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Q7.3 - 14E

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.]${{\mathbf{e}}}^{\mathbf{7}\mathbf{t}}{{\mathbf{sin}}}^{{\mathbf{2}}}{\mathbf{t}}$

The Laplace transform for the given equation is $\frac{2}{{\left(s-7\right)}^{3}+4s-28}$.

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 ${e}^{7t}{\mathrm{sin}}^{2}t$,

Let $f\left(t\right)={\mathrm{sin}}^{2}t$

Find the Laplace transform of $f\left(t\right)={e}^{-t}\mathrm{sin}2t$ using ${s}{i}{{n}}^{{2}}{a}{=}\frac{1}{2}{\left(}{1}{-}{c}{o}{s}{2}{a}{\right)}$, ${\mathcal{L}}{\left\{}{a}{f}{\left(}{x}{\right)}{±}{b}{g}{\left(}{x}{\right)}{\right\}}{=}{a}{\mathcal{L}}{\left\{}{f}{\right\}}{±}{b}{\mathcal{L}}{\left\{}{g}{\left(}{t}{\right)}{\right\}}$, localid="1662723884112" ${\mathcal{L}}{\left\{}{1}{\right\}}{=}\frac{1}{\mathrm{s}}$ and ${\mathcal{L}}{\left\{}{c}{o}{s}{b}{t}{\right\}}{=}\frac{s}{{s}^{2}+{b}^{2}}$ as:

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

Simplify the equation as follows:

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

Find the Laplace transform of the given function ${e}^{7t}{\mathrm{sin}}^{2}t$ using ${\mathcal{L}}\left\{{e}^{at}f\left(t\right)\right\}{=}{\mathcal{L}}{\left\{}{f}{\right\}}{\left(}{s}{-}{a}{\right)}$ as follows:

$\begin{array}{rcl}\mathcal{L}\left\{{e}^{7t}{\mathrm{sin}}^{2}t\right\}& =& \mathcal{L}\left\{{\mathrm{sin}}^{2}\right\}\left(s-7\right)\\ & =& \frac{2}{{\left(s-7\right)}^{3}+4\left(s-7\right)}\\ & =& \frac{2}{{\left(s-7\right)}^{3}+4s-28}\end{array}$

Therefore, the Laplace transform for the given equation is $\frac{2}{{\left(s-7\right)}^{3}+4s-28}$.