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Chapter 8: Problem 25

\(\mathcal{L}\left\\{t^{2} e^{-t / 2}\right\\}\)

Short Answer

Expert verified
\[ \mathcal{L}{\left\{t^{2} e^{-t / 2}\right\}} = \frac{2}{(s + 1/2)^{3}} \]

Step by step solution

01

Identify the components of the function

The function to be transformed is \(t^2 e^{-t / 2}\). Observe that it is a product of \(t^2\) and \(e^{-t / 2}\).
02

Use the Time Shifting Property of Laplace Transform

Recall the time-shifting property which is \(\text{ if } \ \mathcal{L} \ {f(t)} = F(s) \text{ and if } \mathcal{L} \ {e^{at} f(t)} = F(s-a)\). Here \(a = - \frac{1}{2}\).
03

Find Laplace Transform of simpler function

First consider \(f(t) = t^2\). Use the standard Laplace Transform: \( \mathcal{L} {t^n} = \frac{n!}{s^{n+1}}\). For \t^2\, it will be \ \mathcal{L} {t^2} = \ \frac{2!}{s^{3}} = \ \frac{2}{s^3}.\
04

Apply the Time Shifting Property

Next, apply the time-shifting property to get the final Laplace Transform. According to the property, \mathcal{L} {t^2 e^{-t/2}}\ is given by \mathcal{L} {t^2} \ evaluated at \(s + \frac{1}{2}\). Hence, \ \mathcal{L} {t^2 e^{-t/2}}\ = \frac{2}{(s + 1/2)^3}.\

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Key Concepts

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

Time-Shifting Property
The time-shifting property is a powerful tool in Laplace Transformations. It helps to handle exponential functions that multiply another function. This property states: if \(\mathcal{L} \{f(t)\} = F(s) \)and if \(\mathcal{L} \{e^{at} f(t)\} = F(s-a)\).
Here, multiplying a function by an exponential \(e^{at}\) shifts its Laplace Transform on the s-axis by \(a\).
For instance, if the original Laplace Transform is \(F(s)\) and \(a = -\frac{1}{2}\), shifting occurs by adding \(\frac{1}{2}\) to \(s\).
This gives a new transform \(F(s + 1/2)\).
In our problem, applying this to \(e^{-t/2} t^2\) leads to transforming \(t^2\) first, then shifting it, yielding \(\frac{2}{(s+1/2)^3}\).
Transform of Polynomial Functions
Polynomial functions are among the most common types of functions addressed using Laplace Transforms. For a polynomial function of the form \(t^n\), there exists a standard transform formula: \(\mathcal{L} \{t^n\} = \frac{n!}{s^{n+1}}\).
This makes it straightforward to transform any polynomial.
For our function \(t^2\), using this formula: \(\mathcal{L} \{t^2\} = \frac{2!}{s^{3}} = \frac{2}{s^3}\).

This standard solution serves as the building block before integrating other properties, like time-shifting, to derive the final result for more complex functions.
Laplace Transform Table
A Laplace Transform Table is a valuable reference that lists commonly used transforms and their corresponding results. This saves considerable time and effort when working on Laplace Transforms.
Key entries in such a table include:
  • \(\mathcal{L}\{1\} = \frac{1}{s}\)
  • \(\mathcal{L}\{t^n\} = \frac{n!}{s^{n+1}}\)
  • \(\mathcal{L}\{e^{at}\} = \frac{1}{s-a}\)
Using this table, you can quickly identify and apply necessary transforms to different parts of a function.
A combination of entries from the table, like the ones for polynomials and exponentials, often solves complicated transforms much faster, enhancing efficiency in computations.

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Most popular questions from this chapter

Solve the \(L-R-C\) equation for \(I(t)\) if \(L=\) \(1 \mathrm{H}, R=6 \Omega, C=1 / 9 \mathrm{~F}, E(t)=100 \sin t \mathrm{~V}\), and \(I(0)=0\).

Solve the problem of the forced coupled spring-mass system with \(m_{1}=m_{2}=1, k_{1}=\) 3 , and \(k_{2}=2\) if the forcing functions are \(F_{1}(t)=1\) and \(F_{2}(t)=\sin t\) and the initial conditions are \(x(0)=y^{\prime}(0)=0, x^{\prime}(0)=\) \(y(0)=1\). Graph the solution parametrically as well as simultaneously. How does the motion differ from that of Example 8.40? What eventually happens to this system? Will the objects eventually come to rest?

Suppose that an object with mass \(m=1\) is attached to the end of a spring with spring constant 16. If there is no damping and the spring is subjected to the forcing function \(f(t)=\sin t\), determine the motion of the spring if at \(t=1\), the spring is supplied with an upward shock of 4 units.

\(\frac{3+e^{4 s}}{s e^{6 s}}\)

\(x^{\prime}=x-y+f(t), y^{\prime}=2 x-y, x(0)=y(0)=0\), where \(f(t)=\left\\{\begin{array}{l}1,0 \leq t<\pi \\ 0, \pi \leq t<2 \pi\end{array} \quad\right.\) and \(f(t)=\) \(f(t-2 \pi)\) if \(t \geq 2 \pi\)
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