Question

In Problems 3 - 24 fill in the blanks or answer true or false.

11.$\mathit{L}\left\{tsin 2t\right\}$ =………

Short answer

$L\left\{t\sin 2t\right\}=\frac{4s}{{\left(s^2+4\right)}^2}$

Step-by-step solution

Consider the function

The function is ,

$\begin{array}{l}L\left\{t\sin 2t\right\}\\ Thes\tan dardformulaisgivenas,\\ L\left\{\sin \left(at\right)\right\}=\frac{a}{s^2+a^2}\end{array}$

Compute the Laplace transform

consider function is,

$\begin{array}{l}L\left\{t\sin 2t\right\}\\ L\left\{t^{k}f\left(t\right)\right\}={\left(-1\right)}^k\frac{d^k}{d{s}^k}\left(L\left\{f\left(t\right)\right\}\right)\\ \mathrm{For} t\sin \left(2t\right):f\left(t\right)=\sin \left(2t\right), k=1\\ Here,a=2\\ AndL\left\{\sin \left(2t\right)\right\}=\frac{2}{s^2+4}\\ L\left\{t\sin \left(2t\right)\right\}={\left(-1\right)}^1\frac{d}{ds}\left(L\left\{\sin \left(2t\right)\right\}\right)\\ Computethedifferentiation.\\ \frac{d}{ds}\left(\frac{2}{s^2+4}\right)=-\frac{4s}{{\left(s^2+4\right)}^2}\\ SubstitutethevalueandfindtheLaplaceTransform,\\ L\left\{t\sin \left(2t\right)\right\}={\left(-1\right)}^1\left(-\frac{4s}{{\left(s^2+4\right)}^2}\right)\\ \Rightarrow L\left\{t\sin \left(2t\right)\right\}=\frac{4s}{{\left(s^2+4\right)}^2}\\ Therefore,theLaplacetransformisL\left\{t\sin \left(2t\right)\right\}\\ \therefore L\left\{t\sin 2t\right\}=\frac{4s}{{\left(s^2+4\right)}^2}\\ \end{array}$