91Ó°ÊÓ

The given function is$\frac{1+p}{{(p+2)}^2}$

A transformation of a function $f\left(x\right)$into the function $g\left(t\right)$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$F\left(s\right)$is the piecewise-continuous and exponentially-restricted real function$f\left(t\right)$
.

These properties are used to solve the function:

$$\begin{gathered} L^{-1}\left(\frac{1}{(p+a)}\right)=e^{- \frac{\text{at}}{}}    \left(L2\right) \\ {L}^{-1}\left(\frac{1}{{(p+a)}^{i+1}}\right)=\frac{t{e}^k}{k}        \left(L6\right) \end{gathered}$$

Consider the given function.

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

Now, evaluate the inverse transformation as shown.

$$\begin{gathered} L^{-1}\left(\frac{1+p}{{(p+2)}^2}\right)=L^{-1}\left(\frac{1+p+1-1}{{(p+2)}^2}\right) \\ =L^{-1}\left(\frac{(1+p+1)-1}{{(p+2)}^2}\right) \\ =L^{-1}\left(\frac{(p+2)-1}{{(p+2)}^2}\right) \\ =L^{-1}\left(\frac{(p+2)}{{(p+2)}^2}-\frac{1}{{(p+2)}^2}\right) \end{gathered}$$

$$\begin{gathered} =L^{-1}\left(\frac{1}{(p+2)}-\frac{1}{{(p+2)}^2}\right) \\ =L^{-1}\left(\frac{1}{(p+2)}\right)-L^{-1}\left(\frac{1}{{(p+2)}^2}\right) \end{gathered}$$

By the use Property (L2) and Property $\left(L6\right)$in above equation as,

$$\begin{gathered} L^{-1}\left(\frac{1+p}{{(p+2)}^2}\right)=L^{-1}\left(\frac{1}{(p+2)}\right)-L^{-1}\left(\frac{1}{{(p+2)}^2}\right) \\ =e^{-2t}-t{e}^{-2t} \end{gathered}$$

Hence

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