Question

Find the inverse Laplace transform of ${\mathit{e}}^{\mathbf{-}\mathbf{2}\mathbf{P}}\mathit{l}\mathbf{}{\mathit{P}}^{\mathbf{2}}$in the following ways:

(a) Using L5 and L27 and the convolution integral of Section 10;

(b) Using L28.

Short answer

The Inverse Laplace of given statements are

(a) The $L_5$and $L_{27}$is $\frac{t^4}{k!},\delta (t-a)$.

(b) The $L_{28}$is $\frac{t^2}{1!}$.

Step-by-step solution

Given information

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

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)

Find the inverse Laplace by use of L5 and L27

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

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

$$\begin{gathered} =L^{-1}\left[e^{-2p}.\frac{1}{p^2}\right]n \\ =L^{-1}\left[G\left(p\right).H\left(p\right)\right] \\ =g*h \\ =\int g\left(t-\tau \right).h\left(\tau \right)dr \\ =\int h(t-\tau )\cdot g\left(\tau \right)\cdot d\tau =\int \dots \dots \left(1\right)\hspace{0.33em} \end{gathered}$$

By the definition of integral

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

Again by (1),

$$\begin{gathered} H\left(p\right)=\frac{1}{p^2} \\ L\left(h\left(t\right)\right)=\frac{1}{p^2} \\ h\left(t\right)=\frac{1}{p^2} \\ h\left(t\right)=\frac{I^1}{1!} \\ h\left(\tau \right)=t-\tau \hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\dots \dots .\left(4\right) \end{gathered}$$

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

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

Find the inverse Laplace by use of L28

Laplace transform of$L_{25}$

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

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

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

But

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

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