logo1

Transforms and New Formulas A Model The Initial Value Problem Interpretation Double Check

The Laplace Transform of The DiracDelta Function

Bernd Schroder

Bernd Schroder Louisiana Tech University, College of Engineering and Science

The Laplace Transform of The Dirac Delta Function

logo1

Transforms and New Formulas A Model The Initial Value Problem Interpretation 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

The Laplace Transform of The Dirac Delta Function

logo1

Transforms and New Formulas A Model The Initial Value Problem Interpretation 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

The Laplace Transform of The Dirac Delta Function

logo1

Transforms and New Formulas A Model The Initial Value Problem Interpretation 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

The Laplace Transform of The Dirac Delta Function

logo1

Transforms and New Formulas A Model The Initial Value Problem Interpretation 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

The Laplace Transform of The Dirac Delta Function

logo1

Transforms and New Formulas A Model The Initial Value Problem Interpretation 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

The Laplace Transform of The Dirac Delta Function

logo1

Transforms and New Formulas A Model The Initial Value Problem Interpretation 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

The Laplace Transform of The Dirac Delta Function

logo1

Transforms and New Formulas A Model The Initial Value Problem Interpretation 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

The Laplace Transform of The Dirac Delta Function

logo1

Transforms and New Formulas A Model The Initial Value Problem Interpretation 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

The Laplace Transform of The Dirac Delta Function

logo1

Transforms and New Formulas A Model The Initial Value Problem Interpretation 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

The Laplace Transform of The Dirac Delta Function

logo1

Transforms and New Formulas A Model The Initial Value Problem Interpretation 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

The Laplace Transform of The Dirac Delta Function

logo1

Transforms and New Formulas A Model The Initial Value Problem Interpretation 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

The Laplace Transform of The Dirac Delta Function

logo1

Transforms and New Formulas A Model The Initial Value Problem Interpretation Double Check

What is the Delta Function?

1. δ (x) = 0 for all x 6= 0.

2. Sifting property:∫

∞

−∞

f (x)δ (x−a) dx = f (a) ????

Bernd Schroder Louisiana Tech University, College of Engineering and Science

The Laplace Transform of The Dirac Delta Function

logo1

Transforms and New Formulas A Model The Initial Value Problem Interpretation Double Check

What is the Delta Function?1. δ (x) = 0 for all x 6= 0.

2. Sifting property:∫

∞

−∞

f (x)δ (x−a) dx = f (a) ????

Bernd Schroder Louisiana Tech University, College of Engineering and Science

The Laplace Transform of The Dirac Delta Function

logo1

Transforms and New Formulas A Model The Initial Value Problem Interpretation Double Check

What is the Delta Function?1. δ (x) = 0 for all x 6= 0.

2. Sifting property:∫

∞

−∞

f (x)δ (x−a) dx = f (a)

????

Bernd Schroder Louisiana Tech University, College of Engineering and Science

The Laplace Transform of The Dirac Delta Function

logo1

Transforms and New Formulas A Model The Initial Value Problem Interpretation Double Check

What is the Delta Function?1. δ (x) = 0 for all x 6= 0.

2. Sifting property:∫

∞

−∞

f (x)δ (x−a) dx = f (a) ????

Bernd Schroder Louisiana Tech University, College of Engineering and Science

The Laplace Transform of The Dirac Delta Function

logo1

Transforms and New Formulas A Model The Initial Value Problem Interpretation Double Check

What is the Delta Function?1. δ (x) = 0 for all x 6= 0.

2. Sifting property:∫

∞

−∞

f (x)δ (x−a) dx = f (a)

3. The delta function is used to model “instantaneous” energytransfers.

4. L

δ (t−a)

= e−as

Bernd Schroder Louisiana Tech University, College of Engineering and Science

The Laplace Transform of The Dirac Delta Function

logo1

Transforms and New Formulas A Model The Initial Value Problem Interpretation Double Check

What is the Delta Function?1. δ (x) = 0 for all x 6= 0.

2. Sifting property:∫

∞

−∞

f (x)δ (x−a) dx = f (a)

3. The delta function is used to model “instantaneous” energytransfers.

4. L

δ (t−a)

= e−as

Bernd Schroder Louisiana Tech University, College of Engineering and Science

The Laplace Transform of The Dirac Delta Function

logo1

Transforms and New Formulas A Model The Initial Value Problem Interpretation Double Check

What is the Delta Function?1. δ (x) = 0 for all x 6= 0.

2. Sifting property:∫

∞

−∞

f (x)δ (x−a) dx = f (a)

3. The delta function is used to model “instantaneous” energytransfers.

4. L

δ (t−a)

= e−as

Bernd Schroder Louisiana Tech University, College of Engineering and Science

The Laplace Transform of The Dirac Delta Function

logo1

Transforms and New Formulas A Model The Initial Value Problem Interpretation Double Check

A Possible Application

(Dimensions are fictitious.)

In an LRC circuit with L = 1H, R = 8Ω and C =115

F, thecapacitor initially carries a charge of 1C and no currents areflowing. There is no external voltage source. At time t = 2s, apower surge instantaneously applies an impulse of 4δ (t−2)into the system.Describe the charge of the capacitor over time.

Lq′′+Rq′+qC

= E(t)

q′′+8q′+15q = 4δ (t−2)

Bernd Schroder Louisiana Tech University, College of Engineering and Science

The Laplace Transform of The Dirac Delta Function

logo1

Transforms and New Formulas A Model The Initial Value Problem Interpretation Double Check

A Possible Application(Dimensions are fictitious.)

In an LRC circuit with L = 1H, R = 8Ω and C =115

F, thecapacitor initially carries a charge of 1C and no currents areflowing. There is no external voltage source. At time t = 2s, apower surge instantaneously applies an impulse of 4δ (t−2)into the system.Describe the charge of the capacitor over time.

Lq′′+Rq′+qC

= E(t)

q′′+8q′+15q = 4δ (t−2)

Bernd Schroder Louisiana Tech University, College of Engineering and Science

The Laplace Transform of The Dirac Delta Function

logo1

Transforms and New Formulas A Model The Initial Value Problem Interpretation Double Check

A Possible Application(Dimensions are fictitious.)

In an LRC circuit with L = 1H, R = 8Ω and C =115

F, thecapacitor initially carries a charge of 1C and no currents areflowing. There is no external voltage source. At time t = 2s, apower surge instantaneously applies an impulse of 4δ (t−2)into the system.

Describe the charge of the capacitor over time.

Lq′′+Rq′+qC

= E(t)

q′′+8q′+15q = 4δ (t−2)

Bernd Schroder Louisiana Tech University, College of Engineering and Science

The Laplace Transform of The Dirac Delta Function

logo1

Transforms and New Formulas A Model The Initial Value Problem Interpretation Double Check

A Possible Application(Dimensions are fictitious.)

In an LRC circuit with L = 1H, R = 8Ω and C =115

F, thecapacitor initially carries a charge of 1C and no currents areflowing. There is no external voltage source. At time t = 2s, apower surge instantaneously applies an impulse of 4δ (t−2)into the system.Describe the charge of the capacitor over time.

Lq′′+Rq′+qC

= E(t)

q′′+8q′+15q = 4δ (t−2)

Bernd Schroder Louisiana Tech University, College of Engineering and Science

The Laplace Transform of The Dirac Delta Function

logo1

Transforms and New Formulas A Model The Initial Value Problem Interpretation Double Check

A Possible Application(Dimensions are fictitious.)

In an LRC circuit with L = 1H, R = 8Ω and C =115

F, thecapacitor initially carries a charge of 1C and no currents areflowing. There is no external voltage source. At time t = 2s, apower surge instantaneously applies an impulse of 4δ (t−2)into the system.Describe the charge of the capacitor over time.

