91Ó°ÊÓ

The given expressions are$e^{-2\mathrm{μ}{p}^2}$

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)

Laplace transform of$L_5$and$L_{27}$

$$\begin{gathered} L_5:L^{-1}\left(\frac{1}{p^{k+1}}\right)=\frac{t^4}{k!}k>-1 \\ {L}_{27}:L^{-1}\left(e^{-pa}\right)=\mathrm{δ}\left(t-a\right);â‰\yen 0L^{-1}\left[\frac{e^{-2p}}{p^2}\right] \end{gathered}$$

$$\begin{gathered} =L^{-1}\left[e^{-2p}.\frac{1}{p^2}\right]n \\ =L^{-1}\left[G\left(p\right).H\left(p\right)\right] \\ =g*h \\ =∫g\left(t-\mathrm{Ï„}\right).h\left(\mathrm{Ï„}\right)dr \\ =∫h(t-\mathrm{Ï„})â‹\dots g\left(\mathrm{Ï„}\right)â‹\dots d\mathrm{Ï„}=∫……\left(1\right)  \end{gathered}$$

By the definition of integral

role="math" localid="1659247072004" $$\begin{gathered} G\left(p\right)=e^{-2p} \\ L\left(g\left(t\right)\right)=e^{-2p} \\ g\left(t\right)=e^{-2p} \\ g\left(t\right)=\mathrm{δ}\left(t-2\right) \\ g\left(\mathrm{τ}\right)=\mathrm{δ}(\mathrm{τ}-2) \end{gathered}$$

Again by (1),

$$\begin{gathered} H\left(p\right)=\frac{1}{p^2} \\ L\left(h\left(t\right)\right)=\frac{1}{p^2} \\ h\left(t\right)=\frac{1}{p^2} \\ h\left(t\right)=\frac{I^1}{1!} \\ h\left(\mathrm{τ}\right)=t-\mathrm{τ}   …….\left(4\right) \end{gathered}$$

Now substitute the value of (3) and (4) in (2).

role="math" localid="1659247790900" $$\begin{gathered} L^{-1}\left(\frac{e^{-2p}}{p^2}\right)={∫}_0^1\mathrm{δ}\left(t-2\right)d\mathrm{τ} \\ ={∫}_0^1\left(t-\mathrm{τ}\right)\mathrm{δ}\left(\mathrm{τ}-2\right)d\mathrm{τ}n \\ ={\left(t-\mathrm{τ}\right)}_{r=2},0<2<t=0, \\ \mathrm{otherwise}{L}^{-1}\left(\frac{e^{-2p}}{p^2}\right)=t-2,t>2>0=0,t<2 \end{gathered}$$

Laplace transform of$L_{25}$

$L_{28}:L^{-1}[e^{-pu}â‹\dots G(p\left)\right]=g(t-2),t>2>0=0,t<2$

Put$a=2,\mathrm{G}\left(\mathrm{p}\right)=\frac{1}{p^2}$in the above transform

$L^{-1}\left(e^{-2}â‹\dots \frac{1}{p^2}\right)=g(t-2),t>2>0=0,t<2$

But

$$\begin{gathered} G\left(p\right)=\frac{1}{p^2} \\ L\left(g\left(t\right)\right)=\frac{1}{p^2} \\ g\left(t\right)=\frac{1}{p^2} \\ g\left(t\right)=\frac{I^1}{1!} \end{gathered}$$

$$\begin{gathered} g\left(t-2\right)=t-2 \\ By\left(1\right) \\ {L}^{-1}\left(e^{-2p}.\frac{1}{p^2}\right)=t-2,t>2 \\ =0,t<2 \end{gathered}$$