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The given function is$\frac{5-2p}{p^2+p-2}$

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)

These properties are used to solve the function:

$$\begin{gathered} \text{L7:}{L}^{-1}\left(\frac{1}{(p+a)(p+b)}\right)=\frac{e^{-at}-e^{-b}}{b-a} \\ L8:L^{-1}\left(\frac{p}{(p+a)(p+b)}\right)=\frac{a{e}^{-at}-b{e}^{-b}}{a-b} \end{gathered}$$

Consider the given function.

$F\left(p\right)=\frac{5-2p}{p^2+p-2}$

Now, evaluate the inverse transformation as shown.

$$\begin{gathered} L^{-1}\left(\frac{5-2p}{p^2+p-2}\right)=L^{-1}\left(\frac{5-2p}{p^2+(2-1)p-2}\right) \\ =L^{-1}\left(\frac{3-2p}{p^2+2p-p-2}\right) \\ =L^{-1}\left(\frac{5-2p}{p(p+2)-1(p+2)}\right) \\ =L^{-1}\left(\frac{5-2p}{(p+2)p-1)}\right) \end{gathered}$$

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

By the use of Property$\left(L8\right)$ and Property (L8) in above equation as,

$$\begin{gathered} L^{-1}\left(\frac{5-2p}{p^2+p-2}\right)=5L^{-1}\left(\frac{1}{(p-1)(p+2)}\right)-2L^{-1}\left(\frac{p}{(p-1)(p+2)}\right) \\ =5L^{-1}\left(\frac{e^{-2}-e^{-2}}{2-(-1)}\right)-2\left(\frac{\left(-1e^2-\left(2e^2-2\right.\right.}{-1-2}\right) \\ =5\left(\frac{e^{-r-2蟺}}{3}\right)-2\left(\frac{-e^{-2-2}}{-3}\right) \\ =\frac{3e^2-5e-2}{3}-\frac{2+44-2e}{3} \end{gathered}$$

$$\begin{gathered} =\frac{3e^3-5e^{-2}-2e^4-2e}{3} \\ =\frac{3e^2-2e-2t}{3} \\ =\frac{\left.3{\left(e^{-3}-2\right)}^{-1}\right)}{3} \end{gathered}$$

Hence

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