Lq′′+Rq′+qC

= E(t)

q′′+8q′+15q = 4δ (t−2)

Bernd Schroder Louisiana Tech University, College of Engineering and Science

The Laplace Transform of The Dirac Delta Function

logo1

Transforms and New Formulas A Model The Initial Value Problem Interpretation Double Check

A Possible Application(Dimensions are fictitious.)

In an LRC circuit with L = 1H, R = 8Ω and C =115

F, thecapacitor initially carries a charge of 1C and no currents areflowing. There is no external voltage source. At time t = 2s, apower surge instantaneously applies an impulse of 4δ (t−2)into the system.Describe the charge of the capacitor over time.

Lq′′+Rq′+qC

= E(t)

q′′

+8q′+15q = 4δ (t−2)

Bernd Schroder Louisiana Tech University, College of Engineering and Science

The Laplace Transform of The Dirac Delta Function

logo1

Transforms and New Formulas A Model The Initial Value Problem Interpretation Double Check

A Possible Application(Dimensions are fictitious.)

In an LRC circuit with L = 1H, R = 8Ω and C =115

F, thecapacitor initially carries a charge of 1C and no currents areflowing. There is no external voltage source. At time t = 2s, apower surge instantaneously applies an impulse of 4δ (t−2)into the system.Describe the charge of the capacitor over time.

Lq′′+Rq′+qC

= E(t)

q′′+8q′

+15q = 4δ (t−2)

Bernd Schroder Louisiana Tech University, College of Engineering and Science

The Laplace Transform of The Dirac Delta Function

logo1

Transforms and New Formulas A Model The Initial Value Problem Interpretation Double Check

A Possible Application(Dimensions are fictitious.)

In an LRC circuit with L = 1H, R = 8Ω and C =115

F, thecapacitor initially carries a charge of 1C and no currents areflowing. There is no external voltage source. At time t = 2s, apower surge instantaneously applies an impulse of 4δ (t−2)into the system.Describe the charge of the capacitor over time.

Lq′′+Rq′+qC

= E(t)

q′′+8q′+15q

= 4δ (t−2)

Bernd Schroder Louisiana Tech University, College of Engineering and Science

The Laplace Transform of The Dirac Delta Function

logo1

Transforms and New Formulas A Model The Initial Value Problem Interpretation Double Check

A Possible Application(Dimensions are fictitious.)

In an LRC circuit with L = 1H, R = 8Ω and C =115

F, thecapacitor initially carries a charge of 1C and no currents areflowing. There is no external voltage source. At time t = 2s, apower surge instantaneously applies an impulse of 4δ (t−2)into the system.Describe the charge of the capacitor over time.

Lq′′+Rq′+qC

= E(t)

q′′+8q′+15q = 4δ (t−2)

Bernd Schroder Louisiana Tech University, College of Engineering and Science

The Laplace Transform of The Dirac Delta Function

logo1

Transforms and New Formulas A Model The Initial Value Problem Interpretation Double Check

Solve the Initial Value Problemq′′+8q′+15q = 4δ (t−2), q(0) = 1, q′(0) = 0

q′′+8q′+15q = 4δ (t−2)s2Q− s+8sQ−8+15Q = 4e−2s(

s2 +8s+15)

Q = s+8+4e−2s

(keep the exponential separate)

Q =s+8

s2 +8s+15+ e−2s 4

s2 +8s+15

Q =s+8

(s+3)(s+5)+ e−2s 4

(s+3)(s+5)

Bernd Schroder Louisiana Tech University, College of Engineering and Science

The Laplace Transform of The Dirac Delta Function

logo1

Transforms and New Formulas A Model The Initial Value Problem Interpretation Double Check

Solve the Initial Value Problemq′′+8q′+15q = 4δ (t−2), q(0) = 1, q′(0) = 0

q′′+8q′+15q = 4δ (t−2)

s2Q− s+8sQ−8+15Q = 4e−2s(s2 +8s+15

)Q = s+8+4e−2s

(keep the exponential separate)

Q =s+8

s2 +8s+15+ e−2s 4

s2 +8s+15

Q =s+8

(s+3)(s+5)+ e−2s 4

(s+3)(s+5)

Bernd Schroder Louisiana Tech University, College of Engineering and Science

The Laplace Transform of The Dirac Delta Function

logo1

Transforms and New Formulas A Model The Initial Value Problem Interpretation Double Check

Solve the Initial Value Problemq′′+8q′+15q = 4δ (t−2), q(0) = 1, q′(0) = 0

q′′+8q′+15q = 4δ (t−2)s2Q− s

+8sQ−8+15Q = 4e−2s(s2 +8s+15

)Q = s+8+4e−2s

(keep the exponential separate)

Q =s+8

s2 +8s+15+ e−2s 4

s2 +8s+15

Q =s+8

(s+3)(s+5)+ e−2s 4

(s+3)(s+5)

Bernd Schroder Louisiana Tech University, College of Engineering and Science

The Laplace Transform of The Dirac Delta Function

logo1

Transforms and New Formulas A Model The Initial Value Problem Interpretation Double Check

Solve the Initial Value Problemq′′+8q′+15q = 4δ (t−2), q(0) = 1, q′(0) = 0

q′′+8q′+15q = 4δ (t−2)s2Q− s+8sQ−8

+15Q = 4e−2s(s2 +8s+15

)Q = s+8+4e−2s

(keep the exponential separate)

Q =s+8

s2 +8s+15+ e−2s 4

s2 +8s+15

Q =s+8

(s+3)(s+5)+ e−2s 4

(s+3)(s+5)

Bernd Schroder Louisiana Tech University, College of Engineering and Science

The Laplace Transform of The Dirac Delta Function

logo1

Transforms and New Formulas A Model The Initial Value Problem Interpretation Double Check

Solve the Initial Value Problemq′′+8q′+15q = 4δ (t−2), q(0) = 1, q′(0) = 0

q′′+8q′+15q = 4δ (t−2)s2Q− s+8sQ−8+15Q

= 4e−2s(s2 +8s+15

)Q = s+8+4e−2s

(keep the exponential separate)

Q =s+8

s2 +8s+15+ e−2s 4

s2 +8s+15

Q =s+8

(s+3)(s+5)+ e−2s 4

(s+3)(s+5)

Bernd Schroder Louisiana Tech University, College of Engineering and Science

The Laplace Transform of The Dirac Delta Function

logo1

Transforms and New Formulas A Model The Initial Value Problem Interpretation Double Check

Solve the Initial Value Problemq′′+8q′+15q = 4δ (t−2), q(0) = 1, q′(0) = 0

q′′+8q′+15q = 4δ (t−2)s2Q− s+8sQ−8+15Q = 4e−2s

(s2 +8s+15

)Q = s+8+4e−2s

(keep the exponential separate)

Q =s+8

s2 +8s+15+ e−2s 4

s2 +8s+15

Q =s+8

(s+3)(s+5)+ e−2s 4

(s+3)(s+5)

Bernd Schroder Louisiana Tech University, College of Engineering and Science

The Laplace Transform of The Dirac Delta Function

logo1

Transforms and New Formulas A Model The Initial Value Problem Interpretation Double Check

Solve the Initial Value Problemq′′+8q′+15q = 4δ (t−2), q(0) = 1, q′(0) = 0

q′′+8q′+15q = 4δ (t−2)s2Q− s+8sQ−8+15Q = 4e−2s(

s2 +8s+15)

Q = s+8+4e−2s

(keep the exponential separate)

Q =s+8

s2 +8s+15+ e−2s 4

s2 +8s+15

Q =s+8

(s+3)(s+5)+ e−2s 4

(s+3)(s+5)

