# Mahesh's question via email about Laplace Transforms (1)

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In summary, Laplace Transforms are mathematical tools used to transform a function of time into a function of complex frequency. They are calculated using an integral formula and have many applications in engineering, physics, and mathematics. However, there are limitations to using them and further learning can be done through various resources and consulting with experts.
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$\displaystyle x\left( t \right)$ and $\displaystyle y\left( t \right)$ satisfy the following system of differential equations:

$\displaystyle \begin{cases} \frac{\mathrm{d}x}{\mathrm{d}t} + x + 6\,y = 6 \\ \frac{\mathrm{d}y}{\mathrm{d}t} + 9\,x + y = 0 \end{cases}, \quad x \left( 0 \right) = y \left( 0 \right) = 0$

Find the Laplace Transform of $\displaystyle y\left( t \right)$.

Start by taking the Laplace Transform of both equations, which gives

$\displaystyle \begin{cases} s\,X\left( s \right) - s\,x\left( 0 \right) + X\left( s \right) + 6\,Y\left( s \right) = \frac{6}{s} \\ s\,Y\left( s \right) - s\,y\left( 0 \right) + 9\,X\left( s \right) + Y\left( s \right) = 0 \end{cases}$

$\displaystyle \begin{cases} \left( s + 1 \right) X\left( s \right) + 6\,Y\left( s \right) = \frac{6}{s} \\ 9\,X\left( s \right) + \left( s + 1 \right) Y\left( s \right) = 0 \end{cases}$

From the second equation in the system, we have

\displaystyle \begin{align*} 9\,X\left( s \right) &= -\left( s + 1 \right) Y\left( s \right) \\ X\left( s \right) &= -\left( \frac{s + 1}{9} \right) Y\left( s \right) \end{align*}

Substituting into the first equation in the system gives

\displaystyle \begin{align*} \left( s + 1 \right) \left[ -\left( \frac{s + 1}{9} \right) \right] Y\left( s \right) + 6\,Y\left( s \right) &= \frac{6}{s} \\ \left[ 6 -\frac{\left( s + 1 \right) ^2 }{9} \right] Y\left( s \right) &= \frac{6}{s} \\ \left[ \frac{54 - \left( s + 1 \right) ^2 }{9} \right] Y\left( s \right) &= \frac{6}{s} \\ Y\left( s \right) &= \frac{54}{s\left[ 54 - \left( s + 1 \right) ^2 \right]} \end{align*}

In Weblearn this would be entered as

54/( s*( 54 - (s + 1)^2 ) )

benorin and shivajikobardan
Looks like it’s correct to me.

## 1. What are Laplace Transforms and how are they used in science?

Laplace Transforms are mathematical operations that are used to transform a function from the time domain to the frequency domain. They are commonly used in science to solve differential equations, which are used to model and understand various phenomena in nature.

## 2. How do Laplace Transforms differ from other types of mathematical transforms?

Laplace Transforms differ from other types of mathematical transforms in that they are defined for both continuous and discrete functions, while other transforms may only be defined for one or the other. Additionally, Laplace Transforms are particularly useful for solving differential equations, while other transforms may be used for different purposes.

## 3. What are some real-world applications of Laplace Transforms?

Laplace Transforms have numerous real-world applications, including in electrical engineering, control systems, signal processing, and physics. They are often used to model and analyze systems with complex dynamics, such as circuits, mechanical systems, and chemical reactions.

## 4. Can Laplace Transforms be used to solve any type of differential equation?

While Laplace Transforms are a powerful tool for solving many types of differential equations, they may not be suitable for all types. For example, they may not be effective for equations with discontinuous or singular functions. In these cases, other methods may be more appropriate.

## 5. Are there any limitations to using Laplace Transforms in scientific research?

Like any mathematical tool, Laplace Transforms have some limitations. They may not be suitable for all types of differential equations, as mentioned previously. Additionally, they may be computationally intensive for more complex equations, and their application may require specialized knowledge and skills.

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