16
e−2t Really Solve the Initial Value Problem
f ′(t)+ f (t)−2∫ t
0f (z) dz = t, f (0) = 0?
Initial value: Look at f ![−1
2+
13
et +16
e−2t]′
+[−1
2+
13
et +16
e−2t]−2
∫ t
0−1
2+
13
ez +16
e−2z dz
=13
et − 13
e−2t − 12
+13
et +16
e−2t −2[−1
2z+
13
ez− 112
e−2z]t
0
=13
et − 13
e−2t − 12
+13
et +16
e−2t + t−0− 23
et +23
+16
e−2t − 16
= et(
13
+13− 2
3
)
+ e−2t(−1
3+
16
+16
)+ t +
(−1
2+
23− 1
6
)= t
√
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Laplace Transforms and Integral Equations
logo1
Transforms and New Formulas An Example Double Check
Does f (t) = −12
+13
et +16
e−2t Really Solve the Initial Value Problem
f ′(t)+ f (t)−2∫ t
0f (z) dz = t, f (0) = 0?
Initial value: Look at f ![−1
2+
13
et +16
e−2t]′
+[−1
2+
13
et +16
e−2t]−2
∫ t
0−1
2+
13
ez +16
e−2z dz
=13
et − 13
e−2t − 12
+13
et +16
e−2t −2[−1
2z+
13
ez− 112
e−2z]t
0
=13
et − 13
e−2t − 12
+13
et +16
e−2t + t−0− 23
et +23
+16
e−2t − 16
= et(
13
+13− 2
3
)+ e−2t
(−1
3+
16
+16
)
+ t +(−1
2+
23− 1
6
)= t
√
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Laplace Transforms and Integral Equations
logo1
Transforms and New Formulas An Example Double Check
Does f (t) = −12
+13
et +16
e−2t Really Solve the Initial Value Problem
f ′(t)+ f (t)−2∫ t
0f (z) dz = t, f (0) = 0?
Initial value: Look at f ![−1
2+
13
et +16
e−2t]′
+[−1
2+
13
et +16
e−2t]−2
∫ t
0−1
2+
13
ez +16
e−2z dz
=13
et − 13
e−2t − 12
+13
et +16
e−2t −2[−1
2z+
13
ez− 112
e−2z]t
0
=13
et − 13
e−2t − 12
+13
et +16
e−2t + t−0− 23
et +23
+16
e−2t − 16
= et(
13
+13− 2
3
)+ e−2t
(−1
3+
16
+16
)+ t
+(−1
2+
23− 1
6
)= t
√
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Laplace Transforms and Integral Equations
logo1
Transforms and New Formulas An Example Double Check
Does f (t) = −12
+13
et +16
e−2t Really Solve the Initial Value Problem
f ′(t)+ f (t)−2∫ t
0f (z) dz = t, f (0) = 0?
Initial value: Look at f ![−1
2+
13
et +16
e−2t]′
+[−1
2+
13
et +16
e−2t]−2
∫ t
0−1
2+
13
ez +16
e−2z dz
=13
et − 13
e−2t − 12
+13
et +16
e−2t −2[−1
2z+
13
ez− 112
e−2z]t
0
=13
et − 13
e−2t − 12
+13
et +16
e−2t + t−0− 23
et +23
+16
e−2t − 16
= et(
13
+13− 2
3
)+ e−2t
(−1
3+
16
+16
)+ t +
(−1
2+
23− 1
6
)
= t√
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Laplace Transforms and Integral Equations
logo1
Transforms and New Formulas An Example Double Check
Does f (t) = −12
+13
et +16
e−2t Really Solve the Initial Value Problem
f ′(t)+ f (t)−2∫ t
0f (z) dz = t, f (0) = 0?
Initial value: Look at f ![−1
2+
13
et +16
e−2t]′
+[−1
2+
13
et +16
e−2t]−2
∫ t
0−1
2+
13
ez +16
e−2z dz
=13
et − 13
e−2t − 12
+13
et +16
e−2t −2[−1
2z+
13
ez− 112
e−2z]t
0
=13
et − 13
e−2t − 12
+13
et +16
e−2t + t−0− 23
et +23
+16
e−2t − 16
= et(
13
+13− 2
3
)+ e−2t
(−1
3+
16
+16
)+ t +
(−1
2+
23− 1
6
)= t
√
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Laplace Transforms and Integral Equations
logo1
Transforms and New Formulas An Example Double Check
Does f (t) = −12
+13
et +16
e−2t Really Solve the Initial Value Problem
f ′(t)+ f (t)−2∫ t
0f (z) dz = t, f (0) = 0?
Initial value: Look at f ![−1
2+
13
et +16
e−2t]′
+[−1
2+
13
et +16
e−2t]−2
∫ t
0−1
2+
13
ez +16
e−2z dz
=13
et − 13
e−2t − 12
+13
et +16
e−2t −2[−1
2z+
13
ez− 112
e−2z]t
0
=13
et − 13
e−2t − 12
+13
et +16
e−2t + t−0− 23
et +23
+16
e−2t − 16
= et(
13
+13− 2
3
)+ e−2t
(−1
3+
16
+16
)+ t +
(−1
2+
23− 1
6
)= t
√
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Laplace Transforms and Integral Equations
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