Chapter 8: Q1P (page 438)

For integral $k$ , verify $L 5$ and $L 6$ in the Laplace transform table. Hint: From $L 2$ you can write: $\int_{0}^{\infty} e^{- p t} e^{- a t} d t = 1 / (p + a)$ . Differentiate this equation repeatedly with respect to $p$ . (See Chapter 4, Section 12, Example 4, page 235.) Also note $L 32$ for the $Γ$ function results in $L 5$ and $L 6$ , see Chapter 11, Problem 5.7.

Short Answer

Expert verified

The final value is $L (t^{k}) = \frac{k!}{p^{k + 1}}$ or $\frac{Γ (k + 1)}{p^{k + 1}}$ for $L 5$ .

The final value is $L (t^{k} e^{- a t}) = \frac{k!}{{(p + a)}^{k + 1}}$ or $\frac{Γ (k + 1)}{{(p + a)}^{k + 1}}$ for $L 6$ .

Step by step solution

Given information

The given expressions are $\int_{0}^{\infty} e^{- p t} e^{- a t} d t = \frac{1}{(p + a)}$ .

Differential equation

An equation that contains the variables and the derivatives is called a differential equation.

Verify L5 from the Laplace transform table

Laplace transform is an important concept that is helpful to solve a differential equation.

Verify $L 5$ from the Laplace transform table.

$L (t^{k}) = \int_{0}^{\infty} e^{- p t} t^{k} d t = \int_{0}^{\infty} e^{- s} {(\frac{s}{p})}^{k} \frac{d s}{p} = \frac{1}{p^{k + 1}} \int_{0}^{\infty} e^{- s} s^{k} d s = \frac{Γ (k + 1)}{p^{k + 1}}$

Hence, the final value is $L (t^{k}) = \frac{k!}{p^{k + 1}}$ or $\frac{Γ (k + 1)}{p^{k + 1}}$ .

Verify L6 from the Laplace transform table.

Verify $L 6$ from the Laplace transform table.

$L (t^{k} e^{- a t}) = \int_{0}^{\infty} e^{- p t} t^{k} e^{- a t} d t$

= \int_{0}^{\infty} e^{- t (a + p)} t^{k} d t = \frac{Γ (k + 1)}{{(p + a)}^{k + 1}}

Hence, the final value is $L (t^{k} e^{- a t}) = \frac{k!}{{(p + a)}^{k + 1}}$ or $\frac{Γ (k + 1)}{{(p + a)}^{k + 1}} . .$

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Short Answer

Step by step solution

Given information

Differential equation

Verify L5 from the Laplace transform table

Verify L6 from the Laplace transform table.

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