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Answers without the blur. Sign up and see all textbooks for free! Q7.3 - 23E

Expert-verified Found in: Page 365 ### 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 Theorem 4 on page362 to show how entry 32 follows from entry 31 in the Laplace transform table on the inside back cover of the text.

It is proved that, $\mathcal{L}\left\{t\mathrm{sin}bt+bt\mathrm{cos}bt\right\}=\frac{2b{s}^{2}}{{\left({s}^{2}+{b}^{2}\right)}^{2}}$ from the Laplace transform table.

See the step by step solution

## Define Laplace transform

When specific initial conditions are supplied, especially when the initial values are zero, the Laplace transform is a handy method of solving certain types of differential equations. Laplace transform $\mathcal{L}$of a function f(t) is defined as:

${\mathcal{L}}{\left\{}{f}{\left(}{t}{\right)}{\right\}}{=}{{\int }}_{{0}}^{\frac{<}{>}}{{e}}^{-st}{f}{\left(}{t}{\right)}{d}{t}$

In words, we can describe this expression as the Laplace transform of f(t) equals function F of s, that is, $\mathcal{L}\left\{f\left(t\right)\right\}=F\left(s\right)$.

## Show that L{tsinbt+btcosbt}=2bs2s2+b22

Consider the expression $\mathcal{L}\left\{t\mathrm{sin}bt+bt\mathrm{cos}bt\right\}$

Let $f\left(t\right)=t\mathrm{sin}bt$

Then, $f\text{'}\left(t\right)=t\mathrm{sin}bt+bt\mathrm{cos}bt$

Find, $\mathcal{L}\left\{t\mathrm{sin}bt+bt\mathrm{cos}bt\right\}$ using $\mathcal{L}\left\{f\text{'}\right\}\left(s\right)=s\mathcal{L}\left\{f\right\}\left(s\right)-f\left(0\right)$ and $\mathcal{L}\left\{t\mathrm{sin}bt\right\}=\frac{2bs}{{\left({s}^{2}+{b}^{2}\right)}^{2}}$ as:

$\begin{array}{rcl}\mathcal{L}\left\{t\mathrm{sin}bt+bt\mathrm{cos}bt\right\}& =& s\mathcal{L}\left\{\left(t\mathrm{sin}bt\right)\right\}-f\left(0\right)\\ & =& s\frac{2bs}{{\left({s}^{2}+{b}^{2}\right)}^{2}}-0\\ & =& \frac{2b{s}^{2}}{{\left({s}^{2}+{b}^{2}\right)}^{2}}\end{array}$

Hence, it is proved that $\mathcal{L}\left\{t\mathrm{sin}bt+bt\mathrm{cos}bt\right\}=\frac{2b{s}^{2}}{{\left({s}^{2}+{b}^{2}\right)}^{2}}$. ### Want to see more solutions like these? 