# Laplace Transforms - Convolution Theorem

In summary, the conversation discusses using the Laplace transform to solve an initial-value problem involving a second-order differential equation and a cosine function. The conversation includes relevant equations and the attempt at a solution, with the final answer being y = 2cos(3t) + 5/3sin(3t) + 1/6tsin(3t).
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## Homework Statement

Use the Laplace transform to solve the given initial-value problem.

$$y'' + 9y = cos(3t), y(0)=2, y'(0)=5$$

## Homework Equations

$$L\left\{y''\right\} = s^{2}Y(s)-sy(0)-y'(0)$$

$$L\left\{y\right\} = Y(s)$$

$$L\left\{cos(kt)\right\} = \frac{s^{2}}{s^{2} + k^{2}}, s>0$$

$$L\left\{sin(kt)\right\} = \frac{k}{s^{2} + k^{2}}, s>0$$

$$L^{-1}\left\{F(s-a)\right\} = e^{at}f(t)$$ (shifting property)

$$L\left\{F(s)G(s)\right\} = f \ast g$$ (convolution of f and g)

## The Attempt at a Solution

$$y'' + 9y = cos(3t), y(0)=2, y'(0)=5$$

$$L\left\{y''\right\} + 9L\left\{y\right\} = L\left\{cos(3t)\right\}$$

$$\left[s^{2}Y(s)-sy(0)-y'(0)\right] + 9Y(s) = \frac{s}{s^{2}+9}$$

$$Y(s)\left[s^{2} - 2s + 9\right] = 5 + \frac{s}{s^{2} + 9}$$

$$Y(s) = \frac{5}{s^{2} - 2s + 9} + \frac{s}{(s^{2} - 2s + 9)(s^{2} + 9)}$$

$$\frac{5}{(s-1)^{2} + 8} + \frac{s}{\left[(s-1)^{2} + 8)\right](s^{2} + 9)}$$

$$y(t) = \frac{5}{2\sqrt{2}}L^{-1}\left\{\frac{2\sqrt{2}}{s^{2} + (2\sqrt{2})^{2}}_{s\rightarrow(s-1)}\right\} + L\left\{F(s)G(s)\right\}$$ where $$F(s) = \frac{s}{(s-1)^{2} + 8}$$ and $$G(s) = \frac{1}{s^{2} + 9}$$

$$f(t) = L^{-1}\left\{\frac{(s-1) + 1}{(s-1)^{2} + (2\sqrt{2})^{2}}\right\} = L^{-1}\left\{\frac{s}{s^{2} + (2\sqrt{2})^{2}}_{s\rightarrow(s-1)}\right\} + \frac{1}{2\sqrt{2}}L^{-1}\left\{\frac{2\sqrt{2}}{s^{2} + (2\sqrt{2})^{2}}_{s\rightarrow(s-1)}\right\} = e^{t}cos(2\sqrt{2}t) + \frac{1}{2\sqrt{2}}\left[e^{t}sin(2\sqrt{2}t)\right]$$

$$g(t) = \frac{1}{3}L^{-1}\left\{\frac{3}{s^{2} + 3^{2}}\right\} = \frac{1}{3}sin(3t)$$

$$L^{-1}\left\{F(s)G(s)\right\} = f \ast g$$

And I get stuck here. The book's answer is:

$$y = 2cos(3t) + \frac{5}{3}sin(3t) + \frac{1}{6}tsin(3t)$$

Nevermind... Got it.

It was a good exercise in Latex, though. Count your positives! Happy Easter everyone.

## 1. What is the Laplace transform?

The Laplace transform is a mathematical tool used to transform a function from the time domain to the frequency domain. It is defined as the integral of a function multiplied by an exponential function, and is often used to solve differential equations.

## 2. What is the Convolution Theorem?

The Convolution Theorem states that the Laplace transform of a convolution of two functions is equal to the product of their individual Laplace transforms. In other words, it allows us to simplify the Laplace transform of a convolution to a multiplication of transforms.

## 3. How is the Convolution Theorem used in practice?

The Convolution Theorem is used in various engineering and scientific applications, such as signal processing and control systems. It allows us to easily solve differential equations by transforming them into algebraic equations, which can then be solved using standard mathematical techniques.

## 4. What are the advantages of using Laplace transforms and the Convolution Theorem?

Laplace transforms and the Convolution Theorem provide a powerful and efficient way to solve complex problems involving differential equations. They allow us to analyze systems in the frequency domain, which can provide insights into the behavior of a system that may not be apparent in the time domain.

## 5. Are there any limitations to using Laplace transforms and the Convolution Theorem?

While Laplace transforms and the Convolution Theorem have many advantages, they are not suitable for all types of functions. The function must be well-defined and have a finite Laplace transform for the theorem to be applicable. Additionally, care must be taken when dealing with discontinuous or non-causal functions, as the convolution may not exist in these cases.

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