Short Answer

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

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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:

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

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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Fundamentals Of Differential Equations And Boundary Value Problems

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Short Answer

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.

Step by Step Solution

Define Laplace transform

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

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Fundamentals Of Differential Equations And Boundary Value Problems

Answers without the blur.

Short Answer

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.

Step by Step Solution

Define Laplace transform

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

Most popular questions for Math Textbooks

Want to see more solutions like these?

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