91Ó°ÊÓ

The equation is$\mathrm{c}\frac{\mathrm{p}}{{({\mathrm{p}}^2-1)}^2}=\frac{\mathrm{p}}{{\mathrm{p}}^2-1}.\frac{\mathrm{p}}{{\mathrm{p}}^2-1}$ .

The piecewise-continuous and exponentially-restricted real function f(t) is the inverse Laplace transform of a function F(s), and it has the property:

$$\begin{gathered} \mathbf{L}\left\{f\right\}\left(s\right)\mathbf{=}\mathbf{L}\left\{f\left(t\right)\right\}\left(s\right) \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{F}\left(s\right) \end{gathered}$$

where L is the Laplace transform.

As per Convolution theorem, if we have two functions, taking their convolution and then Laplace is the same as taking the Laplace first (of the two functions separately) and then multiplying the two Laplace Transforms.

Consider the equation.

$\mathrm{c}\frac{\mathrm{p}}{{({\mathrm{p}}^2-1)}^2}=\frac{\mathrm{p}}{{\mathrm{p}}^2-1}.\frac{\mathrm{p}}{{\mathrm{p}}^2-1}$

As per the convolution theorem.

role="math" localid="1659268906381" $$\begin{gathered} L^{-1}\left[\frac{p}{p^2-1}.\frac{p}{p^2-1}\right]={∫}_0{g}_1{g}_{2}dx \\ ={∫}_0^t{g}_1\left(x\right)g_2\left(t-x\right)dx \\ ={∫}_0^t\cos h\left(x\right)\sin h\left(t-x\right)dx \\ ={∫}_0^t\cos h\left(x\right)\left(\sin h\cos hx-\cos ht\sin x\right)dx \end{gathered}$$

Further solve,

role="math" localid="1659269317585" $$\begin{gathered} L^{-1}\left[\frac{p}{p^2-1}.\frac{p}{p^2-1}\right]=\sin ht{∫}_0^t{\left(\cos hx\right)}^{2}dx-\cos ht{∫}_0^t\cos hx\sin hxdx \\ =\sin ht{∫}_0^t{\left(\cos hx\right)}^{2}dx-\frac{\cos ht}{2}{∫}_0^t\cos hx\sin hxdx \\ =\sin ht{∫}_0^t{\left(\cos hx\right)}^{2}dx-\cos ht{∫}_0^t\cos hx\sin hxdx \\ =\frac{t\sin ht}{2}+\frac{\sin ht}{2}.\frac{\sin h2t}{2}-\frac{\cos ht}{2}.\frac{\cos h2t}{2}+\frac{1}{2}\frac{\cos ht}{2} \end{gathered}$$

Further solve,

role="math" localid="1659269600381" $$\begin{gathered} L^{-1}\left[\frac{p}{p^2-1}.\frac{p}{p^2-1}\right]=\frac{t\sin ht}{2}+\frac{1}{4}\left(\cos h\left(2t-t\right)\right)\frac{\cos ht}{4} \\ =\frac{t\sin ht}{2}-\frac{\cos ht}{4}+\frac{\cos ht}{4} \\ =\frac{t\sin ht}{2} \end{gathered}$$

Thus, the inverse transform of given equation is $=\frac{t\sin ht}{2}$.