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Chapter 4: Problem 44

The Laplace transform of the given equation is \\[ \mathscr{L}\\{t\\}-2 \mathscr{L}\\{f\\}=\mathscr{L}\left\\{e^{t}-e^{-t}\right\\} \mathscr{L}\\{f\\} \\] Solving for \(\mathscr{L}\\{f\\}\) we obtain \\[ \mathscr{L}\\{f\\}=\frac{s^{2}-1}{2 s^{4}}=\frac{1}{2} \frac{1}{s^{2}}-\frac{1}{12} \frac{3 !}{s^{4}} \\] Thus \\[ f(t)=\frac{1}{2} t-\frac{1}{12} t^{3} \\]

Short Answer

Expert verified
The function is \( f(t) = \frac{1}{2} t - \frac{1}{12} t^3 \).

Step by step solution

01

Analyze the Given Equation

We start with the given equation involving the Laplace transforms: \[ \mathscr{L}\{t\} - 2 \mathscr{L}\{f\} = \mathscr{L}\{e^{t} - e^{-t}\} \mathscr{L}\{f\} \]. Identify the goal, which is to solve for \( \mathscr{L}\{f\} \).
02

Simplify Each Term

Identify the Laplace transform for each component: \( \mathscr{L}\{t\} \) is \( \frac{1}{s^2} \) and \( \mathscr{L}\{e^{t} - e^{-t}\} \) is \( \frac{1}{s-1} - \frac{1}{s+1} \). Expand the equation in these terms.
03

Rearrange the Equation

Plug the simplified Laplace transforms into the equation: \( \frac{1}{s^2} - 2L = \left( \frac{1}{s-1} - \frac{1}{s+1} \right) L \). Rearrange to isolate \( \mathscr{L}\{f\} \) (denoted as \( L \)) on one side.
04

Factor the Equation

Combine and factor the terms: \[ L \left(2 + \frac{1}{s-1} - \frac{1}{s+1}\right) = \frac{1}{s^2} \]. Simplify further to isolate \( L \).
05

Solve for L

Simplify the factor: \( L (2 + \frac{2s}{s^2 - 1}) = \frac{1}{s^2} \). Solve for \( L \) by dividing both sides by \( 2 + \frac{2s}{s^2 - 1} \). Ultimately, solve \( L = \frac{s^2 - 1}{2s^4} \).
06

Break Down Partial Fractions

Decompose \( \frac{s^2 - 1}{2s^4} \) using partial fractions. This gives: \( \frac{1}{2} \left( \frac{1}{s^2} \right) - \frac{1}{12} \frac{3!}{s^4} \).
07

Perform Inverse Laplace Transform

Apply the inverse Laplace transform to \( \frac{1}{2}\frac{1}{s^2} - \frac{1}{12} \frac{3!}{s^4} \). The inverse results are: \( \frac{1}{2} t \) from \( \frac{1}{s^2} \) and \( -\frac{1}{12} t^3 \) from \( \frac{3!}{s^4} \).
08

