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Chapter 4: Problem 16

In Problems, find either \(F(s)\) or \(f(t)\), as indicated. $$ \mathscr{L}^{-1}\left\\{\frac{2 s+5}{s^{2}+6 s+34}\right\\} $$

Short Answer

Expert verified
The inverse Laplace is \(2e^{-3t}\cos(5t) - \dfrac{1}{5}e^{-3t}\sin(5t)\).

Step by step solution

01

Simplify the Denominator

First, we simplify the denominator of the Laplace transform, which is \(s^2 + 6s + 34\). This can be written in the form \((s-a)^2 + b^2\). To complete the square, we focus on \(s^2 + 6s\). Half of 6 is 3, so add and subtract 9 inside the square: \(s^2 + 6s + 9 - 9 + 34 = (s+3)^2 + 25\).
02

Identify Form Suitable for Inverse Transform

In the rearranged form \((s+3)^2 + 5^2\), this is a standard form needed for inverse Laplace transforms associated with \(e^{-at}\sin(bt)\) and \(e^{-at}\cos(bt)\).
03

Express Numerator Appropriate for Known Transform

Modify the numerator \(2s+5\) in terms of \((s+3) + ")\circleddash{\text{term for \ensuremath{A}}\")\) to fit the formulas for inverse Laplace transforms of \(e^{-at}\sin(bt)\) and \(e^{-at}\cos(bt)\). Rewrite it as \(2(s+3) - 1\).
04

Break Down the Fraction

Break down the fraction into two separate terms: \({\dfrac{2(s+3)}{(s+3)^2 + 5^2}} - \dfrac{1}{(s+3)^2 + 5^2}\).
05

Apply Inverse Laplace Transform

Apply the inverse Laplace transform using known transforms: for \(\dfrac{s-a}{(s-a)^2 + b^2}\) the transform is \(e^{-at}\cos(bt)\), and for \(\dfrac{1}{(s-a)^2 + b^2}\), it is \(\dfrac{1}{b}e^{-at}\sin(bt)\). Thus, we have \(2e^{-3t}\cos(5t) - \dfrac{1}{5}e^{-3t}\sin(5t)\).
06

Combine Results

Combine individual inverse transforms to get the final result \(\mathscr{L}^{-1}\{\dfrac{2s+5}{s^2+6s+34}\}= 2e^{-3t}\cos(5t) - \dfrac{1}{5}e^{-3t}\sin(5t)\).

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Key Concepts

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

Completing the Square
The process of completing the square is a useful technique in algebra, especially when dealing with quadratic expressions. This method can transform a quadratic equation into a perfect square plus or minus a constant. The general form is taking an expression like \(s^2 + bs\) and turning it into \((s + c)^2 + d\). Here's how it works step by step.
  • First, focus on the linear and quadratic terms: \(s^2 + 6s\).

  • Next, take half of the coefficient of \(s\), in this case, 6, which is 3, and square it, giving us 9.

  • Add and subtract this square within the expression \(s^2 + 6s\) to maintain the balance, making it \((s^2 + 6s + 9 - 9)\).

  • The expression \(s^2 + 6s + 9\) becomes \((s+3)^2\).
  • Lastly, don't forget the constants: \(s^2 + 6s + 34 = (s+3)^2 + 25\).

This transformation is crucial when seeking forms suitable for inverse Laplace transforms used in engineering and science fields.
Cosine Function
When dealing with inverse Laplace transforms, the cosine function plays a critical role. Particularly, the term \(e^{-at}\cos(bt)\) arises often from specific algebraic forms.
  • The standard term in inverse Laplace related to cosine is \(\frac{s-a}{(s-a)^2 + b^2}\).

  • In the exercise, once we have \((s+3)^2 + 5^2\), the whole process fits the form leading to \(e^{-3t}\cos(5t)\).

  • It is important that the transformation aligns with this pattern, ensuring that the result accurately reflects a cosine function multiplied by an exponential decay factor.

The cosine function is crucial in oscillatory systems, describing periodic behavior such as vibrations, waves, and alternating circuits, providing insight into the system's amplitude and frequency behavior.
Sine Function
The sine function is another pivotal component in inverse Laplace transforms, particularly through \(e^{-at}\sin(bt)\). This results from expressions where the numerator is constant relative to the denominator.
  • For the transform \(\frac{1}{(s-a)^2 + b^2}\), the solution becomes \(\frac{1}{b}e^{-at}\sin(bt)\)."

  • In our specific example, it manifests as \(-\frac{1}{5}e^{-3t}\sin(5t)\), derived from the second term \(\frac{1}{(s+3)^2 + 5^2}\).

  • The sine function helps describe systems where energy is dissipated over time, effectively modeling phenomena such as damping and resonance.

Through implementing inverse transforms, systems analysis becomes possible, aiding in understanding how different inputs can affect the overall behavior of real-world dynamical systems.

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Most popular questions from this chapter

In Problems, use the Laplace transform to solve the given initial-value problem. $$ y^{\prime}+y=f(t), \quad y(0)=0, \text { where } f(t)=\left\\{\begin{array}{lr} 0, & 0 \leq t<1 \\ 5, & t \geq 1 \end{array}\right. $$

Use the Laplace transform to solve the given differential equation subject to the indicated initial conditions. $$ y^{\prime \prime}+y=\delta(t-2 \pi)+\delta(t-4 \pi), \quad y(0)=1, y^{\prime}(0)=0 $$

Use Theorem to evaluate the given Laplace transform. Do not evaluate the integral before transforming. $$ \mathscr{L}\left\\{e^{-t} * e^{t} \cos t\right\\} $$

Use the Laplace transform to solve the given system of differential equations. $$ \begin{aligned} &\frac{d x}{d t}+x-\frac{d y}{d t}+y=0 \\ &\frac{d x}{d t}+\frac{d y}{d t}+2 y=0 \\ &x(0)=0, y(0)=1 \end{aligned} $$

In Problems, find either \(F(s)\) or \(f(t)\), as indicated. $$ \mathscr{L}^{-1}\left\\{\frac{e^{-s}}{s(s+1)}\right\\} $$
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