Bernd Schroder Louisiana Tech University, College of Engineering and Science

The Laplace Transform of The Dirac Delta Function

logo1

Transforms and New Formulas A Model The Initial Value Problem Interpretation Double Check

Solve the Initial Value Problemq′′+8q′+15q = 4δ (t−2), q(0) = 1, q′(0) = 0

q′′+8q′+15q = 4δ (t−2)s2Q− s+8sQ−8+15Q = 4e−2s(

s2 +8s+15)

Q = s+8+4e−2s

(keep the exponential separate)

Q =s+8

s2 +8s+15+ e−2s 4

s2 +8s+15

Q =s+8

(s+3)(s+5)+ e−2s 4

(s+3)(s+5)

Bernd Schroder Louisiana Tech University, College of Engineering and Science

The Laplace Transform of The Dirac Delta Function

logo1

Transforms and New Formulas A Model The Initial Value Problem Interpretation Double Check

Solve the Initial Value Problemq′′+8q′+15q = 4δ (t−2), q(0) = 1, q′(0) = 0

q′′+8q′+15q = 4δ (t−2)s2Q− s+8sQ−8+15Q = 4e−2s(

s2 +8s+15)

Q = s+8+4e−2s

(keep the exponential separate)

Q =s+8

s2 +8s+15+ e−2s 4

s2 +8s+15

Q =s+8

(s+3)(s+5)+ e−2s 4

(s+3)(s+5)

Bernd Schroder Louisiana Tech University, College of Engineering and Science

The Laplace Transform of The Dirac Delta Function

logo1

Transforms and New Formulas A Model The Initial Value Problem Interpretation Double Check

Solve the Initial Value Problemq′′+8q′+15q = 4δ (t−2), q(0) = 1, q′(0) = 0

q′′+8q′+15q = 4δ (t−2)s2Q− s+8sQ−8+15Q = 4e−2s(

s2 +8s+15)

Q = s+8+4e−2s

(keep the exponential separate)

Q =s+8

s2 +8s+15+ e−2s 4

s2 +8s+15

Q =s+8

(s+3)(s+5)+ e−2s 4

(s+3)(s+5)

Bernd Schroder Louisiana Tech University, College of Engineering and Science

The Laplace Transform of The Dirac Delta Function

logo1

Transforms and New Formulas A Model The Initial Value Problem Interpretation Double Check

Solve the Initial Value Problemq′′+8q′+15q = 4δ (t−2), q(0) = 1, q′(0) = 0

Partial fraction decompositions.s+8

(s+3)(s+5)=

As+3

+B

s+5s+8 = A(s+5)+B(s+3)

s =−3 : 5 = 2A, A =52

s =−5 3 =−2B, B =−32

s+8(s+3)(s+5)

=52

1s+3

− 32

1s+5

Bernd Schroder Louisiana Tech University, College of Engineering and Science

The Laplace Transform of The Dirac Delta Function

logo1

Transforms and New Formulas A Model The Initial Value Problem Interpretation Double Check

Solve the Initial Value Problemq′′+8q′+15q = 4δ (t−2), q(0) = 1, q′(0) = 0

Partial fraction decompositions.

s+8(s+3)(s+5)

=A

s+3+

Bs+5

s+8 = A(s+5)+B(s+3)

s =−3 : 5 = 2A, A =52

s =−5 3 =−2B, B =−32

s+8(s+3)(s+5)

=52

1s+3

− 32

1s+5

Bernd Schroder Louisiana Tech University, College of Engineering and Science

The Laplace Transform of The Dirac Delta Function

logo1

Transforms and New Formulas A Model The Initial Value Problem Interpretation Double Check

Solve the Initial Value Problemq′′+8q′+15q = 4δ (t−2), q(0) = 1, q′(0) = 0

Partial fraction decompositions.s+8

(s+3)(s+5)=

As+3

+B

s+5

s+8 = A(s+5)+B(s+3)

s =−3 : 5 = 2A, A =52

s =−5 3 =−2B, B =−32

s+8(s+3)(s+5)

=52

1s+3

− 32

1s+5

Bernd Schroder Louisiana Tech University, College of Engineering and Science

The Laplace Transform of The Dirac Delta Function

logo1

Transforms and New Formulas A Model The Initial Value Problem Interpretation Double Check

Solve the Initial Value Problemq′′+8q′+15q = 4δ (t−2), q(0) = 1, q′(0) = 0

Partial fraction decompositions.s+8

(s+3)(s+5)=

As+3

+B

s+5s+8 = A(s+5)+B(s+3)

s =−3 : 5 = 2A, A =52

s =−5 3 =−2B, B =−32

s+8(s+3)(s+5)

=52

1s+3

− 32

1s+5

Bernd Schroder Louisiana Tech University, College of Engineering and Science

The Laplace Transform of The Dirac Delta Function

logo1

Transforms and New Formulas A Model The Initial Value Problem Interpretation Double Check

Solve the Initial Value Problemq′′+8q′+15q = 4δ (t−2), q(0) = 1, q′(0) = 0

Partial fraction decompositions.s+8

(s+3)(s+5)=

As+3

+B

s+5s+8 = A(s+5)+B(s+3)

s =−3 :

5 = 2A, A =52

s =−5 3 =−2B, B =−32

s+8(s+3)(s+5)

=52

1s+3

− 32

1s+5

Bernd Schroder Louisiana Tech University, College of Engineering and Science

The Laplace Transform of The Dirac Delta Function

logo1

Transforms and New Formulas A Model The Initial Value Problem Interpretation Double Check

Solve the Initial Value Problemq′′+8q′+15q = 4δ (t−2), q(0) = 1, q′(0) = 0

Partial fraction decompositions.s+8

(s+3)(s+5)=

As+3

+B

s+5s+8 = A(s+5)+B(s+3)

s =−3 : 5 = 2A

, A =52

s =−5 3 =−2B, B =−32

s+8(s+3)(s+5)

=52

1s+3

− 32

1s+5

Bernd Schroder Louisiana Tech University, College of Engineering and Science

The Laplace Transform of The Dirac Delta Function

logo1

Transforms and New Formulas A Model The Initial Value Problem Interpretation Double Check

Solve the Initial Value Problemq′′+8q′+15q = 4δ (t−2), q(0) = 1, q′(0) = 0

Partial fraction decompositions.s+8

(s+3)(s+5)=

As+3

+B

s+5s+8 = A(s+5)+B(s+3)

s =−3 : 5 = 2A, A =52

s =−5 3 =−2B, B =−32

s+8(s+3)(s+5)

=52

1s+3

− 32

1s+5

Bernd Schroder Louisiana Tech University, College of Engineering and Science

The Laplace Transform of The Dirac Delta Function

logo1

Transforms and New Formulas A Model The Initial Value Problem Interpretation Double Check

Solve the Initial Value Problemq′′+8q′+15q = 4δ (t−2), q(0) = 1, q′(0) = 0

Partial fraction decompositions.s+8

(s+3)(s+5)=

As+3

+B

s+5s+8 = A(s+5)+B(s+3)

s =−3 : 5 = 2A, A =52

s =−5

3 =−2B, B =−32

s+8(s+3)(s+5)

=52

1s+3

− 32

1s+5

Bernd Schroder Louisiana Tech University, College of Engineering and Science

The Laplace Transform of The Dirac Delta Function

logo1

Transforms and New Formulas A Model The Initial Value Problem Interpretation Double Check

Solve the Initial Value Problemq′′+8q′+15q = 4δ (t−2), q(0) = 1, q′(0) = 0

Partial fraction decompositions.s+8

(s+3)(s+5)=

As+3

+B

s+5s+8 = A(s+5)+B(s+3)

s =−3 : 5 = 2A, A =52

s =−5 3 =−2B

, B =−32

s+8(s+3)(s+5)

