Use the results which you have obtained in Problems 21 and 22 to find the inverse transform of$\left(p^2+2p-1\right)\mathbf{}\mathit{I}\mathbf{}{\left(p^2+4p+5\right)}^{\mathbf{2}}$.

The given function is$F\left(p\right)=({\mathrm{p}}^2+2\mathrm{p}-1)I{({\mathrm{p}}^2+4\mathrm{p}+5)}^2$

A transformation of a function f(x) into the function g(t) that is useful especially in reducing the solution of an ordinary linear differential equation with constant coefficients to the solution of a polynomial equation.

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

Formula used for a given function

$$\begin{gathered} L\left\{t{e}^{-at}\sin bt\right\}=\frac{2b\left(p+a\right)}{{\left({\left(p+a\right)}^2+b^2\right)}^2} \\ L\left\{t{e}^{-a\mathrm{l}}\cos bt\right\}=\frac{{\left(p+a\right)}^2-b^2}{{\left({\left(p+a\right)}^2+b^2\right)}^2} \end{gathered}$$

Simplify the given function as shown:

$\frac{p^2+2p-1}{{\left(\left(p^2+4p+4\right)+1\right)}^2}=\frac{p^2+2p-1}{{\left({\left(p+2\right)}^2+1^2\right)}^2}$

This implies,

$\frac{p^2+2p-1}{{\left(p^2+4p+5\right)}^2}=\frac{p^2+2p-1}{{\left({\left(p+2\right)}^2+1^2\right)}^2}.................\left(3\right)$

Compare the denominator of equation (3) with the denominator of equation (1) and (2),

Then, a=2 and b = 1

Put these values in equation (1) and (2)

And

Then subtract equation (1) from equation (2) as,

$$\begin{gathered} L\left\{t{e}^{-at}\cos bt\right\}-L\left\{t{e}^{-at}\sin bt\right\}=\frac{{\left(p+a\right)}^2-b^2}{{\left({\left(p+a\right)}^2+b^2\right)}^2}\frac{A\left(p+a\right)}{{\left({\left(p+a\right)}^2+b^2\right)}^2} \\ L\left\{t{e}^{-2t}\cos bt\right\}-L\left\{t{e}^{-2t}\sin t\right\}=\frac{{\left(p+2\right)}^2-1^2}{\{{\left(p+2\right)}^2+1^2{)}^2}\frac{A\left(p+a\right)}{\left({\left(p+2\right)}^2+1^2\right)}n \\ L\left\{t{e}^{-2t}\cos t-t{e}^{-2t}\sin t\right\}=\frac{p^2+4p+4-1}{\left({\left(p+2\right)}^2+1^2\right)}-\frac{2p+4}{\left({\left(p+2\right)}^2+1^2\right)}n \\ L\left\{t{e}^{-2y}\left(\cos t-\sin t\right)\right\}=\frac{p^2+4p+3-2p-4}{{\left({\left(p+2\right)}^2+1^2\right)}^2} \end{gathered}$$

Solving further,

$$\begin{gathered} L\left\{t{e}^{-2}\left(\cos t-\sin t\right)\right\}=\frac{p^2+2p-1}{{\left({\left(p+2\right)}^2+1^2\right)}^2} \\ =\frac{p^2+2p-1}{{\left(p^2+4p+5\right)}^2} \\ =L\left\{L^{-2x}\left(\cos t-\sin t\right)\right\}{L}^{-1}\left\{\frac{p^2+2p-1}{{\left(p^2+4p+5\right)}^2}\right\} \\ =t{e}^{-2u}\left(\cos t-\sin t\right) \end{gathered}$$

$\mathrm{Thus},L^{\mathit{-}\mathit{1}}\left\{\frac{{\mathrm{p}}^2+2\mathrm{p}-1}{{\left({\mathrm{p}}^2+4\mathrm{p}+5\right)}^2}\right\}=t{e}^{\mathit{-}\mathit{2}}\left(cost-sint\right)$