Conclude with the Final Form

Combine the results from the inverse Laplace transform to obtain the final form of the function: \( f(t) = \frac{1}{2} t - \frac{1}{12} t^3 \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Partial Fractions
Partial fraction decomposition is a technique often used in calculus and algebra to simplify complex rational expressions, making them easier to manage, especially when applying the Laplace Transform. When given a fraction, the goal is to express it as a sum of simpler fractions that are easier to integrate or transform. For example, when dealing with the fraction \( \frac{s^2 - 1}{2s^4} \), we can decompose it into separate terms using partial fraction decomposition. Here's how it works:
  • Write the complex fraction as a combination of simpler fractions whose denominators are factors of the original denominator.
  • In our exercise, \( \frac{s^2 - 1}{2s^4} \) is decomposed into \( \frac{1}{2} \cdot \frac{1}{s^2} - \frac{1}{12} \cdot \frac{3!}{s^4} \).
  • The coefficients for these simpler fractions are determined by equating coefficients from both sides of the equation after clearing the denominators.
Partial fractions are particularly useful in inverse Laplace transforms, as they help separate and individually address each part of the expression.
Inverse Laplace Transform
The inverse Laplace transform is a powerful tool used to revert a function from its Laplace transform back to its original time-domain form. This is particularly useful in solving differential equations and system dynamics problems where finding the original function description of the system is desired. To perform an inverse Laplace transform, follow these steps:
  • Identify the terms of the transformed function. Each term corresponds to a familiar form whose inverse is known.
  • For instance, in our problem, the terms \( \frac{1}{2}\cdot\frac{1}{s^2} \) and \( -\frac{1}{12} \cdot \frac{3!}{s^4} \) are each separately transformed back to their time-domain equivalents.
  • The inverse Laplace of \( \frac{1}{s^2} \) is known to be \( t \) based on standard Laplace transform tables.
  • Thus, \( \frac{1}{2}\cdot\frac{1}{s^2} \) inverts to \( \frac{1}{2}t \).
  • Similarly, \( \frac{3!}{s^4} \) inverts to \( t^3 \), giving us \( -\frac{1}{12}t^3 \) from \( -\frac{1}{12}\cdot\frac{3!}{s^4} \).
Combining these results, we find the original function \( f(t) = \frac{1}{2}t - \frac{1}{12}t^3 \). This step is key in converting a system response from the frequency domain back to the time domain.
Differential Equations
Differential equations represent a fundamental aspect of mathematics and engineering, used to describe systems and phenomena that change over time. A differential equation involves functions and their derivatives, expressing the rate of change in dependent variables concerning independent variables. There are several key techniques for solving differential equations:
  • The Laplace transform is one particularly powerful technique, allowing us to transform a differential equation into an algebraic equation.
  • This step greatly simplifies solving thanks to the ease of algebraic manipulations compared to differential calculus, especially for systems with initial conditions.
  • In our example, we transformed the differential equation into the s-domain (Laplace domain) using known transforms of time functions like \(t\) and exponential functions.
  • Once solved for \( \mathscr{L}\{f\} \), we used the inverse Laplace transform to revert it back into its time representation \( f(t) = \frac{1}{2}t - \frac{1}{12}t^3 \).
This process illustrates how Laplace transforms pair with differential equations to streamline the resolution process, providing clear and often more comprehensible solutions for dynamic systems.

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Most popular questions from this chapter

The Laplace transform of the initial-value problem is \\[s \mathscr{L}\\{y\\}-y(0)+6 \mathscr{L}\\{y\\}=\frac{1}{s-4}\\]. Solving for \(\mathscr{L}\\{y\\}\) we obtain \\[\mathscr{L}\\{y\\}=\frac{1}{(s-4)(s+6)}+\frac{2}{s+6}=\frac{1}{10} \cdot \frac{1}{s-4}+\frac{19}{10} \cdot \frac{1}{s+6}\\] Thus \\[y=\frac{1}{10} e^{4 t}+\frac{19}{10} e^{-6 t}\\].

$$\mathscr{L}-1\left\\{\frac{0.9 s}{(s-0.1)(s+0.2)}\right\\}=\mathscr{L}^{-1}\left\\{(0.3) \cdot \frac{1}{s-0.1}+(0.6) \cdot \frac{1}{s+0.2}\right\\}=0.3 e^{0.1 t}+0.6 e^{-0.2 t}$$

