91Ó°ÊÓ

The given function is$\frac{2p-1}{p^2-2p+10}$

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 functiion

$$\begin{gathered} L13:L\left(e^{-at}\sin bt\right)=\frac{b}{{(p+a)}^2+b^2},Re(p+a)>|Imb|, \\ L14:L\left(e^{-at}\cos bt\right)=\frac{p+a}{{(p+a)}^2+b^2},Re(p+a)>|Imb| \end{gathered}$$

Consider the given function.

$F\left(p\right)=\frac{2p-1}{p^2-2p+10}$

Rewrite the given function as;

$$\begin{gathered} \frac{2p-1}{p^2-2p+10}=\frac{2p-2+2-1}{p^2-2p+1+9} \\ =\frac{(2p-2)+2-1}{\left(p^2-2p+1\right)+9} \\ =\frac{2(p-1)+1}{{(p-1)}^2+3^2} \\ =\frac{2(p-1)}{{(p-1)}^2+3^2}+\frac{1}{{(p-1)}^2+3^2} \end{gathered}$$

Apply inverse Laplace operator both the side of the form of$F\left(p\right)$ and use L13 and L14 rule

$$\begin{gathered} L^{-1}\left[\frac{2p-1}{p^2-2p+10}\right]=L^{-1}\left[\frac{2(p-1)}{{(p-1)}^2+3^2}\right]+L^{-1}\left[\frac{1}{{(p-1)}^2+3^2}\right] \\ =2L^{-1}\left[\frac{(p-1)}{{(p-1)}^2+3^2}\right]+\frac{3}{3}{L}^{-1}\left[\frac{1}{{(p-1)}^2+3^2}\right] \\ =2L^{-1}\left[\frac{(p-1)}{{(p-1)}^2+3^2}\right]+\frac{1}{3}{L}^{-1}\left[\frac{3}{{(p-1)}^2+3^2}\right] \end{gathered}$$

Now use inverse Laplace from the above defined rule and get,

$$\begin{gathered} L^{-1}\left[\frac{2p-1}{p^2-2p+10}\right]=2e^{-(-1)t}\cos 3t+\frac{1}{3}{e}^{-(-1)t}\sin 3t \\ =2e^t\cos 3t+\frac{1}{3}{e}^t\sin 3t \end{gathered}$$

Hence,$L^{-1}\left\{\frac{2p-1}{p^2-2p+10}\right\}=2e^t\cos 3t+\frac{1}{3}{e}^t\sin 3t$