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### Fundamentals Of Differential Equations And Boundary Value Problems

Book edition 9th
Author(s) R. Kent Nagle, Edward B. Saff, Arthur David Snider
Pages 616 pages
ISBN 9780321977069

# Use the convolution theorem to find the inverse Laplace transform of the given function.

The inverse Laplace transform for the given function by using the convolution theorem is.$y\left(t\right)=\frac{2{e}^{-2t}}{3}+\frac{{e}^{t}}{3}$

See the step by step solution

## Step 1: Define convolution theorem

Let and be piecewise continuous on ${\left[}{0}{,}{\infty }{\right)}$and of exponential orderand set,${\mathbit{F}}{\mathbf{\left(}}{\mathbit{s}}{\mathbf{\right)}}{\mathbf{=}}{\mathsc{L}}{\mathbf{\left\{}}{\mathbit{f}}{\mathbf{\right\}}}{\mathbf{\left(}}{\mathbit{s}}{\mathbf{\right)}}{\mathbf{\text{and}}}{\mathbit{G}}{\mathbf{\left(}}{\mathbit{s}}{\mathbf{\right)}}{\mathbf{=}}{\mathsc{L}}{\mathbf{\left\{}}{\mathbit{g}}{\mathbf{\right\}}}{\mathbf{\left(}}{\mathbit{s}}{\mathbf{\right)}}$ then,

${\mathsc{L}}{\mathbf{\left\{}}{\mathbit{f}}{\mathbf{\ast }}{\mathbit{g}}{\mathbf{\right\}}}{\mathbf{\left(}}{\mathbit{s}}{\mathbf{\right)}}{\mathbf{=}}{\mathbit{F}}{\mathbf{\left(}}{\mathbit{s}}{\mathbf{\right)}}{\mathbit{G}}{\mathbf{\left(}}{\mathbit{s}}{\mathbf{\right)}}{\mathbf{,}}$or

${{\mathcal{L}}}^{-1}{\left\{}{F}{\left(}{s}{\right)}{G}{\left(}{s}{\right)}{\right\}}{\left(}{t}{\right)}{=}{\left(}{f}{\ast }{g}{\right)}{\left(}{t}{\right)}$

## Step 2: Use the convolution theorem to obtain the inverse Laplace transform

Consider the given equation,

$\frac{s}{\left(s-1\right)\left(s+2\right)}$

Let,

$\begin{array}{c}y\left(s\right)=\frac{s}{\left(s-1\right)\left(s+2\right)}\\ =\frac{1}{s+2}\left[1+\frac{1}{s-1}\right]\\ =\frac{1}{s+2}+\frac{1}{\left(s-1\right)\left(s+2\right)}\end{array}$

Take inverse Laplace transform on both sides,

$\to \left(1\right)$

Hence, the convolution formula is,${\mathcal{L}}^{-1}\left[f\left(s\right)\cdot g\left(s\right)\right]=f\star g={\int }_{0}^{t}f\left(t-v\right)g\left(v\right)dv$, where

$f\left(s\right)=\frac{1}{s-1}$and $f\left(t\right)={e}^{t}$

$g\left(s\right)=\frac{1}{s+2}$and $g\left(t\right)={e}^{-2t}$

Thus, the equationcan be written as,

$\begin{array}{c}y\left(t\right)={e}^{-2t}+{\int }_{0}^{t}{e}^{t-v}\cdot {e}^{-2v}dv\\ ={e}^{-2t}+{e}^{t}{\int }_{0}^{t}{e}^{-3v}dv\\ ={e}^{-2t}+{e}^{t}{\left[\frac{{e}^{-3v}}{-3}\right]}_{0}^{t}\\ ={e}^{-2t}-\frac{{e}^{t}}{3}\left[{e}^{-3t}-1\right]\end{array}$

$\begin{array}{c}y\left(t\right)={e}^{-2t}-\frac{{e}^{-2t}}{3}+\frac{{e}^{t}}{3}\\ =\frac{2{e}^{-2t}}{3}+\frac{{e}^{t}}{3}\end{array}$

Therefore, the inverse Laplace transform for the given function is. $y\left(t\right)=\frac{2{e}^{-2t}}{3}+\frac{{e}^{t}}{3}$