75.1 LaPlace Transforms Problem Set

LAPLACE TRANSFORMS | PRACTICE PROBLEMS Complete the following to reinforce your understanding of the concept covered in this module.

PROBLEM 1: The Laplace Transform of the given function is best written as:

𝑓 𝑡 = 𝑒 !! + 2

A. B. C.

! !!!

! !

! !!! ! !!!

!

+!+1

!

!

!

!!!

D. +



+

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PROBLEM 2: The Laplace Transform of the given function is best represented as:

𝑓 𝑡 = (1 + 2𝑡)𝑒 !!

A. B. C. D.



! !!!

!

+

! !!! !

! !!! ! ! !!! ! ! !!! !

−

! !!! !

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PROBLEM 3: The Laplace Transform of the defined function can be written as:

𝑓 𝑡 = 8 + 4𝑡 + 𝑡 ! − 4𝑡 !

A.

! !!!

+

!

! (!!!)! !

+

! !!

!

+

+ !! + !! + !! !!!

C.

! !

!!

!

!

!

!!

D. +



!

+ +

! !!

−

!!

!"

B.

+

!

!" !!

! (!!!)!

+

! !!

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LAPLACE TRANSFORMS | SOLUTIONS SOLUTION 1: The GENERAL FORMULA for the LAPLACE TRANSFORM can be referenced under the topic of MATHEMATICS on page 34 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. Throughout our studies, we have learned a number of techniques to deploy when solving various forms of DIFFERENTIAL EQUATIONS. These methods are solid, very useful, but at times, fall short depending on the structure or complexity of the DRIVING FUNCTION. Time is always of the essence when it comes to performing on the FE Exam, so quickly realizing the most efficient route to take when working DIFFERENTIAL EQUATIONS, is extremely crucial. In most cases, our typical methods of solving DIFFERENTIAL EQUATIONS will suffice. However, there are two instances where the LAPLACE TRANFORM will either significantly cut down on our effort, or flat make it possible to find a solution. These two scenarios are: 1. When the DRIVING FUNCTION of a HIGHER ORDER NONHOMOGENEOUS DIFFERENTIAL EQUATION is so complex that a starting PARTICULAR



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FORMULA can not be found in the table that is provided to us on page 31 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. 2. When the DRIVING FUNCTION is a PIECEWISE FUNCTION of a defined INTERVAL. In this problem, we are given the function: 𝑓 𝑡 = 𝑒 !! + 2 We are not presented this function as a DRIVING FUNCTION of a HIGHER ORDER NONHOMOGENEOUS DIFFERENTIAL EQUATION, but rather, just simply as a function. We will carry out the process of determining the LAPLACE TRANSFORM, but let’s not lose sight of what we are doing. Carrying out the process will be equivalent to defining the PARTICULAR SOLUTION of HIGHER ORDER NONHOMOGENEOUS DIFFERENTIAL EQUATION if that’s what we are given. Although the term itself may incite some uncomfortable feelings, LAPLACE TRANSFORMS can essentially break any DIFFERENTIAL EQUATION down in to a simple ALGEBRAIC expression for us to solve. The algebra of deriving LAPLACE TRANSFORMS can be very gritty and convoluted, we could spend time on doing each derivation, but it wouldn’t be good use for our time.



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Fortunately, we are provided a TABLE of the most common LAPLACE TRANFORMS, and the ones that we should expect to need and use on the exam, on page 34 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. A LAPLACE TRANSFORM can be presented using the NOMENCLATURE: ℒ 𝑓(𝑡) = 𝐹 𝑠 Or in our case: ℒ 𝑒 !! + 2 = 𝐹 𝑠 The LAPLACE TRANSFORM will produce a new function of the variable “s”, where all the “t” variables in the original function will drop out through the INTEGRATION process. This INTEGRATION, or more properly stated, the LAPLACE TRANFORM, is generally written as: !

𝑓 𝑡 𝑒 !!" 𝑑𝑡

𝐹 𝑠 = !!

And in our case: !

(𝑒 !! + 2)𝑒 !!" 𝑑𝑡

𝐹 𝑠 = !!



