91Ó°ÊÓ

• 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).

Given that $\mathrm{tsin}2\mathrm{tsin}5\mathrm{t}$,

Let $f\left(t\right)=\sin 2t\sin 5t$

Find the Laplace transform of $f\left(t\right)=\sin 2t\sin 5t$using $\sin a\sin b=\frac{1}{2}\left[\cos \right(a-b)-\cos (a+b\left)\right]$, $\cos (-x)=\cos x$, $\mathcal{L}\left\{af\right(x)Â\pm bg(x\left)\right\}=a\mathcal{L}\left\{f\right\}Â\pm b\mathcal{L}\left\{g\right(t\left)\right\}$, $\frac{a}{c}Â\pm \frac{b}{d}=\frac{daÂ\pm cb}{cd}$and $\mathcal{L}\left\{\cos bt\right\}=\frac{s}{s^2+b^2}$as:

$\begin{array}{rcl}F\left(s\right)& =& \mathcal{L}\left\{\sin 2t\sin 5t\right\}\\ & =& \mathcal{L}\left\{\frac{1}{2}\left[\cos (2t-5t)-\cos (2t+5t)\right]\right\}\\ & =& \frac{1}{2}\left[\mathcal{L}\left\{\cos 3t\right\}-\mathcal{L}\left\{\cos 7t\right\}\right]\\ & =& \frac{1}{2}\left[\frac{s}{s^2+9}-\frac{s}{s^2+49}\right]\end{array}$

Simplify the equation as:

$\begin{array}{rcl}F\left(s\right)& =& \frac{1}{2}\left[\frac{\left(s^2+49\right)·s-\left(s^2+9\right)·s}{\left(s^2+9\right)\left(s^2+49\right)}\right]\\ & =& \frac{1}{2}\left[\frac{s^3+49s-s^3-9s}{\left(s^2+9\right)\left(s^2+49\right)}\right]\\ & =& \frac{1}{2}\left[\frac{40s}{\left(s^2+9\right)\left(s^2+49\right)}\right]\\ & =& \frac{20s}{\left(s^2+9\right)\left(s^2+49\right)}\end{array}$

Find the Laplace transform of the given function $e^{-t}t\sin 2t$using $\mathcal{L}\left\{tf\right(t\left)\right\}=(-1)\frac{d}{ds}F\left(s\right)$and ${\left(\frac{g}{f}\right)}^{\text{'}}=\frac{fg\text{'}-gf\text{'}}{f^2}$as follows:

$\begin{array}{rcl}\mathcal{L}\left\{t{e}^{2t}\cos 5t\right\}& =& \mathcal{L}\left\{tF\right(s\left)\right\}\\ & =& (-1)\frac{d}{ds}F\left(s\right)\\ & =& -\frac{d}{ds}\left[\frac{20s}{\left(s^4+58s^2+441\right)}\right]\\ & =& -\frac{\left(s^4+58s^2+441\right)·20-20s·\left(4s^3+116s\right)}{{\left(s^4+58s^2+441\right)}^2}\end{array}$

Simplify the equation as:

$\begin{array}{rcl}\mathcal{L}\left\{t{e}^{2t}\cos 5t\right\}& =& -\frac{20s^4+1160s^2+8820-80s^4-2320s^2}{{\left(s^4+58s^2+441\right)}^2}\\ & =& -\frac{-60s^4-1160s^2+8820}{{\left(s^4+58s^2+441\right)}^2}\\ & =& \frac{{\displaystyle \stackrel{20common}{\stackrel{Áåœ}{60s^4+1160s^2-8820}}}}{{\displaystyle {\left(s^4+58s^2+441\right)}^2}}\\ & =& \frac{20\left(3s^4+58s^2-441\right)}{{\left(s^4+58s^2+441\right)}^2}\end{array}$

Hence, the Laplace transform is$\frac{20\left(3s^4+58s^2-441\right)}{{\left(s^4+58s^2+441\right)}^2}$.