Question

Use L28 and L4 to find the inverse transform of$\mathit{p}{\mathit{e}}^{\mathbf{-}\mathbf{p}\mathbf{t}}\mathbf{}\mathit{I}\left(p^2+1\right)$.

Short answer

Answer

The inverse Laplace transforms of is

$L^{-1}\left[\frac{p^{-ez}}{p^2+1}\right]=f\left(t\right)=\left\{\begin{array}{l}\cos \left(t-\mathrm{\pi }\right),\\ 0,\end{array}\right.\left.\begin{array}{r}t>\mathrm{\pi }\\ t<\mathrm{\pi }\end{array}\right\}$

Step-by-step solution

Given information

The given function is $p{e}^{-p2}I\left(p^2+1\right)$

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)

Inverse Transformation of the given function

From$\left(L28\right)$

$L\left[f\left(t\right)=\left\{\begin{array}{l}g\left(t-a\right),\\ 0,\end{array}\right.\left.\begin{array}{r}t>0\\ 0\end{array}\right\}\right]=e^{-pw}G\left(p\right)$

From $\left(L4\right)$

$L\left(\cos f\right)=\frac{p}{w^2+a^2}$

From above equation

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

By use of $\left(L28\right);$

$L\left[f\left(t\right)=\left\{\begin{array}{l}g\left(t-a\right),\\ 0,\end{array}\left.\begin{array}{r}t>a>0\\ 0\end{array}\right\}\right.\right]=e^{-pu}G\left(p\right)$

And consider the

$$\begin{gathered} L^{-1}\left[e^{-px}G\left(p\right)\right]=f\left(t\right)=\left\{\begin{array}{l}g\left(t-a\right),\\ 0,\end{array}\left.\begin{array}{r}t>a>0\\ 0\end{array}\right\}\right. \\ {L}^{-1}\left[\frac{px}{p^2+1}\right]=f\left(t\right)=\left\{\begin{array}{l}g\left(t-\mathrm{\pi }\right),\\ 0,\end{array}\left.\begin{array}{r}t>\mathrm{\pi }\\ t<\mathrm{\pi }\end{array}\right\}\right. \end{gathered}$$

Thus, the inverse Laplace transforms of $\frac{p\times r}{p^2+1}$is$L^{-1}\left[\frac{p^{-ez}}{p^2+1}\right]=f\left(t\right)=\left\{\begin{array}{l}\cos \left(t-\mathrm{\pi }\right),\\ 0,\end{array}\left.\begin{array}{r}t>\mathrm{\pi }\\ t<\mathrm{\pi }\end{array}\right\}\right.$.