Question

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.]

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

Short answer

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

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

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

Find the Laplace transform of $f\left(t\right)=e^{-t}\sin 2t$using $si{n}^{2}a=\frac{1}{2}(1-cos2a)$, $\mathcal{L}\left\{af\right(x)\pm bg(x\left)\right\}=a\mathcal{L}\left\{f\right\}\pm b\mathcal{L}\left\{g\right(t\left)\right\}$, localid="1662723884112" $\mathcal{L}\left\{1\right\}=\frac{1}{\mathrm{s}}$and $\mathcal{L}\left\{cosbt\right\}=\frac{s}{s^2+b^2}$as:

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

Simplify the equation as follows:

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

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

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

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