The differential equation is $$E I \frac{d^{4} y}{d x^{4}}=w_{0}[1-\mathscr{U}(x-L / 2)].$$ Taking the Laplace transform of both sides and using \(y(0)=y^{\prime}(0)=0\) we obtain $$s^{4} \mathscr{L}\\{y\\}-s y^{\prime \prime}(0)-y^{\prime \prime \prime}(0)=\frac{w_{0}}{F I} \frac{1}{s}\left(1-e^{-L s / 2}\right).$$ Letting \(y^{\prime \prime}(0)=c_{1}\) and \(y^{\prime \prime \prime}(0)=c_{2}\) we have $$\mathscr{L}\\{y\\}=\frac{c_{1}}{s^{3}}+\frac{c_{2}}{s^{4}}+\frac{w_{0}}{E I} \frac{1}{s^{5}}\left(1-e^{-L s / 2}\right).$$ so that $$y(x)=\frac{1}{2} c_{1} x^{2}+\frac{1}{6} c_{2} x^{3}+\frac{1}{24} \frac{w_{0}}{E I}\left[x^{4}-\left(x-\frac{L}{2}\right)^{4} q\left(x-\frac{L}{2}\right)\right].$$ To find \(c_{1}\) and \(c_{2}\) we compute $$y^{\prime \prime}(x)=c_{1}+c_{2} x+\frac{1}{2} \frac{w_{0}}{E I}\left[x^{2}-\left(x-\frac{L}{2}\right)^{2} \mathscr{U}\left(x-\frac{L}{2}\right)\right].$$ Then \(y(L)=y^{\prime \prime}(L)=0\) yields the system $$\begin{aligned} \frac{1}{2} c_{1} L^{2}+\frac{1}{6} c_{2} L^{3}+\frac{1}{24} \frac{w_{0}}{E I}\left[L^{4}-\left(\frac{L}{2}\right)^{4}\right]=\frac{1}{2} c_{1} L^{2}+\frac{1}{6} c_{2} L^{3}+\frac{5 w_{0}}{128 E I} L^{4}=0 \\ c_{1}+c_{2} L+\frac{1}{2} \frac{w_{0}}{E I}\left[L^{2}-\left(\frac{L}{2}\right)^{2}\right]=c_{1}+c_{2} L+\frac{3 w_{0}}{8 E I} L^{2}=0. \end{aligned}$$ Solving for \(c_{1}\) and \(c_{2}\) we obtain \(c_{1}=\frac{9}{128} w_{0} L^{2} / E I\) and \(c_{2}=-\frac{57}{128} w_{0} L / E I .\) Thus $$y(x)=\frac{w_{0}}{E I}\left[\frac{9}{256} L^{2} x^{2}-\frac{19}{256} L x^{3}+\frac{1}{24} x^{4}-\frac{1}{24}\left(x-\frac{L}{2}\right)^{4} q\left(x-\frac{L}{2}\right)\right].$$

$$\mathscr{L}\\{\sinh k t\\}=\frac{1}{2} \mathscr{L}\left\\{e^{k t}-e^{-k t}\right\\}=\frac{1}{2}\left[\frac{1}{s-k}-\frac{1}{s+k}\right]=\frac{k}{s^{2}-k^{2}}$$

From Problem 13 we know that $$y=\frac{1}{2} y^{\prime \prime}(0) x^{2}+\frac{1}{6} y^{\prime \prime \prime}(0) x^{3}+\frac{1}{6} \frac{P_{0}}{E I}\left(x-\frac{L}{2}\right)^{3} q\left(x-\frac{L}{2}\right)$$ Using \(y(L)=0\) and \(y^{\prime}(L)=0\) we obtain $$y=\frac{1}{16} \frac{P_{0} L}{E I} x^{2}-\frac{1}{12} \frac{P_{0}}{E I} x^{3}+\frac{1}{6} \frac{P_{0}}{E I}\left(x-\frac{L}{2}\right)^{3} \mathscr{U}\left(x-\frac{L}{2}\right)$$ $$=\left\\{\begin{array}{ll}\frac{P_{0}}{E I}\left(\frac{L}{16} x^{2}-\frac{1}{12} x^{3}\right), & 0 \leq x<\frac{L}{2} \\\\\frac{P_{0}}{E I}\left(\frac{L}{16} x^{2}-\frac{1}{12} x^{3}\right)+\frac{1}{6} \frac{P_{0}}{E I}\left(x-\frac{L}{2}\right)^{3}, & \frac{L}{2} \leq x \leq L\end{array}\right.$$
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