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Q7.4 - 8E

Expert-verified
Found in: Page 374

### 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.$\frac{\mathbf{1}}{{\mathbf{s}}^{\mathbf{5}}}$.

The inverse laplace transform of the given function is $\frac{{\mathrm{t}}^{4}}{24}$.

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
• localid="1662725690417" ${\mathrm{f}}{=}{{\mathcal{L}}}^{-1}{\left\{}{\mathrm{F}}{\right\}}$
• localid="1662725700279" ${{\mathcal{L}}}^{-1}\left\{\frac{\mathrm{n}!}{{\left(\mathrm{s}-\mathrm{a}\right)}^{\mathrm{n}+1}}\right\}{=}{{\mathrm{e}}}^{{\mathrm{at}}}{{\mathrm{t}}}^{{\mathrm{n}}}{,}{\mathrm{n}}{=}{1}{,}{2}{,}{\dots }$
• Constant multiplication property of inverse laplace transform, for function and constant.
• localid="1662725707425" ${{\mathrm{L}}}^{-1}{\left\{}{\mathrm{a}}{·}{\mathrm{f}}{\left(}{\mathrm{t}}{\right)}{\right\}}{=}{\mathrm{a}}{·}{{\mathrm{L}}}^{-1}{\left\{}{\mathrm{f}}{\left(}{\mathrm{t}}{\right)}{\right\}}$

## Find inverse laplace transform for the given function

The given function is $\frac{1}{{s}^{5}}$.

Find the inverse laplace transform of $\frac{1}{{s}^{5}}$ using localid="1662725728091" ${\mathrm{L}}^{-1}\left\{\mathrm{a}·\mathrm{f}\left(\mathrm{t}\right)\right\}=\mathrm{a}·{\mathrm{L}}^{-1}\left\{\mathrm{f}\left(\mathrm{t}\right)\right\}$ and ${L}^{-1}\left\{\frac{\left(n\right)!}{{s}^{n+1}}\right\}={t}^{n}$ as:

localid="1662725718604" $\begin{array}{rcl}{\mathcal{L}}^{-1}\left\{\frac{1}{{s}^{5}}\right\}& =& {\mathcal{L}}^{-1}\left\{\frac{1}{24}·\frac{24}{{s}^{5}}\right\}\\ & =& \frac{1}{24}{\mathcal{L}}^{-1}\left\{\frac{24}{{s}^{5}}\right\}\\ & =& \frac{{t}^{4}}{24}\end{array}$

Therefore, the inverse laplace transform of the given function is $\frac{{t}^{4}}{24}$.