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This GENERAL FORMULA for the LAPLACE TRANSFORM can be referenced under the topic of MATHEMATICS on page 34 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. The LAPLACE TRANSFORM is a LINEAR OPERATOR, so when given a function with multiple terms, as we are here, we can break it down term by term to determine what the complete transform will look like. Therefore, we can rewrite our LAPLACE TRANSFORM as: ℒ 𝑒 !! + ℒ 2 = 𝐹 𝑠 Or in the INTEGRATION form: !

!

(𝑒 !! )𝑒 !!" 𝑑𝑡 +

𝐹 𝑠 = !!

(2)𝑒 !!" 𝑑𝑡 !!

From this point, we can just go down the route of assessing each of these INTEGRALS, which we are capable of doing, and determine what the solution would be for this particular function. However, the algebra can get a bit convoluted, and very very time consuming…so we will rely on the TABLES…at least to the extent which we can. The TABLE of COMMON LAPLACE TRANSFORMS can be referenced under the topic of MATHEMATICS on page 34 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing.

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This table provides us the most common LAPLACE TRANSFORMS that we will need and encounter on the day of our exam. Some useful LAPLACE TRANSFORMS are represented as: 𝑓(𝑡)

𝐹(𝑠)

𝛿 𝑡 , 𝐼𝑚𝑝𝑢𝑙𝑠𝑒 𝑎𝑡 𝑡 = 0

1

𝑢 𝑡 , 𝑆𝑡𝑒𝑝 𝑎𝑡 𝑡 = 0

1/𝑠

𝑡 𝑢 𝑡 , 𝑅𝑎𝑚𝑝 𝑎𝑡 𝑡 = 0

1/𝑠 !

𝑒 !!"

1/(𝑠 + 𝛼)

𝑡𝑒 !!"

1/ 𝑠 + 𝛼

!

𝑒 !!" sin 𝛽𝑡

𝛽/ 𝑠 + 𝛼

𝑒 !!" cos 𝛽𝑡

(𝑠 + 𝛼)/ 𝑠 + 𝛼

!

+ 𝛽! !

+ 𝛽!

!!! !

𝑠 𝐹 𝑠 −

𝑑! 𝑓 𝑡 𝑑𝑡 !

𝑠

!!!!!

!!!

!

𝑑 ! 𝑓(0) 𝑑𝑡 !

(1/𝑠)𝐹(𝑠)

𝑓 𝜏 𝑑𝜏 ! !

𝑥 𝑡 − 𝜏 ℎ(𝜏) 𝑑𝜏

𝐻 𝑠 𝑋(𝑠)

!

𝑓 𝑡 − 𝜏 𝑢(𝑡 − 𝜏)

𝑒 !!" 𝐹(𝑠)

limit 𝑓(𝑡)

limit 𝑠𝐹(𝑠)

limit 𝑓(𝑡)

limit 𝑠𝐹(𝑠)

!→! !→!



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Revisiting our function written in LAPLACE TRANSFORM terms, we have: ℒ 𝑒 !! + ℒ 2 = 𝐹 𝑠 We will cycle term by term creating each TRANSFORM and combining them along the way. Let’s start with: ℒ 𝑒 !! Referencing the table, we match this term up with the row: 𝑓(𝑡)

𝐹(𝑠)

𝑒 !!"

1/(𝑠 + 𝛼)

This allows us to conclude that: ℒ 𝑒 !! = 1/(𝑠 − 3) Cycling to our next term, we have: ℒ 2 Which we can rewrite as: 2ℒ 1

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Now this is just a step function, and referencing the table, we match this term up with the row: 𝑓(𝑡)

𝐹(𝑠)

𝑢 𝑡 , 𝑆𝑡𝑒𝑝 𝑎𝑡 𝑡 = 0

1/𝑠

This allows us to conclude that: 2ℒ 1 = 2(1/𝑠) Or: ℒ 2 = 2/𝑠 In this case, we had to tweak our original term a little bit to make it fit in to a form that we had defined within our table. This is fine, it remains equivalent throughout, but we have to be quick to recognize this come exam day so we aren’t wasting any time twiddling our thumbs wondering where we should go if it isn’t given to us verbatim. Now the final step is to gather all of our LAPLACE TRANFORMS and combine them together in to a single solution, giving us:

𝐹 𝑠 =

1 2 + 𝑠−3 𝑠

The correct answer choice is A.



