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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

# Find the Laplace transform of .${\mathbit{f}}\mathbf{\left(}\mathbf{t}\mathbf{\right)}{\mathbf{=}}\underset{0}{\overset{t}{\int }}{e}^{v}sin\left(t-v\right)dv$

The Laplace transformation of ${\int }_{0}^{t}{e}^{v}\mathrm{sin}\left(t-v\right)dv$is $\frac{1}{\left({s}^{2}+1\right)\left(s-1\right)}$.

See the step by step solution

## Step 1: Given that,

$f\left(t\right)={\int }_{0}^{t}{e}^{v}\mathrm{sin}\left(t-v\right)dv$

## Step 2: Applying the Convolution Theorem

$g\star h={\int }_{0}^{t}g\left(t-v\right)h\left(v\right)dv\phantom{\rule{0ex}{0ex}}g\left(t\right)=\mathrm{sin}t\phantom{\rule{0ex}{0ex}}g\left(s\right)=\frac{1}{{s}^{2}+1}\phantom{\rule{0ex}{0ex}}h\left(t\right)={e}^{t}\phantom{\rule{0ex}{0ex}}g\left(s\right)=\frac{1}{s-1}$

## Step 3: Simplify

$\begin{array}{c}\mathcal{L}\left[g\star h\right]=g\left(s\right)\cdot h\left(s\right)\\ =\frac{1}{{s}^{2}+1}\cdot \frac{1}{s-1}\\ =\frac{1}{\left({s}^{2}+1\right)\left(s-1\right)}\end{array}$

Hence,

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