=52

1s+3

− 32

1s+5

Bernd Schroder Louisiana Tech University, College of Engineering and Science

The Laplace Transform of The Dirac Delta Function

logo1

Transforms and New Formulas A Model The Initial Value Problem Interpretation Double Check

Solve the Initial Value Problemq′′+8q′+15q = 4δ (t−2), q(0) = 1, q′(0) = 0

Partial fraction decompositions.s+8

(s+3)(s+5)=

As+3

+B

s+5s+8 = A(s+5)+B(s+3)

s =−3 : 5 = 2A, A =52

s =−5 3 =−2B, B =−32

s+8(s+3)(s+5)

=52

1s+3

− 32

1s+5

Bernd Schroder Louisiana Tech University, College of Engineering and Science

The Laplace Transform of The Dirac Delta Function

logo1

Transforms and New Formulas A Model The Initial Value Problem Interpretation Double Check

Solve the Initial Value Problemq′′+8q′+15q = 4δ (t−2), q(0) = 1, q′(0) = 0

Partial fraction decompositions.s+8

(s+3)(s+5)=

As+3

+B

s+5s+8 = A(s+5)+B(s+3)

s =−3 : 5 = 2A, A =52

s =−5 3 =−2B, B =−32

s+8(s+3)(s+5)

=52

1s+3

− 32

1s+5

Bernd Schroder Louisiana Tech University, College of Engineering and Science

The Laplace Transform of The Dirac Delta Function

logo1

Transforms and New Formulas A Model The Initial Value Problem Interpretation Double Check

Solve the Initial Value Problemq′′+8q′+15q = 4δ (t−2), q(0) = 1, q′(0) = 0

Partial fraction decompositions.4

(s+3)(s+5)=

As+3

+B

s+54 = A(s+5)+B(s+3)

s =−3 : 4 = 2A, A = 2s =−5 : 4 =−2B, B =−2

4(s+3)(s+5)

= 21

s+3−2

1s+5

Bernd Schroder Louisiana Tech University, College of Engineering and Science

The Laplace Transform of The Dirac Delta Function

logo1

Transforms and New Formulas A Model The Initial Value Problem Interpretation Double Check

Solve the Initial Value Problemq′′+8q′+15q = 4δ (t−2), q(0) = 1, q′(0) = 0

Partial fraction decompositions.

4(s+3)(s+5)

=A

s+3+

Bs+5

4 = A(s+5)+B(s+3)s =−3 : 4 = 2A, A = 2s =−5 : 4 =−2B, B =−2

4(s+3)(s+5)

= 21

s+3−2

1s+5

Bernd Schroder Louisiana Tech University, College of Engineering and Science

The Laplace Transform of The Dirac Delta Function

logo1

Transforms and New Formulas A Model The Initial Value Problem Interpretation Double Check

Solve the Initial Value Problemq′′+8q′+15q = 4δ (t−2), q(0) = 1, q′(0) = 0

Partial fraction decompositions.4

(s+3)(s+5)=

As+3

+B

s+5

4 = A(s+5)+B(s+3)s =−3 : 4 = 2A, A = 2s =−5 : 4 =−2B, B =−2

4(s+3)(s+5)

= 21

s+3−2

1s+5

Bernd Schroder Louisiana Tech University, College of Engineering and Science

The Laplace Transform of The Dirac Delta Function

logo1

Transforms and New Formulas A Model The Initial Value Problem Interpretation Double Check

Solve the Initial Value Problemq′′+8q′+15q = 4δ (t−2), q(0) = 1, q′(0) = 0

Partial fraction decompositions.4

(s+3)(s+5)=

As+3

+B

s+54 = A(s+5)+B(s+3)

s =−3 : 4 = 2A, A = 2s =−5 : 4 =−2B, B =−2

4(s+3)(s+5)

= 21

s+3−2

1s+5

Bernd Schroder Louisiana Tech University, College of Engineering and Science

The Laplace Transform of The Dirac Delta Function

logo1

Transforms and New Formulas A Model The Initial Value Problem Interpretation Double Check

Solve the Initial Value Problemq′′+8q′+15q = 4δ (t−2), q(0) = 1, q′(0) = 0

Partial fraction decompositions.4

(s+3)(s+5)=

As+3

+B

s+54 = A(s+5)+B(s+3)

s =−3 :

4 = 2A, A = 2s =−5 : 4 =−2B, B =−2

4(s+3)(s+5)

= 21

s+3−2

1s+5

Bernd Schroder Louisiana Tech University, College of Engineering and Science

The Laplace Transform of The Dirac Delta Function

logo1

Transforms and New Formulas A Model The Initial Value Problem Interpretation Double Check

Solve the Initial Value Problemq′′+8q′+15q = 4δ (t−2), q(0) = 1, q′(0) = 0

Partial fraction decompositions.4

(s+3)(s+5)=

As+3

+B

s+54 = A(s+5)+B(s+3)

s =−3 : 4 = 2A

, A = 2s =−5 : 4 =−2B, B =−2

4(s+3)(s+5)

= 21

s+3−2

1s+5

Bernd Schroder Louisiana Tech University, College of Engineering and Science

The Laplace Transform of The Dirac Delta Function

logo1

Transforms and New Formulas A Model The Initial Value Problem Interpretation Double Check

Solve the Initial Value Problemq′′+8q′+15q = 4δ (t−2), q(0) = 1, q′(0) = 0

Partial fraction decompositions.4

(s+3)(s+5)=

As+3

+B

s+54 = A(s+5)+B(s+3)

s =−3 : 4 = 2A, A = 2

s =−5 : 4 =−2B, B =−24

(s+3)(s+5)= 2

1s+3

−21

s+5

Bernd Schroder Louisiana Tech University, College of Engineering and Science

The Laplace Transform of The Dirac Delta Function

logo1

Transforms and New Formulas A Model The Initial Value Problem Interpretation Double Check

Solve the Initial Value Problemq′′+8q′+15q = 4δ (t−2), q(0) = 1, q′(0) = 0

Partial fraction decompositions.4

(s+3)(s+5)=

As+3

+B

s+54 = A(s+5)+B(s+3)

s =−3 : 4 = 2A, A = 2s =−5 :

4 =−2B, B =−24

(s+3)(s+5)= 2

1s+3

−21

s+5

Bernd Schroder Louisiana Tech University, College of Engineering and Science

The Laplace Transform of The Dirac Delta Function

logo1

Transforms and New Formulas A Model The Initial Value Problem Interpretation Double Check

Solve the Initial Value Problemq′′+8q′+15q = 4δ (t−2), q(0) = 1, q′(0) = 0

Partial fraction decompositions.4

(s+3)(s+5)=

As+3

+B

s+54 = A(s+5)+B(s+3)

s =−3 : 4 = 2A, A = 2s =−5 : 4 =−2B

, B =−24

(s+3)(s+5)= 2

1s+3

−21

s+5

Bernd Schroder Louisiana Tech University, College of Engineering and Science

The Laplace Transform of The Dirac Delta Function

logo1

Transforms and New Formulas A Model The Initial Value Problem Interpretation Double Check

Solve the Initial Value Problemq′′+8q′+15q = 4δ (t−2), q(0) = 1, q′(0) = 0

Partial fraction decompositions.4

(s+3)(s+5)=

As+3

+B

s+54 = A(s+5)+B(s+3)

s =−3 : 4 = 2A, A = 2s =−5 : 4 =−2B, B =−2

4(s+3)(s+5)