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SOLUTION 2: The GENERAL FORMULA for the LAPLACE TRANSFORM can be referenced under the topic of MATHEMATICS on page 34 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. Throughout our studies, we have learned a number of techniques to deploy when solving various forms of DIFFERENTIAL EQUATIONS. These methods are solid, very useful, but at times, fall short depending on the structure or complexity of the DRIVING FUNCTION. Time is always of the essence when it comes to performing on the FE Exam, so quickly realizing the most efficient route to take when working DIFFERENTIAL EQUATIONS, is extremely crucial. In most cases, our typical methods of solving DIFFERENTIAL EQUATIONS will suffice. However, there are two instances where the LAPLACE TRANFORM will either significantly cut down on our effort, or flat make it possible to find a solution. These two scenarios are: 1. When the DRIVING FUNCTION of a HIGHER ORDER NONHOMOGENEOUS DIFFERENTIAL EQUATION is so complex that a starting PARTICULAR FORMULA can not be found in the table that is provided to us on page 31 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing.



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2. When the DRIVING FUNCTION is a PIECEWISE FUNCTION of a defined INTERVAL. In this problem, we are given the function: 𝑓 𝑡 = (1 + 2𝑡)𝑒 !! We are not presented this function as a DRIVING FUNCTION of a HIGHER ORDER NONHOMOGENEOUS DIFFERENTIAL EQUATION, but rather, just simply as a function. We will carry out the process of determining the LAPLACE TRANSFORM, but let’s not lose sight of what we are doing. Carrying out the process will be equivalent to defining the PARTICULAR SOLUTION of HIGHER ORDER NONHOMOGENEOUS DIFFERENTIAL EQUATION if that’s what we are given. Although the term itself may incite some uncomfortable feelings, LAPLACE TRANSFORMS can essentially break any DIFFERENTIAL EQUATION down in to a simple ALGEBRAIC expression for us to solve. The algebra of deriving LAPLACE TRANSFORMS can be very gritty and convoluted, we could spend time on doing each derivation, but it wouldn’t be good use for our time. Fortunately, we are provided a TABLE of the most common LAPLACE TRANFORMS, and the ones that we should expect to need and use on the exam, on page 34 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing.



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A LAPLACE TRANSFORM can be presented using the NOMENCLATURE: ℒ 𝑓(𝑡) = 𝐹 𝑠 Or in our case: ℒ (1 + 2𝑡)𝑒 !! = 𝐹 𝑠 The LAPLACE TRANSFORM will produce a new function of the variable “s”, where all the “t” variables in the original function will drop out through the INTEGRATION process. This INTEGRATION, or more properly stated, the LAPLACE TRANFORM, is generally written as: !

𝑓 𝑡 𝑒 !!" 𝑑𝑡

𝐹 𝑠 = !!

And in our case: !

(1 + 2𝑡)𝑒 !! 𝑒 !!" 𝑑𝑡

𝐹 𝑠 = !!

This GENERAL FORMULA for the LAPLACE TRANSFORM can be referenced under the topic of MATHEMATICS on page 34 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing.

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Before we move forward, let’s distribute our exponential function through so that we have a LAPLACE TRANSFORM that reads: ℒ 𝑒 !! + 2𝑡𝑒 !! = 𝐹 𝑠 The LAPLACE TRANSFORM is a LINEAR OPERATOR, so when given a function with multiple terms, as we are here, we can break it down term by term to determine what the complete transform will look like. Therefore, we can rewrite our LAPLACE TRANSFORM as: ℒ 𝑒 !! + ℒ 2𝑡𝑒 !! = 𝐹 𝑠 Which can further be broken down as: ℒ 𝑒 !! + 2ℒ 𝑡𝑒 !! = 𝐹 𝑠 The TABLE of COMMON LAPLACE TRANSFORMS can be referenced under the topic of MATHEMATICS on page 34 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. This table provides us the most common LAPLACE TRANSFORMS that we will need and encounter on the day of our exam.



