91Ó°ÊÓ

The given pair of linear equation is .

${∫}_0^{∞}{\mathrm{e}}^{-\mathrm{t}}(1-\cos 2\mathrm{t})\mathrm{dt}$

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$\mathbf{L}\left(1-\cos \mathrm{at}\right)\mathbf{=}\frac{{\mathbf{a}}^{\mathbf{2}}}{\mathbf{p}\left(p^2+a^2\right)}$

The given expression is,

${∫}_0^{∞}{\mathrm{e}}^{-\mathrm{pt}}\left(1-\cos \mathrm{at}\right)\mathrm{dt}$

From definition of Laplace transforms,

${∫}_0^{∞}{\mathrm{e}}^{-\mathrm{pt}}\left(1-\cos \mathrm{at}\right)\mathrm{dtt}=\mathrm{L}\left(1-\cos \mathrm{at}\right)$

Use this property, and Laplace transform $\mathrm{L}\left(1-\cos \mathrm{at}\right)=\frac{{\mathrm{a}}^2}{\mathrm{p}\left({\mathrm{p}}^2+{\mathrm{a}}^2\right)}$(Property (L 15) ) as,

$$\begin{gathered} {∫}_0^{∞}{\mathrm{e}}^{-\mathrm{pt}}\left(1-\cos \mathrm{at}\right)\mathrm{dt}=\mathrm{L}\left(1-\cos \mathrm{at}\right) \\ {∫}_0^{∞}{\mathrm{e}}^{-\mathrm{pt}}\left(1-\cos \mathrm{at}\right)\mathrm{dt}=\frac{2^2}{\mathrm{p}\left({\mathrm{p}}^2+2^2\right)} \end{gathered}$$

Substitute 1 for in equation (1).

role="math" localid="1659265838954" $$\begin{gathered} {∫}_0^{∞}{\mathrm{e}}^{-\mathrm{pt}}\left(1-\cos \mathrm{at}\right)\mathrm{dt}=\frac{4}{\mathrm{p}\left({\mathrm{p}}^2+4\right)} \\ {∫}_0^{∞}{\mathrm{e}}^{-\mathrm{pt}}\left(1-\cos 2\mathrm{t}\right)\mathrm{dt}=\frac{4}{\mathrm{p}\left({\mathrm{p}}^2+4\right)} \end{gathered}$$

Solve the equation further

$$\begin{gathered} {∫}_0^{∞}{\mathrm{e}}^{-\mathrm{pt}}\left(1-\cos 2t\right)\mathrm{dt}=\frac{4}{1\left(1^2+4\right)} \\ =\frac{4}{1\left(1^2+4\right)} \\ =\frac{4}{1\left(1+4\right)} \\ =\frac{4}{5} \end{gathered}$$

Hence, the Laplace transforms is${∫}_0^{∞}{\mathrm{e}}^{-\mathrm{pt}}\left(1-\cos 2\mathrm{t}\right)\mathrm{dt}=\frac{4}{5}$.