= 21

s+3−2

1s+5

Bernd Schroder Louisiana Tech University, College of Engineering and Science

The Laplace Transform of The Dirac Delta Function

logo1

Transforms and New Formulas A Model The Initial Value Problem Interpretation Double Check

Solve the Initial Value Problemq′′+8q′+15q = 4δ (t−2), q(0) = 1, q′(0) = 0

Partial fraction decompositions.4

(s+3)(s+5)=

As+3

+B

s+54 = A(s+5)+B(s+3)

s =−3 : 4 = 2A, A = 2s =−5 : 4 =−2B, B =−2

4(s+3)(s+5)

= 21

s+3−2

1s+5

Bernd Schroder Louisiana Tech University, College of Engineering and Science

The Laplace Transform of The Dirac Delta Function

logo1

Transforms and New Formulas A Model The Initial Value Problem Interpretation Double Check

Solve the Initial Value Problemq′′+8q′+15q = 4δ (t−2), q(0) = 1, q′(0) = 0

Q =s+8

(s+3)(s+5)+ e−2s 4

(s+3)(s+5)

Q =52

1s+3

− 32

1s+5

+ e−2s[

21

s+3−2

1s+5

]q =

52

e−3t− 32

e−5t +U (t−2)[2e−3t−2e−5t

]t→t−2

=52

e−3t− 32

e−5t +U (t−2)[2e−3(t−2)−2e−5(t−2)

]

Bernd Schroder Louisiana Tech University, College of Engineering and Science

The Laplace Transform of The Dirac Delta Function

logo1

Transforms and New Formulas A Model The Initial Value Problem Interpretation Double Check

Solve the Initial Value Problemq′′+8q′+15q = 4δ (t−2), q(0) = 1, q′(0) = 0

Q =s+8

(s+3)(s+5)+ e−2s 4

(s+3)(s+5)

Q =52

1s+3

− 32

1s+5

+ e−2s[

21

s+3−2

1s+5

]q =

52

e−3t− 32

e−5t +U (t−2)[2e−3t−2e−5t

]t→t−2

=52

e−3t− 32

e−5t +U (t−2)[2e−3(t−2)−2e−5(t−2)

]

Bernd Schroder Louisiana Tech University, College of Engineering and Science

The Laplace Transform of The Dirac Delta Function

logo1

Transforms and New Formulas A Model The Initial Value Problem Interpretation Double Check

Solve the Initial Value Problemq′′+8q′+15q = 4δ (t−2), q(0) = 1, q′(0) = 0

Q =s+8

(s+3)(s+5)+ e−2s 4

(s+3)(s+5)

Q =52

1s+3

− 32

1s+5

+ e−2s[

21

s+3−2

1s+5

]

q =52

e−3t− 32

e−5t +U (t−2)[2e−3t−2e−5t

]t→t−2

=52

e−3t− 32

e−5t +U (t−2)[2e−3(t−2)−2e−5(t−2)

]

Bernd Schroder Louisiana Tech University, College of Engineering and Science

The Laplace Transform of The Dirac Delta Function

logo1

Transforms and New Formulas A Model The Initial Value Problem Interpretation Double Check

Solve the Initial Value Problemq′′+8q′+15q = 4δ (t−2), q(0) = 1, q′(0) = 0

Q =s+8

(s+3)(s+5)+ e−2s 4

(s+3)(s+5)

Q =52

1s+3

− 32

1s+5

+ e−2s[

21

s+3−2

1s+5

]q =

52

e−3t− 32

e−5t +U (t−2)[2e−3t−2e−5t

]t→t−2

=52

e−3t− 32

e−5t +U (t−2)[2e−3(t−2)−2e−5(t−2)

]

Bernd Schroder Louisiana Tech University, College of Engineering and Science

The Laplace Transform of The Dirac Delta Function

logo1

Transforms and New Formulas A Model The Initial Value Problem Interpretation Double Check

Solve the Initial Value Problemq′′+8q′+15q = 4δ (t−2), q(0) = 1, q′(0) = 0

Q =s+8

(s+3)(s+5)+ e−2s 4

(s+3)(s+5)

Q =52

1s+3

− 32

1s+5

+ e−2s[

21

s+3−2

1s+5

]q =

52

e−3t

− 32

e−5t +U (t−2)[2e−3t−2e−5t

]t→t−2

=52

e−3t− 32

e−5t +U (t−2)[2e−3(t−2)−2e−5(t−2)

]

Bernd Schroder Louisiana Tech University, College of Engineering and Science

The Laplace Transform of The Dirac Delta Function

logo1

Transforms and New Formulas A Model The Initial Value Problem Interpretation Double Check

Solve the Initial Value Problemq′′+8q′+15q = 4δ (t−2), q(0) = 1, q′(0) = 0

Q =s+8

(s+3)(s+5)+ e−2s 4

(s+3)(s+5)

Q =52

1s+3

− 32

1s+5

+ e−2s[

21

s+3−2

1s+5

]q =

52

e−3t− 32

e−5t

+U (t−2)[2e−3t−2e−5t

]t→t−2

=52

e−3t− 32

e−5t +U (t−2)[2e−3(t−2)−2e−5(t−2)

]

Bernd Schroder Louisiana Tech University, College of Engineering and Science

The Laplace Transform of The Dirac Delta Function

logo1

Transforms and New Formulas A Model The Initial Value Problem Interpretation Double Check

Solve the Initial Value Problemq′′+8q′+15q = 4δ (t−2), q(0) = 1, q′(0) = 0

Q =s+8

(s+3)(s+5)+ e−2s 4

(s+3)(s+5)

Q =52

1s+3

− 32

1s+5

+ e−2s[

21

s+3−2

1s+5

]q =

52

e−3t− 32

e−5t +U (t−2)

[2e−3t−2e−5t

]t→t−2

=52

e−3t− 32

e−5t +U (t−2)[2e−3(t−2)−2e−5(t−2)

]

Bernd Schroder Louisiana Tech University, College of Engineering and Science

The Laplace Transform of The Dirac Delta Function

logo1

Transforms and New Formulas A Model The Initial Value Problem Interpretation Double Check

Solve the Initial Value Problemq′′+8q′+15q = 4δ (t−2), q(0) = 1, q′(0) = 0

Q =s+8

(s+3)(s+5)+ e−2s 4

(s+3)(s+5)

Q =52

1s+3

− 32

1s+5

+ e−2s[

21

s+3−2

1s+5

]q =

52

e−3t− 32

e−5t +U (t−2)[2e−3t−2e−5t

]

t→t−2

=52

e−3t− 32

e−5t +U (t−2)[2e−3(t−2)−2e−5(t−2)

]

Bernd Schroder Louisiana Tech University, College of Engineering and Science

The Laplace Transform of The Dirac Delta Function

logo1

Transforms and New Formulas A Model The Initial Value Problem Interpretation Double Check

Solve the Initial Value Problemq′′+8q′+15q = 4δ (t−2), q(0) = 1, q′(0) = 0

Q =s+8

(s+3)(s+5)+ e−2s 4

(s+3)(s+5)

Q =52

1s+3

− 32

1s+5

+ e−2s[

21

s+3−2

1s+5

]q =

52

e−3t− 32

e−5t +U (t−2)[2e−3t−2e−5t

]t→t−2

=52

e−3t− 32

e−5t +U (t−2)[2e−3(t−2)−2e−5(t−2)

]

Bernd Schroder Louisiana Tech University, College of Engineering and Science

The Laplace Transform of The Dirac Delta Function

logo1

Transforms and New Formulas A Model The Initial Value Problem Interpretation Double Check

Solve the Initial Value Problemq′′+8q′+15q = 4δ (t−2), q(0) = 1, q′(0) = 0

Q =s+8

(s+3)(s+5)+ e−2s 4

(s+3)(s+5)