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Some useful LAPLACE TRANSFORMS are represented as: 𝑓(𝑡)

𝐹(𝑠)

𝛿 𝑡 , 𝐼𝑚𝑝𝑢𝑙𝑠𝑒 𝑎𝑡 𝑡 = 0

1

𝑢 𝑡 , 𝑆𝑡𝑒𝑝 𝑎𝑡 𝑡 = 0

1/𝑠

𝑡 𝑢 𝑡 , 𝑅𝑎𝑚𝑝 𝑎𝑡 𝑡 = 0

1/𝑠 !

𝑒 !!"

1/(𝑠 + 𝛼)

𝑡𝑒 !!"

1/ 𝑠 + 𝛼

!

𝑒 !!" sin 𝛽𝑡

𝛽/ 𝑠 + 𝛼

𝑒 !!" cos 𝛽𝑡

(𝑠 + 𝛼)/ 𝑠 + 𝛼

!

+ 𝛽! !

+ 𝛽!

!!! !

𝑠 𝐹 𝑠 −

𝑑! 𝑓 𝑡 𝑑𝑡 !

𝑠

!!!!!

!!!

!

𝑑 ! 𝑓(0) 𝑑𝑡 !

(1/𝑠)𝐹(𝑠)

𝑓 𝜏 𝑑𝜏 ! !

𝑥 𝑡 − 𝜏 ℎ(𝜏) 𝑑𝜏

𝐻 𝑠 𝑋(𝑠)

!

𝑓 𝑡 − 𝜏 𝑢(𝑡 − 𝜏)

𝑒 !!" 𝐹(𝑠)

limit 𝑓(𝑡)

limit 𝑠𝐹(𝑠)

limit 𝑓(𝑡)

limit 𝑠𝐹(𝑠)

!→! !→!

!→!

!→!

Revisiting our function written in LAPLACE TRANSFORM terms, we have: ℒ 𝑒 !! + 2ℒ 𝑡𝑒 !! = 𝐹 𝑠



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We will cycle term by term creating each TRANSFORM and combining them along the way. Let’s start with: ℒ 𝑒 !! Referencing the table, we match this term up with the row: 𝑓(𝑡)

𝐹(𝑠)

𝑒 !!"

1/(𝑠 + 𝛼)

This allows us to conclude that: ℒ 𝑒 !! = 1/(𝑠 − 2) Cycling to our next term, we have: 2ℒ 𝑡𝑒 !! Referencing the table, we match this term up with the row: 𝑓(𝑡) 𝑡𝑒 !!"



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𝐹(𝑠) 1/ 𝑠 + 𝛼

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This allows us to conclude that: 2ℒ 𝑡𝑒 !! = 2(1/ 𝑠 − 2 ! ) In this case, we had to tweak our original term a little bit to make it fit in to a form that we had defined within our table. This is fine, it remains equivalent throughout, but we have to be quick to recognize this come exam day so we aren’t wasting any time twiddling our thumbs wondering where we should go if it isn’t given to us verbatim. Now the final step is to gather all of our LAPLACE TRANFORMS and combine them together in to a single solution, giving us:

𝐹 𝑠 =

1 2 + 𝑠−2 𝑠−2

!

Let’s reduce it down to a single term by determining the common denominator, which gives us:

𝐹 𝑠 =

𝑠−2 +2 𝑠−2 !

Which simplifies in to the single term:

𝐹 𝑠 =

𝑠 𝑠−2

!

The correct answer choice is B.

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SOLUTION 3: The GENERAL FORMULA for the LAPLACE TRANSFORM can be referenced under the topic of MATHEMATICS on page 34 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. Throughout our studies, we have learned a number of techniques to deploy when solving various forms of DIFFERENTIAL EQUATIONS. These methods are solid, very useful, but at times, fall short depending on the structure or complexity of the DRIVING FUNCTION. Time is always of the essence when it comes to performing on the FE Exam, so quickly realizing the most efficient route to take when working DIFFERENTIAL EQUATIONS, is extremely crucial. In most cases, our typical methods of solving DIFFERENTIAL EQUATIONS will suffice. However, there are two instances where the LAPLACE TRANFORM will either significantly cut down on our effort, or flat make it possible to find a solution. These two scenarios are: 1. When the DRIVING FUNCTION of a HIGHER ORDER NONHOMOGENEOUS DIFFERENTIAL EQUATION is so complex that a starting PARTICULAR FORMULA can not be found in the table that is provided to us on page 31 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing.



