# Are these two inverse Laplace transform solutions equivalent?

• Jrlinton
In summary, the conversation is discussing the equivalence of two solutions for the equation Y=(8s-4)/(s²-4). The speaker has rearranged the right side and used the Laplace transform chart to identify the transforms of hyperbolic sine and hyperbolic cosine. They have obtained a solution of y=8cosh(2t)-2sinh(2t), while the instructor's solution is y=3e^(2t)+5e(-2t). The speaker is wondering if these two solutions are equivalent or if there was a mistake. The expert suggests working it out for oneself to learn.
Jrlinton

Y=(8s-4)/(s²-4)

## The Attempt at a Solution

I rearranged the right side as:
8*(s/(s²-2²))-2*(2/(s²-2²))

Using the Laplace transform chart given in the class I was able to identify these as the transforms of hyperbolic sine and hyperbolic cosine making the inverse Laplace:
y=8cosh(2t)-2sinh(2t)
I checked this answer by working backwards and it all checks out but the worked out problem given by the instructor states the final answer as:
y=3e^(2t)-5e(-2t)=3[e][/2t]-5[e[/-2t]

I am wondering if these two solutions are equivalent?

Correction: The given solution was y=3e^(2t)+5e(-2t).
I understand how he got this by using partial fractions, I would just like to know if the two functions are equivalent or if there was a mistake.

Jrlinton said:
Correction: The given solution was y=3e^(2t)+5e(-2t).
I understand how he got this by using partial fractions, I would just like to know if the two functions are equivalent or if there was a mistake.

You do not need to ask us; you can work it out for yourself---and you should: that is how you will learn. Just expand sinh and cosh in terms of exp and see what you get.

Orodruin

## 1. What is an inverse Laplace transform?

An inverse Laplace transform is a mathematical operation that is used to find the original function (or signal) from its Laplace transform. It is the inverse operation of the Laplace transform, which transforms a function from the time domain to the frequency domain.

## 2. How is the inverse Laplace transform calculated?

The inverse Laplace transform is calculated using the following formula:f(t) = (1/2πi) ∫F(s)e^(st)dswhere f(t) is the original function, F(s) is its Laplace transform, and the integral is taken along a vertical line in the complex plane. This formula is also known as the Bromwich integral.

## 3. What are some common properties of the inverse Laplace transform?

Some common properties of the inverse Laplace transform include linearity, time shifting, differentiation, and integration. These properties can be used to simplify the calculation of the inverse Laplace transform in certain cases.

## 4. What are some applications of the inverse Laplace transform?

The inverse Laplace transform has numerous applications in mathematics, physics, engineering, and other fields. It is commonly used to solve differential equations, analyze signals and systems, and model physical phenomena such as heat transfer and fluid flow.

## 5. Is there a way to calculate the inverse Laplace transform without using the Bromwich integral?

Yes, there are alternative methods for calculating the inverse Laplace transform, such as the Fourier-Mellin transform, the convolution theorem, and the partial fraction expansion. These methods can be used to solve specific types of problems or to simplify the calculation in certain cases.

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