Q =52

1s+3

− 32

1s+5

+ e−2s[

21

s+3−2

1s+5

]q =

52

e−3t− 32

e−5t +U (t−2)[2e−3t−2e−5t

]t→t−2

=52

e−3t− 32

e−5t +U (t−2)[2e−3(t−2)−2e−5(t−2)

]

Bernd Schroder Louisiana Tech University, College of Engineering and Science

The Laplace Transform of The Dirac Delta Function

logo1

Transforms and New Formulas A Model The Initial Value Problem Interpretation Double Check

What Happens in the Physical System?

q =52

e−3t− 32

e−5t +U (t−2)[2e−3(t−2)−2e−5(t−2)

]

Bernd Schroder Louisiana Tech University, College of Engineering and Science

The Laplace Transform of The Dirac Delta Function

logo1

Transforms and New Formulas A Model The Initial Value Problem Interpretation Double Check

What Happens in the Physical System?

q =52

e−3t− 32

e−5t +U (t−2)[2e−3(t−2)−2e−5(t−2)

]Bernd Schroder Louisiana Tech University, College of Engineering and Science

The Laplace Transform of The Dirac Delta Function

logo1

Transforms and New Formulas A Model The Initial Value Problem Interpretation Double Check

What Happens in the Physical System?

q′ =−152

e−3t +152

e−5t +U (t−2)[−6e−3(t−2) +10e−5(t−2)

]

Bernd Schroder Louisiana Tech University, College of Engineering and Science

The Laplace Transform of The Dirac Delta Function

logo1

Transforms and New Formulas A Model The Initial Value Problem Interpretation Double Check

What Happens in the Physical System?

q′ =−152

e−3t +152

e−5t +U (t−2)[−6e−3(t−2) +10e−5(t−2)

]Bernd Schroder Louisiana Tech University, College of Engineering and Science

The Laplace Transform of The Dirac Delta Function

logo1

Transforms and New Formulas A Model The Initial Value Problem Interpretation Double Check

Does q =52

e−3t− 32

e−5t +U (t−2)[2e−3(t−2)−2e−5(t−2)

]Solve the Initial

Value Problem q′′+8q′+15q = 4δ (t−2), q(0) = 1, q′(0) = 0?

First consider q1 =52

e−3t− 32

e−5t.

15(

52

e−3t− 32

e−5t)

+8(−15

2e−3t +

152

e−5t)

+(

452

e−3t− 752

e−5t)

=(

752− 120

2+

452

)e−3t +

(−45

2+

1202− 75

2

)e−5t

= 0√

q1(0) =52

e−3·0− 32

e−5·0 = 1√

q′1(0) = −152

e−3·0 +152

e−5·0 = 0√

Bernd Schroder Louisiana Tech University, College of Engineering and Science

The Laplace Transform of The Dirac Delta Function

logo1

Transforms and New Formulas A Model The Initial Value Problem Interpretation Double Check

Does q =52

e−3t− 32

e−5t +U (t−2)[2e−3(t−2)−2e−5(t−2)

]Solve the Initial

Value Problem q′′+8q′+15q = 4δ (t−2), q(0) = 1, q′(0) = 0?

First consider q1 =52

e−3t− 32

e−5t.

15(

52

e−3t− 32

e−5t)

+8(−15

2e−3t +

152

e−5t)

+(

452

e−3t− 752

e−5t)

=(

752− 120

2+

452

)e−3t +

(−45

2+

1202− 75

2

)e−5t

= 0√

q1(0) =52

e−3·0− 32

e−5·0 = 1√

q′1(0) = −152

e−3·0 +152

e−5·0 = 0√

Bernd Schroder Louisiana Tech University, College of Engineering and Science

The Laplace Transform of The Dirac Delta Function

logo1

Transforms and New Formulas A Model The Initial Value Problem Interpretation Double Check

Does q =52

e−3t− 32

e−5t +U (t−2)[2e−3(t−2)−2e−5(t−2)

]Solve the Initial

Value Problem q′′+8q′+15q = 4δ (t−2), q(0) = 1, q′(0) = 0?

First consider q1 =52

e−3t− 32

e−5t.

15(

52

e−3t− 32

e−5t)

+8(−15

2e−3t +

152

e−5t)

+(

452

e−3t− 752

e−5t)

=(

752− 120

2+

452

)e−3t +

(−45

2+

1202− 75

2

)e−5t

= 0√

q1(0) =52

e−3·0− 32

e−5·0 = 1√

q′1(0) = −152

e−3·0 +152

e−5·0 = 0√

Bernd Schroder Louisiana Tech University, College of Engineering and Science

The Laplace Transform of The Dirac Delta Function

logo1

Transforms and New Formulas A Model The Initial Value Problem Interpretation Double Check

Does q =52

e−3t− 32

e−5t +U (t−2)[2e−3(t−2)−2e−5(t−2)

]Solve the Initial

Value Problem q′′+8q′+15q = 4δ (t−2), q(0) = 1, q′(0) = 0?

First consider q1 =52

e−3t− 32

e−5t.

15(

52

e−3t− 32

e−5t)

+8(−15

2e−3t +

152

e−5t)

+(

452

e−3t− 752

e−5t)

=(

752− 120

2+

452

)e−3t +

(−45

2+

1202− 75

2

)e−5t

= 0√

q1(0) =52

e−3·0− 32

e−5·0 = 1√

q′1(0) = −152

e−3·0 +152

e−5·0 = 0√

Bernd Schroder Louisiana Tech University, College of Engineering and Science

The Laplace Transform of The Dirac Delta Function

logo1

Transforms and New Formulas A Model The Initial Value Problem Interpretation Double Check

Does q =52

e−3t− 32

e−5t +U (t−2)[2e−3(t−2)−2e−5(t−2)

]Solve the Initial

Value Problem q′′+8q′+15q = 4δ (t−2), q(0) = 1, q′(0) = 0?

First consider q1 =52

e−3t− 32

e−5t.

15(

52

e−3t− 32

e−5t)

+8(−15

2e−3t +

152

e−5t)

+(

452

e−3t− 752

e−5t)

=(

752− 120

2+

452

)e−3t +

(−45

2+

1202− 75

2

)e−5t

= 0√

q1(0) =52

e−3·0− 32

e−5·0 = 1√

q′1(0) = −152

e−3·0 +152

e−5·0 = 0√

Bernd Schroder Louisiana Tech University, College of Engineering and Science

The Laplace Transform of The Dirac Delta Function

logo1

Transforms and New Formulas A Model The Initial Value Problem Interpretation Double Check

Does q =52

e−3t− 32

e−5t +U (t−2)[2e−3(t−2)−2e−5(t−2)

]Solve the Initial

Value Problem q′′+8q′+15q = 4δ (t−2), q(0) = 1, q′(0) = 0?

First consider q1 =52

e−3t− 32

e−5t.

15(

52

e−3t− 32

e−5t)

+8(−15

2e−3t +

152

e−5t)

+(

452

e−3t− 752

e−5t)

=(

752− 120

2+

452

)e−3t +

(−45

2+

1202− 75

2

)e−5t

= 0√

q1(0) =52

e−3·0− 32

e−5·0 = 1√

q′1(0) = −152

e−3·0 +152

e−5·0 = 0√

Bernd Schroder Louisiana Tech University, College of Engineering and Science

The Laplace Transform of The Dirac Delta Function

logo1

Transforms and New Formulas A Model The Initial Value Problem Interpretation Double Check

Does q =52

e−3t− 32

e−5t +U (t−2)[2e−3(t−2)−2e−5(t−2)

]Solve the Initial

Value Problem q′′+8q′+15q = 4δ (t−2), q(0) = 1, q′(0) = 0?