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2. When the DRIVING FUNCTION is a PIECEWISE FUNCTION of a defined INTERVAL. In this problem, we are given the function: 𝑓 𝑡 = 8 + 4𝑡 + 𝑡 ! − 4𝑡 ! We are not presented this function as a DRIVING FUNCTION of a HIGHER ORDER NONHOMOGENEOUS DIFFERENTIAL EQUATION, but rather, just simply as a function. We will carry out the process of determining the LAPLACE TRANSFORM, but let’s not lose sight of what we are doing. Carrying out the process will be equivalent to defining the PARTICULAR SOLUTION of HIGHER ORDER NONHOMOGENEOUS DIFFERENTIAL EQUATION if that’s what we are given. Although the term itself may incite some uncomfortable feelings, LAPLACE TRANSFORMS can essentially break any DIFFERENTIAL EQUATION down in to a simple ALGEBRAIC expression for us to solve. The algebra of deriving LAPLACE TRANSFORMS can be very gritty and convoluted, we could spend time on doing each derivation, but it wouldn’t be good use for our time. Fortunately, we are provided a TABLE of the most common LAPLACE TRANFORMS, and the ones that we should expect to need and use on the exam, on page 34 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing.



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A LAPLACE TRANSFORM can be presented using the NOMENCLATURE: ℒ 𝑓(𝑡) = 𝐹 𝑠 Or in our case: ℒ 8 + 4𝑡 + 𝑡 ! − 4𝑡 ! = 𝐹 𝑠 The LAPLACE TRANSFORM will produce a new function of the variable “s”, where all the “t” variables in the original function will drop out through the INTEGRATION process. This INTEGRATION, or more properly stated, the LAPLACE TRANFORM, is generally written as: !

𝑓 𝑡 𝑒 !!" 𝑑𝑡

𝐹 𝑠 = !!

And in our case: !

(8 + 4𝑡 + 𝑡 ! − 4𝑡 ! )𝑒 !!" 𝑑𝑡

𝐹 𝑠 = !!

This GENERAL FORMULA for the LAPLACE TRANSFORM can be referenced under the topic of MATHEMATICS on page 34 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing.

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The LAPLACE TRANSFORM is a LINEAR OPERATOR, so when given a function with multiple terms, as we are here, we can break it down term by term to determine what the complete transform will look like. Therefore, we can rewrite our LAPLACE TRANSFORM as: ℒ 8 + ℒ 4𝑡 + ℒ 𝑡 ! + ℒ −4𝑡 ! = 𝐹 𝑠 Which can also be written as: 8ℒ 1 + 4ℒ 𝑡 + ℒ 𝑡 ! − 4ℒ 𝑡 ! = 𝐹 𝑠 Remember, we might need to tweak our terms to fit in to the mold we are originally used to seeing. As long as the integrity of the function and the process as a whole process remains, we can do this all day long. We are essentially skipping past expanding on the INTEGRATION form of a LAPLACE TRANSFORM, however, if called to do so, it’s as simple as plugging and playing the appropriate terms in to the general layout. The only time we will need write out the INTEGRATION is if we are deriving the LAPLACE TRANSFORM from scratch. This algebra can get a bit convoluted, and very very time consuming…so we will rely on the TABLES…at least to the extent which we can.



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The TABLE of COMMON LAPLACE TRANSFORMS can be referenced under the topic of MATHEMATICS on page 34 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. This table provides us the most common LAPLACE TRANSFORMS that we will need and encounter on the day of our exam. Some useful LAPLACE TRANSFORMS are represented as: 𝑓(𝑡)

𝐹(𝑠)

𝛿 𝑡 , 𝐼𝑚𝑝𝑢𝑙𝑠𝑒 𝑎𝑡 𝑡 = 0

1

𝑢 𝑡 , 𝑆𝑡𝑒𝑝 𝑎𝑡 𝑡 = 0

1/𝑠

𝑡 𝑢 𝑡 , 𝑅𝑎𝑚𝑝 𝑎𝑡 𝑡 = 0

1/𝑠 !

