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Q7.3 - 16E

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.]${{\mathbf{tsin}}}^{{\mathbf{2}}}{\mathbf{t}}$

The Laplace transform for the given equation is $\left[\frac{6{s}^{2}+8}{{\left({s}^{3}+4s\right)}^{2}}\right]$.

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

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

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

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

Simplifying 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 $t{\mathrm{sin}}^{2}t$ using role="math" localid="1655785948196" ${\left(\frac{f}{g}\right)}^{{\text{'}}}{=}\frac{f{g}^{\text{'}}-g{f}^{\text{'}}}{{f}^{2}}$ and ${\mathcal{L}}\left\{{t}^{n}f\left(t\right)\right\}{=}\left(-1{\right)}^{n}\frac{{d}^{n}}{d{s}^{n}}{\left[}{\mathcal{L}}{\left\{}{f}{\right\}}{\left(}{s}{\right)}{\right]}$ as follows:

$\begin{array}{rcl}\mathcal{L}\left\{t{\mathrm{sin}}^{2}t\right\}& =& \left(-1{\right)}^{1}\frac{{d}^{1}}{d{s}^{1}}\left[\mathcal{L}\left\{f\right\}\left(s\right)\right]\\ & =& -\frac{d}{ds}\left[\frac{2}{{s}^{3}+4s}\right]\\ & =& -\left[\frac{\left({s}^{3}+4s\right){\left(2\right)}^{\text{'}}-\left(2\right){\left({s}^{3}+4s\right)}^{\text{'}}}{{\left({s}^{3}+4s\right)}^{2}}\right]\\ & =& -\left[\frac{\left({s}^{3}+4s\right)\left(0\right)-\left(2\right)\left(3{s}^{2}+4\right)}{{\left({s}^{3}+4s\right)}^{2}}\right]\end{array}$

Simplifying the equation as follows:

$\begin{array}{rcl}\mathcal{L}\left\{t{\mathrm{sin}}^{2}t\right\}& =& -\left[\frac{0-\left(2\right)\left(3{s}^{2}+4\right)}{{\left({s}^{3}+4s\right)}^{2}}\right]\\ & =& -\left[\frac{-\left(6{s}^{2}+8\right)}{{\left({s}^{3}+4s\right)}^{2}}\right]\\ & =& \left[\frac{6{s}^{2}+8}{{\left({s}^{3}+4s\right)}^{2}}\right]\end{array}$

Therefore, the Laplace transform for the given equation is $\left[\frac{6{s}^{2}+8}{{\left({s}^{3}+4s\right)}^{2}}\right]$.