Find the inverse transforms of the functions$F\left(p\right)$.

role="math" localid="1664277165358" $\frac{3p+2}{3p^2+5p-2}$

The given function is$\frac{3p+2}{3p^2+5p-2}$

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.

$$\begin{gathered} L^{-1}\left(\frac{3p+2}{3p^2+5p-2}\right)=L^{-1}\left(\frac{3p+2}{3p^2+(6-1)p-2}\right) \\ =L^{-1}\left(\frac{3p+2}{3p^2+6pp-p-2}\right) \\ =L^{-1}\left(\frac{3p+2}{3p(p+2)-1(p+2)}\right) \\ =L^{-1}\left(\frac{3p+2}{(3p-1)(p+2)}\right) \end{gathered}$$

Continue further as shown below:

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

Divide numerator and denominator by 3 in $\left(\frac{3}{(3p-1)(p+2)}\right)$.

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

$$\begin{gathered} =e^{-2t}+\left(\frac{e^{\frac{1}{3}}-e^{-2x}}{2-\left(-\frac{1}{3}\right)}\right) \\ =e^{-2t}+\left(\frac{e^{\frac{t}{3}}-e^{-2t}}{\frac{6+1}{3}}\right) \end{gathered}$$

Simplify the expression further:

$$\begin{gathered} L^{-1}\left(\frac{3p+2}{3p^2+5p-2}\right) \\ =e^{-2t}+\frac{3\left(e^{\frac{1}{3}}-e^{-2t}\right)}{7} \\ =\frac{7e^{-2t}+3\left(e^{\frac{1}{3}}-e^{-2t}\right)}{7}\backslash \mathrm{hfill} \\ =\frac{7e^{-2t}+3e^{\frac{4}{3}}-3e^{-2t}}{7} \\ =\frac{4e^{-2}+3e^{\frac{t}{3}}}{7} \end{gathered}$$

Hence,$L^{-1}\left(\frac{3p+2}{3p^2+5p-2}\right)=\frac{4e^{-2t}+3e^{\frac{t}{3}}}{7}$