𝑒 !!"

1/(𝑠 + 𝛼)

𝑡𝑒 !!"

1/ 𝑠 + 𝛼

!

𝑒 !!" sin 𝛽𝑡

𝛽/ 𝑠 + 𝛼

𝑒 !!" cos 𝛽𝑡

(𝑠 + 𝛼)/ 𝑠 + 𝛼

!

+ 𝛽! !

+ 𝛽!

!!! !

𝑠 𝐹 𝑠 −

𝑑! 𝑓 𝑡 𝑑𝑡 !

𝑠

!!!!!

!!!

!

𝑑 ! 𝑓(0) 𝑑𝑡 !

(1/𝑠)𝐹(𝑠)

𝑓 𝜏 𝑑𝜏 ! !

𝑥 𝑡 − 𝜏 ℎ(𝜏) 𝑑𝜏

𝐻 𝑠 𝑋(𝑠)

!

𝑓 𝑡 − 𝜏 𝑢(𝑡 − 𝜏)

𝑒 !!" 𝐹(𝑠)

limit 𝑓(𝑡)

limit 𝑠𝐹(𝑠)

!→!



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limit 𝑓(𝑡)

limit 𝑠𝐹(𝑠)

!→!

!→!

Revisiting our function written in LAPLACE TRANSFORM terms, we have: 8ℒ 1 + 4ℒ 𝑡 + ℒ 𝑡 ! − 4ℒ 𝑡 ! = 𝐹 𝑠 We will cycle term by term creating each TRANSFORM and combining them along the way. Let’s start with: 8ℒ 1 This is just a step function, and referencing the table, we match this term up with the row: 𝑓(𝑡)

𝐹(𝑠)

𝑢 𝑡 , 𝑆𝑡𝑒𝑝 𝑎𝑡 𝑡 = 0

1/𝑠

This allows us to conclude that: 8ℒ 1 = 8(1/𝑠) Or: ℒ 8 = 8/𝑠

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Cycling to our next term, we have: 4ℒ 𝑡 Referencing the table, we match this term up with the row: 𝑓(𝑡)

𝐹(𝑠)

𝑡 𝑢 𝑡 , 𝑅𝑎𝑚𝑝 𝑎𝑡 𝑡 = 0

1/𝑠 !

This allows us to conclude that: 4ℒ 𝑡 = 4(1/𝑠 ! ) Or: ℒ 4𝑡 = 4/𝑠 ! The next term in our sequence is: ℒ 𝑡! Referencing the table, we can’t directly relate this term to one that is given, however, this is a VERY COMMON transform that should be included…and certainly one we should memorize.



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Generally speaking: ℒ 𝑡 ! = 𝑛!/𝑠 !!! This allows us to conclude: ℒ 𝑡 ! = 2!/𝑠 !!! Or:

ℒ 𝑡! =

2! 𝑠!

Lastly, we have the term: −4ℒ 𝑡 ! Again, referencing the table, we can’t directly relate this term to one that is given, however, this is a VERY COMMON transform and is equivalent to the previous term, despite the exponent being larger. Recall that, generally speaking: ℒ 𝑡 ! = 𝑛!/𝑠 !!!



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With n = 3 in this case, we can conclude that: −4ℒ 𝑡 ! = −4(3!/𝑠 !!! ) Or:

ℒ −4𝑡 ! = −4

3! 𝑠!

Now the final step is to gather all of our LAPLACE TRANFORMS and combine them together in to a single solution, giving us:

𝐹 𝑠 =

8 4 2! 4 3! + + − ! 𝑠 𝑠! 𝑠! 𝑠

Simplifying we get,

𝐹 𝑠 =

8 4 2 24 + !+ !− ! 𝑠 𝑠 𝑠 𝑠 𝟖

𝟒

𝟐

𝟐𝟒

The correct answer choice is C. 𝒔 + 𝒔𝟐 + 𝒔𝟑 + 𝒔𝟒



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by Prepineer | Prepineer.com