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Q7.3 - 15E

Expert-verified
Found in: Page 365

### Fundamentals Of Differential Equations And Boundary Value Problems

Book edition 9th
Author(s) R. Kent Nagle, Edward B. Saff, Arthur David Snider
Pages 616 pages
ISBN 9780321977069

# In Problems 1-20, determine the Laplace transform of the given function using Table 7.1 on page 356 and the properties of the transform given in Table 7.2. [Hint: In Problems 12-20, use an appropriate trigonometric identity.]${{\mathbf{cos}}}^{{\mathbf{3}}}{\mathbf{t}}$

The Laplace transform for the given equation is $\frac{{s}^{3}+7s}{\left({s}^{2}+9\right)\left({s}^{2}+1\right)}$.

See the step by step solution

## Definition of Laplace transform

• The integral transform of a given derivative function with real variable t into a complex function with variable s is known as the Laplace transform.
• Let f(t) be supplied for t(0), and assume that the function meets certain constraints that will be presented subsequently.
• The Laplace transform formula defines the Laplace transform of f(t), which is indicated by $\mathcal{L}\left\{f\left(t\right)\right\}$ or F(s).

## Determine the Laplace transform for the given equation

Given that ${\mathrm{cos}}^{3}t$,

Find the Laplace transform of ${\mathrm{cos}}^{3}t$ using ${c}{o}{{s}}^{{3}}{a}{=}\frac{1}{4}{\left(}{c}{o}{s}{3}{a}{+}{3}{c}{o}{s}{a}{\right)}$, ${\mathcal{L}}{\left\{}{a}{f}{\left(}{x}{\right)}{±}{b}{g}{\left(}{x}{\right)}{\right\}}{=}{a}{\mathcal{L}}{\left\{}{f}{\right\}}{±}{b}{\mathcal{L}}{\left\{}{g}{\left(}{t}{\right)}{\right\}}$, ${\mathcal{L}}{\left\{}{c}{o}{s}{b}{t}{\right\}}{=}\frac{s}{{s}^{2}+{b}^{2}}$ and $\frac{a}{b}{±}\frac{c}{d}{=}\frac{da±bc}{bd}$ as:

$\begin{array}{rcl}\mathcal{L}\left\{{\mathrm{cos}}^{3}t\right\}& =& \mathcal{L}\left\{\frac{1}{4}\left(\mathrm{cos}3t+3\mathrm{cos}t\right)\right\}\\ & =& \frac{1}{4}\mathcal{L}\left\{\mathrm{cos}3t+3\mathrm{cos}t\right\}\\ & =& \frac{1}{4}\left[\mathcal{L}\left\{\mathrm{cos}3t\right\}+3\mathcal{L}\left\{\mathrm{cos}t\right\}\right]\\ & =& \frac{1}{4}\left[\frac{s}{{s}^{2}+{3}^{2}}+3×\frac{s}{{s}^{2}+{1}^{2}}\right]\end{array}$

Simplify the equation as:

$\begin{array}{rcl}\mathcal{L}\left\{{\mathrm{cos}}^{3}t\right\}& =& \frac{1}{4}\left[\frac{s}{{s}^{2}+9}+\frac{3s}{{s}^{2}+1}\right]\\ & =& \frac{1}{4}\left[\frac{s\left({s}^{2}+1\right)+3s\left({s}^{2}+9\right)}{\left({s}^{2}+9\right)\left({s}^{2}+1\right)}\right]\\ & =& \frac{1}{4}\left[\frac{{s}^{3}+s+3{s}^{3}+27s}{\left({s}^{2}+9\right)\left({s}^{2}+1\right)}\right]\\ & =& \frac{1}{4}\left[\frac{{\stackrel{⏞}{4{s}^{3}+28s}}^{4\text{Common}}}{\left({s}^{2}+9\right)\left({s}^{2}+1\right)}\right]\end{array}$

Further simplifying the equation as follows:

$\begin{array}{rcl}\mathcal{L}\left\{{\mathrm{cos}}^{3}t\right\}& =& 4\left[\frac{4\left({s}^{3}+7s\right)}{\left({s}^{2}+9\right)\left({s}^{2}+1\right)}\right]\\ & =& \frac{{s}^{3}+7s}{\left({s}^{2}+9\right)\left({s}^{2}+1\right)}\end{array}$

Therefore, the Laplace transform for the given equation is $\frac{{s}^{3}+7s}{\left({s}^{2}+9\right)\left({s}^{2}+1\right)}$.

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