Q7.3 - 23E Use Theorem 4聽on page362 to sho... [FREE SOLUTION]

91影视

Fundamentals Of Differential Equations And Boundary Value Problems

R. Kent Nagle, Edward B. Saff, Arthur David Snider

$Math Studyset 91影视 Explanations$ Math

9th Edition 路 616 pages

ISBN: 9780321977069

Chapter 7: Q7.3 - 23E (page 365)

Use Theorem 4on page362 to show how entry 32 follows from entry 31 in the Laplace transform table on the inside back cover of the text.

Short Answer

Expert verified

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

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 $L$ of a function f(t) is defined as:

$L {f (t)} = {鈭�}_{0}^{\frac{<}{>}} e^{- s t} f (t) d t$

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

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

Consider the expression $L {t \sin b t + b t \cos b t}$

Let $f (t) = t \sin b t$

Then, $f' (t) = t \sin b t + b t \cos b t$

Find, $L {t \sin b t + b t \cos b t}$ using $L \{f'\} (s) = s L {f} (s) - f (0)$ and $L {t \sin b t} = \frac{2 b s}{{(s^{2} + b^{2})}^{2}}$ as:

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

Hence, it is proved that $L {t \sin b t + b t \cos b t} = \frac{2 b s^{2}}{{(s^{2} + b^{2})}^{2}}$ .

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91影视

Use Theorem 4on page362 to show how entry 32 follows from entry 31 in the Laplace transform table on the inside back cover of the text.

Short Answer

Step by step solution

Define Laplace transform

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

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