First consider q1 =52

e−3t− 32

e−5t.

15(

52

e−3t− 32

e−5t)

+8(−15

2e−3t +

152

e−5t)

+(

452

e−3t− 752

e−5t)

=(

752− 120

2+

452

)e−3t +

(−45

2+

1202− 75

2

)e−5t

= 0

√

q1(0) =52

e−3·0− 32

e−5·0 = 1√

q′1(0) = −152

e−3·0 +152

e−5·0 = 0√

Bernd Schroder Louisiana Tech University, College of Engineering and Science

The Laplace Transform of The Dirac Delta Function

logo1

Transforms and New Formulas A Model The Initial Value Problem Interpretation Double Check

Does q =52

e−3t− 32

e−5t +U (t−2)[2e−3(t−2)−2e−5(t−2)

]Solve the Initial

Value Problem q′′+8q′+15q = 4δ (t−2), q(0) = 1, q′(0) = 0?

First consider q1 =52

e−3t− 32

e−5t.

15(

52

e−3t− 32

e−5t)

+8(−15

2e−3t +

152

e−5t)

+(

452

e−3t− 752

e−5t)

=(

752− 120

2+

452

)e−3t +

(−45

2+

1202− 75

2

)e−5t

= 0√

q1(0) =52

e−3·0− 32

e−5·0 = 1√

q′1(0) = −152

e−3·0 +152

e−5·0 = 0√

Bernd Schroder Louisiana Tech University, College of Engineering and Science

The Laplace Transform of The Dirac Delta Function

logo1

Transforms and New Formulas A Model The Initial Value Problem Interpretation Double Check

Does q =52

e−3t− 32

e−5t +U (t−2)[2e−3(t−2)−2e−5(t−2)

]Solve the Initial

Value Problem q′′+8q′+15q = 4δ (t−2), q(0) = 1, q′(0) = 0?

First consider q1 =52

e−3t− 32

e−5t.

15(

52

e−3t− 32

e−5t)

+8(−15

2e−3t +

152

e−5t)

+(

452

e−3t− 752

e−5t)

=(

752− 120

2+

452

)e−3t +

(−45

2+

1202− 75

2

)e−5t

= 0√

q1(0)

=52

e−3·0− 32

e−5·0 = 1√

q′1(0) = −152

e−3·0 +152

e−5·0 = 0√

Bernd Schroder Louisiana Tech University, College of Engineering and Science

The Laplace Transform of The Dirac Delta Function

logo1

Transforms and New Formulas A Model The Initial Value Problem Interpretation Double Check

Does q =52

e−3t− 32

e−5t +U (t−2)[2e−3(t−2)−2e−5(t−2)

]Solve the Initial

Value Problem q′′+8q′+15q = 4δ (t−2), q(0) = 1, q′(0) = 0?

First consider q1 =52

e−3t− 32

e−5t.

15(

52

e−3t− 32

e−5t)

+8(−15

2e−3t +

152

e−5t)

+(

452

e−3t− 752

e−5t)

=(

752− 120

2+

452

)e−3t +

(−45

2+

1202− 75

2

)e−5t

= 0√

q1(0) =52

e−3·0− 32

e−5·0

= 1√

q′1(0) = −152

e−3·0 +152

e−5·0 = 0√

Bernd Schroder Louisiana Tech University, College of Engineering and Science

The Laplace Transform of The Dirac Delta Function

logo1

Transforms and New Formulas A Model The Initial Value Problem Interpretation Double Check

Does q =52

e−3t− 32

e−5t +U (t−2)[2e−3(t−2)−2e−5(t−2)

]Solve the Initial

Value Problem q′′+8q′+15q = 4δ (t−2), q(0) = 1, q′(0) = 0?

First consider q1 =52

e−3t− 32

e−5t.

15(

52

e−3t− 32

e−5t)

+8(−15

2e−3t +

152

e−5t)

+(

452

e−3t− 752

e−5t)

=(

752− 120

2+

452

)e−3t +

(−45

2+

1202− 75

2

)e−5t

= 0√

q1(0) =52

e−3·0− 32

e−5·0 = 1

√

q′1(0) = −152

e−3·0 +152

e−5·0 = 0√

Bernd Schroder Louisiana Tech University, College of Engineering and Science

The Laplace Transform of The Dirac Delta Function

logo1

Transforms and New Formulas A Model The Initial Value Problem Interpretation Double Check

Does q =52

e−3t− 32

e−5t +U (t−2)[2e−3(t−2)−2e−5(t−2)

]Solve the Initial

Value Problem q′′+8q′+15q = 4δ (t−2), q(0) = 1, q′(0) = 0?

First consider q1 =52

e−3t− 32

e−5t.

15(

52

e−3t− 32

e−5t)

+8(−15

2e−3t +

152

e−5t)

+(

452

e−3t− 752

e−5t)

=(

752− 120

2+

452

)e−3t +

(−45

2+

1202− 75

2

)e−5t

= 0√

q1(0) =52

e−3·0− 32

e−5·0 = 1√

q′1(0) = −152

e−3·0 +152

e−5·0 = 0√

Bernd Schroder Louisiana Tech University, College of Engineering and Science

The Laplace Transform of The Dirac Delta Function

logo1

Transforms and New Formulas A Model The Initial Value Problem Interpretation Double Check

Does q =52

e−3t− 32

e−5t +U (t−2)[2e−3(t−2)−2e−5(t−2)

]Solve the Initial

Value Problem q′′+8q′+15q = 4δ (t−2), q(0) = 1, q′(0) = 0?

First consider q1 =52

e−3t− 32

e−5t.

15(

52

e−3t− 32

e−5t)

+8(−15

2e−3t +

152

e−5t)

+(

452

e−3t− 752

e−5t)

=(

752− 120

2+

452

)e−3t +

(−45

2+

1202− 75

2

)e−5t

= 0√

q1(0) =52

e−3·0− 32

e−5·0 = 1√

q′1(0)

= −152

e−3·0 +152

e−5·0 = 0√

Bernd Schroder Louisiana Tech University, College of Engineering and Science

The Laplace Transform of The Dirac Delta Function

logo1

Transforms and New Formulas A Model The Initial Value Problem Interpretation Double Check

Does q =52

e−3t− 32

e−5t +U (t−2)[2e−3(t−2)−2e−5(t−2)

]Solve the Initial

Value Problem q′′+8q′+15q = 4δ (t−2), q(0) = 1, q′(0) = 0?

First consider q1 =52

e−3t− 32

e−5t.

15(

52

e−3t− 32

e−5t)

+8(−15

2e−3t +

152

e−5t)

+(

452

e−3t− 752

e−5t)

=(

752− 120

2+

452

)e−3t +

(−45

2+

1202− 75

2

)e−5t

= 0√

q1(0) =52

e−3·0− 32

e−5·0 = 1√

q′1(0) = −152

e−3·0 +152

e−5·0

= 0√

Bernd Schroder Louisiana Tech University, College of Engineering and Science

The Laplace Transform of The Dirac Delta Function

logo1

Transforms and New Formulas A Model The Initial Value Problem Interpretation Double Check

Does q =52

e−3t− 32

e−5t +U (t−2)[2e−3(t−2)−2e−5(t−2)

]Solve the Initial

Value Problem q′′+8q′+15q = 4δ (t−2), q(0) = 1, q′(0) = 0?

First consider q1 =52

e−3t− 32

e−5t.

