Solve a Laplace transform puzzle

In summary, there was a discrepancy between two methods used to find the Laplace transform of a given function. The first method used the standard Laplace formula, while the second used the direct integral formula. The second method was found to be incorrect due to a mistake in handling the unit step function in the integral form. The correct answer was found using the first method.
  • #1
davyyao
4
0
Hi~
I recently solve a Laplace transform problem as following
L[int{t,0}cosh(t'-1)U(t'-1)dt']=? U(t'-1) is the unit step function(=1 for t'>1, =zero otherwise)
According the standard Laplace formula :
(1)L[cosh(t-1)U(t-1)]=exp(-s)*s/(s^2-1);
(2)L(int{t,0}f(t')dt')=F(s)/s. where F(s)=L(f(t))

I conclude the answer to be:
exp(-s)/(s^2-1)

But from the direct integral formula:
L[int{t,0}cosh(t'-1)U(t'-1)dt']=L[sinh(t-1)]=(1/2)*(exp(-1)/(s-1)-exp(1)/(s+1)).

This two answers are not consistent!
I totally have no idea what's wrong with the two methods?

I believe the second method should be correct by basic definition of Laplace transformation.

Thanks very much!
 
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  • #2
How did you find the direct integral (the step at the first equal sign)?
 
  • #3
davyyao said:
Hi~
I recently solve a Laplace transform problem as following
L[int{t,0}cosh(t'-1)U(t'-1)dt']=? U(t'-1) is the unit step function(=1 for t'>1, =zero otherwise)
According the standard Laplace formula :
(1)L[cosh(t-1)U(t-1)]=exp(-s)*s/(s^2-1);
(2)L(int{t,0}f(t')dt')=F(s)/s. where F(s)=L(f(t))

I conclude the answer to be:
exp(-s)/(s^2-1)

But from the direct integral formula:
L[int{t,0}cosh(t'-1)U(t'-1)dt']=L[sinh(t-1)]=(1/2)*(exp(-1)/(s-1)-exp(1)/(s+1)).

This two answers are not consistent!
I totally have no idea what's wrong with the two methods?

I believe the second method should be correct by basic definition of Laplace transformation.

Thanks very much!

[tex] \int_0^t \cosh(t'-1) U(t'-1) \, dt' \neq \sinh(t-1) [/tex]
because the integral vanishes if t < 1 but sinh(t-1) does not.
 
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  • #4
[tex]\int_0^t cosh(t'- 1)U(t'-1)dt'= \int_1^t cos(t'- 1)dt'[/tex]
and with the substitution u= t'- 1, du= dt', when t'= 1, u= 0 and when t'= t, u= t- 1 so that becomes
[tex]\int_0^{t-1} cosh(u)du[/tex].
 
  • #5
Thanks Ray Vickson and Hallsoflvy for nice answering.
I think I make mistake to deal with the unit step function in the integral form.
It clear now for me!
 

1. What is a Laplace transform puzzle?

A Laplace transform puzzle is a mathematical problem that involves finding the Laplace transform of a given function or solving an inverse Laplace transform. It is a common topic in calculus and differential equations courses.

2. How do you solve a Laplace transform puzzle?

To solve a Laplace transform puzzle, you need to follow a specific set of steps. First, you need to take the Laplace transform of the given function using the appropriate formula. Then, you can use algebraic manipulation and tables of Laplace transforms to simplify the expression. Finally, you can use inverse Laplace transform formulas to find the solution.

3. What are the benefits of using Laplace transforms?

Laplace transforms are useful in solving differential equations and other mathematical problems because they transform a function from the time domain to the frequency domain. This can make solving complex problems easier and more efficient.

4. Do I need to memorize Laplace transform formulas?

While it is helpful to have some of the most commonly used Laplace transform formulas memorized, it is not necessary. Most Laplace transform puzzles will provide the necessary formulas, and you can also refer to a table of Laplace transforms for reference.

5. Are there any tips for solving Laplace transform puzzles?

One helpful tip for solving Laplace transform puzzles is to practice regularly and become familiar with the common formulas and techniques. It can also be helpful to break down the problem into smaller steps and use algebraic manipulation to simplify the expression before attempting to find the solution.

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