91Ó°ÊÓ

The given pair of linear equation is${∫}_0{\mathrm{te}}^{-\mathrm{t}}\sin 5\mathrm{tdt}$

Laplace transformation is a way for solving differential equations. The differential equation in the time domain is first translated into an algebraic equation in the frequency domain. The outcome of solving the algebraic equation in frequency domain is then translated to time domain form to obtain the differential equation's ultimate solution.

By the use of Laplace transformation table$\mathbf{L}\left(t\sin \mathrm{dt}\right)\mathbf{=}\frac{\mathbf{2}\mathbf{ap}}{{\left(p^2+a^2\right)}^{\mathbf{2}}}$

Consider the following definite integral

${∫}_0^{∞}{\mathrm{te}}^{-\mathrm{t}}\sin 5\mathrm{tdt}$

Object is to evaluate this integral by using the Laplace transforms.

Consider the following result of Laplace transforms:

$\mathrm{L}\left(\mathrm{t}\sin \mathrm{dt}\right)=\frac{2\mathrm{ap}}{{\left({\mathrm{p}}^2+{\mathrm{a}}^2\right)}^2}$

The Laplace transform of f (t) is define by the equation

$$\begin{gathered} \mathrm{L}\left(\mathrm{t}\right)={∫}_0^{∞}\mathrm{f}\left(\mathrm{t}\right){\mathrm{e}}^{-\mathrm{pt}}\mathrm{dt} \\ =\mathrm{F}\left(\mathrm{p}\right) \end{gathered}$$

Compare the given definite integral with the general integral ${∫}_0^{∞}\mathrm{f}\left(\mathrm{t}\right){\mathrm{e}}^{-\mathrm{pt}}\mathrm{dt}$and get

$\mathrm{p}=1,\mathrm{f}\left(\mathrm{t}\right)=\mathrm{t}\sin 5\mathrm{t}$

By the definition of Laplace transform of f (t) ,

${∫}_0^{∞}\mathrm{f}\left(\mathrm{t}\right){\mathrm{e}}^{-\mathrm{pt}}\mathrm{dt}=\mathrm{L}\left\{\mathrm{f}\left(\mathrm{t}\right)\right\}$

Therefore,

$$\begin{gathered} {∫}_0^{∞}{\mathrm{te}}^{-\mathrm{t}}\sin 5\mathrm{tdt}={∫}_0^{∞}(\mathrm{t}\sin 5\mathrm{t})e^{-\mathrm{t}}\mathrm{dt} \\ =\mathrm{L}(\mathrm{t}\sin 5\mathrm{t}) \end{gathered}$$

Since, the expression is,

$$\begin{gathered} \mathrm{L}(\mathrm{t}\sin \mathrm{dt})=\frac{2\mathrm{ap}}{{\left({\mathrm{p}}^2+{\mathrm{a}}^2\right)}^2} \\ \mathrm{L}(\mathrm{t}\sin \mathrm{dt})=\frac{2.5\mathrm{p}}{{\left({\mathrm{p}}^2+5^2\right)}^2} \end{gathered}$$

Substitute 1 For p in above expression.

$$\begin{gathered} \mathrm{L}(\mathrm{t}\sin \mathrm{dt})=\frac{10.1}{c} \\ =\frac{10}{{\left(1^2+25\right)}^2} \\ =\frac{10}{676} \\ =\frac{5}{338} \end{gathered}$$

Hence, the${∫}_0^{∞}{\mathrm{te}}^{-\mathrm{t}}\sin 5\mathrm{tdt}=\frac{5}{338}$${∫}_0^{∞}{\mathrm{te}}^{-\mathrm{t}}\sin 5\mathrm{tdt}=\frac{5}{338}$