15(

52

e−3t− 32

e−5t)

+8(−15

2e−3t +

152

e−5t)

+(

452

e−3t− 752

e−5t)

=(

752− 120

2+

452

)e−3t +

(−45

2+

1202− 75

2

)e−5t

= 0√

q1(0) =52

e−3·0− 32

e−5·0 = 1√

q′1(0) = −152

e−3·0 +152

e−5·0 = 0

√

Bernd Schroder Louisiana Tech University, College of Engineering and Science

The Laplace Transform of The Dirac Delta Function

logo1

Transforms and New Formulas A Model The Initial Value Problem Interpretation Double Check

Does q =52

e−3t− 32

e−5t +U (t−2)[2e−3(t−2)−2e−5(t−2)

]Solve the Initial

Value Problem q′′+8q′+15q = 4δ (t−2), q(0) = 1, q′(0) = 0?

First consider q1 =52

e−3t− 32

e−5t.

15(

52

e−3t− 32

e−5t)

+8(−15

2e−3t +

152

e−5t)

+(

452

e−3t− 752

e−5t)

=(

752− 120

2+

452

)e−3t +

(−45

2+

1202− 75

2

)e−5t

= 0√

q1(0) =52

e−3·0− 32

e−5·0 = 1√

q′1(0) = −152

e−3·0 +152

e−5·0 = 0√

Bernd Schroder Louisiana Tech University, College of Engineering and Science

The Laplace Transform of The Dirac Delta Function

logo1

Transforms and New Formulas A Model The Initial Value Problem Interpretation Double Check

Does q =52

e−3t− 32

e−5t +U (t−2)[2e−3(t−2)−2e−5(t−2)

]Solve the Initial Value

Problem q′′+8q′+15q = 4δ (t−2), q(0) = 1, q′(0) = 0?

Now consider q2 = 2e−3(t−2)−2e−5(t−2).

15(2e−3(t−2)−2e−5(t−2)

)+8

(−6e−3(t−2)+10e−5(t−2)

)+

(18e−3(t−2)−50e−5(t−2)

)= (30−48+18)e−3(t−2) +(−30+80−50)e−5(t−2)

= 0√

Bernd Schroder Louisiana Tech University, College of Engineering and Science

The Laplace Transform of The Dirac Delta Function

logo1

Transforms and New Formulas A Model The Initial Value Problem Interpretation Double Check

Does q =52

e−3t− 32

e−5t +U (t−2)[2e−3(t−2)−2e−5(t−2)

]Solve the Initial Value

Problem q′′+8q′+15q = 4δ (t−2), q(0) = 1, q′(0) = 0?

Now consider q2 = 2e−3(t−2)−2e−5(t−2).

15(2e−3(t−2)−2e−5(t−2)

)+8

(−6e−3(t−2)+10e−5(t−2)

)+

(18e−3(t−2)−50e−5(t−2)

)= (30−48+18)e−3(t−2) +(−30+80−50)e−5(t−2)

= 0√

Bernd Schroder Louisiana Tech University, College of Engineering and Science

The Laplace Transform of The Dirac Delta Function

logo1

Transforms and New Formulas A Model The Initial Value Problem Interpretation Double Check

Does q =52

e−3t− 32

e−5t +U (t−2)[2e−3(t−2)−2e−5(t−2)

]Solve the Initial Value

Problem q′′+8q′+15q = 4δ (t−2), q(0) = 1, q′(0) = 0?

Now consider q2 = 2e−3(t−2)−2e−5(t−2).

15(2e−3(t−2)−2e−5(t−2)

)

+8(−6e−3(t−2)+10e−5(t−2)

)+

(18e−3(t−2)−50e−5(t−2)

)= (30−48+18)e−3(t−2) +(−30+80−50)e−5(t−2)

= 0√

Bernd Schroder Louisiana Tech University, College of Engineering and Science

The Laplace Transform of The Dirac Delta Function

logo1

Transforms and New Formulas A Model The Initial Value Problem Interpretation Double Check

Does q =52

e−3t− 32

e−5t +U (t−2)[2e−3(t−2)−2e−5(t−2)

]Solve the Initial Value

Problem q′′+8q′+15q = 4δ (t−2), q(0) = 1, q′(0) = 0?

Now consider q2 = 2e−3(t−2)−2e−5(t−2).

15(2e−3(t−2)−2e−5(t−2)

)+8

(−6e−3(t−2)+10e−5(t−2)

)

+(18e−3(t−2)−50e−5(t−2)

)= (30−48+18)e−3(t−2) +(−30+80−50)e−5(t−2)

= 0√

Bernd Schroder Louisiana Tech University, College of Engineering and Science

The Laplace Transform of The Dirac Delta Function

logo1

Transforms and New Formulas A Model The Initial Value Problem Interpretation Double Check

Does q =52

e−3t− 32

e−5t +U (t−2)[2e−3(t−2)−2e−5(t−2)

]Solve the Initial Value

Problem q′′+8q′+15q = 4δ (t−2), q(0) = 1, q′(0) = 0?

Now consider q2 = 2e−3(t−2)−2e−5(t−2).

15(2e−3(t−2)−2e−5(t−2)

)+8

(−6e−3(t−2)+10e−5(t−2)

)+

(18e−3(t−2)−50e−5(t−2)

)

= (30−48+18)e−3(t−2) +(−30+80−50)e−5(t−2)

= 0√

Bernd Schroder Louisiana Tech University, College of Engineering and Science

The Laplace Transform of The Dirac Delta Function

logo1

Transforms and New Formulas A Model The Initial Value Problem Interpretation Double Check

Does q =52

e−3t− 32

e−5t +U (t−2)[2e−3(t−2)−2e−5(t−2)

]Solve the Initial Value

Problem q′′+8q′+15q = 4δ (t−2), q(0) = 1, q′(0) = 0?

Now consider q2 = 2e−3(t−2)−2e−5(t−2).

15(2e−3(t−2)−2e−5(t−2)

)+8

(−6e−3(t−2)+10e−5(t−2)

)+

(18e−3(t−2)−50e−5(t−2)

)= (30−48+18)e−3(t−2) +(−30+80−50)e−5(t−2)

= 0√

Bernd Schroder Louisiana Tech University, College of Engineering and Science

The Laplace Transform of The Dirac Delta Function

logo1

Transforms and New Formulas A Model The Initial Value Problem Interpretation Double Check

Does q =52

e−3t− 32

e−5t +U (t−2)[2e−3(t−2)−2e−5(t−2)

]Solve the Initial Value

Problem q′′+8q′+15q = 4δ (t−2), q(0) = 1, q′(0) = 0?

Now consider q2 = 2e−3(t−2)−2e−5(t−2).

15(2e−3(t−2)−2e−5(t−2)

)+8

(−6e−3(t−2)+10e−5(t−2)

)+

(18e−3(t−2)−50e−5(t−2)

)= (30−48+18)e−3(t−2) +(−30+80−50)e−5(t−2)

= 0

√

Bernd Schroder Louisiana Tech University, College of Engineering and Science

The Laplace Transform of The Dirac Delta Function

logo1

Transforms and New Formulas A Model The Initial Value Problem Interpretation Double Check

Does q =52

e−3t− 32

e−5t +U (t−2)[2e−3(t−2)−2e−5(t−2)

]Solve the Initial Value

Problem q′′+8q′+15q = 4δ (t−2), q(0) = 1, q′(0) = 0?

Now consider q2 = 2e−3(t−2)−2e−5(t−2).

15(2e−3(t−2)−2e−5(t−2)

)+8

(−6e−3(t−2)+10e−5(t−2)

)+

(18e−3(t−2)−50e−5(t−2)

)= (30−48+18)e−3(t−2) +(−30+80−50)e−5(t−2)

= 0√

Bernd Schroder Louisiana Tech University, College of Engineering and Science

The Laplace Transform of The Dirac Delta Function