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Q7.4 - 1E

Expert-verified
Found in: Page 375

### 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

# Determine the inverse Laplace transform of the given function.$\frac{\mathbf{6}}{{\mathbf{\left(}\mathbf{s}\mathbf{-}\mathbf{1}\mathbf{\right)}}^{\mathbf{4}}}$.

The inverse laplace transform of the given function is ${e}^{t}{t}^{3}$.

See the step by step solution

## Determining the inverse laplace transform

• For a given transfer function H, the Inverse Laplace Transform takes the output Y(s) and determines what X(s) it is in terms of (s).
• Consider a function F(s), if there is a function f(t) that is continuous on $\left[0,\infty \right)$and satisfies $\mathcal{L}\left\{f\right\}=F$ then we say that f(t) is the inverse Laplace transform of F(s) and employ the notation.
• ${f}{=}{{\mathcal{L}}}^{-1}{\left\{}{F}{\right\}}$
• ${{\mathcal{L}}}^{-1}\left\{\frac{n!}{{\left(s-a\right)}^{n+1}}\right\}{=}{{e}}^{at}{{t}}^{{n}}{,}{n}{=}{1}{,}{2}{,}{\dots }$

## Find inverse laplace transform for the given function

The given function is $\frac{6}{{\left(s-1\right)}^{4}}$.

Find the inverse laplace transform of $\frac{6}{{\left(s-1\right)}^{4}}$ using as ${\mathcal{L}}^{-1}\left\{\frac{n!}{{\left(s-a\right)}^{n+1}}\right\}={e}^{at}{t}^{n},n=1,2,\dots$ as

$\begin{array}{rcl}{\mathcal{L}}^{-1}\left\{\frac{6}{{\left(s-1\right)}^{4}}\right\}& =& {\mathcal{L}}^{-1}\left\{\frac{3!}{{\left(s-1\right)}^{3+1}}\right\}\\ & =& {e}^{t}{t}^{3}\\ & & \end{array}$

Therefore, the inverse laplace transform of the given function is ${e}^{t}{t}^{3}$.