Question

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

$\frac{6-p}{p^2+4p+20}$

Short answer

The inverse transform of function$\frac{6-p}{p^2+4p+20}$ is$L^{-1}\left(\frac{6-p}{n^2+4n+30}\right)=2e^{-2t}\sin 4t-e^{-2t}\cos 4t$

Step-by-step solution

Given information

The given function is$\frac{6-p}{p^2+4p+20}$

Definition of Laplace Transformation

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)

Properties used to find the Laplace Transformation

These properties are used to solve the function:

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

Calculate the Inverse Transformation of given function

Consider the given function.

$F\left(p\right)=\frac{6-p}{p^2+4p+20}$

Now, evaluate the inverse transformation as shown.

$$\begin{gathered} L^{-1}\left(\frac{6-p}{p^2+4p+20}\right)=L^{-1}\left(\frac{6-p}{p^2+4p+4+16}\right) \\ =L^{-1}\left(\frac{6-p}{\left(p^2+4p+4\right)+16}\right) \end{gathered}$$

Use the identity,$x^2+2xy+y^2=(x+y{)}^2$

$\begin{array}{c}{L}^{-1}\left(\frac{6-p}{p^2+4p+20}\right)=L^{-1}\left(\frac{6-p}{(p+2{)}^2+1}\right.\\ =L^{-1}\left(\frac{6-p}{(p+2{)}^2+4^2}\right)\\ =L^{-1}\left(\frac{6-p}{(p+2{)}^2+4^2}\right)\\ =L^{-1}\left(\frac{6-(p+2-2)}{(p+2{)}^2+4^2}\right)\\ \end{array}$

$=L^{-1}\left(\frac{6+2-(p+2)}{{(p+2)}^2+4^2}\right)$

Continue further:

role="math" localid="1664275708067" $$\begin{gathered} L^{-1}\left(\frac{6-p}{p^2+4p+20}\right)=L^{-1}\left(\frac{8}{{(p+2)}^2+4^2}\right)-L^{-1}\left(\frac{(p+2)}{{(p+2)}^2+4^2}\right) \\ =L^{-1}\left(\frac{2\times 4}{{(p+2)}^2+4^2}\right)-L^{-1}\left(\frac{(p+2)}{{(p+2)}^2+4^2}\right) \\ =2L^{-1}\left(\frac{4}{{(p+2)}^2+4^2}\right)-L^{-1}\left(\frac{(p+2)}{{(p+2)}^2+4^2}\right) \\ =2e^{-2t}\sin 4t-e^{-2t}\cos 4t \end{gathered}$$

Hence$L^{-1}\left(\frac{6-p}{n^2+4n+30}\right)=2e^{-2t}\sin 4t-e^{-2t}\cos 4t$