Question

For integral $k$, verify $L5$and $L6$in the Laplace transform table. Hint: From $L2$you can write: ${\int }_0^{\infty }{e}^{-pt}{e}^{-at}dt=1/(p+a)$. Differentiate this equation repeatedly with respect to $p$. (See Chapter 4, Section 12, Example 4, page 235.) Also note$L32$for the$\Gamma$function results in$L5$and$L6$, see Chapter 11, Problem 5.7.

Short answer

The final value is $L\left(t^k\right)=\frac{k!}{p^{k+1}}$or$\frac{\Gamma (k+1)}{p^{k+1}}$ for $L5$.

The final value is$L\left(t^k{e}^{-at}\right)=\frac{k!}{{(p+a)}^{k+1}}$ or$\frac{\Gamma (k+1)}{{(p+a)}^{k+1}}$ for$L6$ .

Step-by-step solution

Given information

The given expressions are ${\int }_0^{\infty }{e}^{-pt}{e}^{-at}dt=\frac{1}{(p+a)}$.

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$L5$ from the Laplace transform table.

$$\begin{gathered} L\left(t^k\right)={\int }_0^{\infty }{e}^{-pt}{t}^{k}dt \\ ={\int }_0^{\infty }{e}^{-s}{\left(\frac{s}{p}\right)}^k\frac{ds}{p} \\ =\frac{1}{p^{k+1}}{\int }_0^{\infty }{e}^{-s}{s}^{k}ds \\ =\frac{\Gamma (k+1)}{p^{k+1}} \end{gathered}$$

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

Verify L6 from the Laplace transform table.

Verify $L6$from the Laplace transform table.

$L\left(t^k{e}^{-at}\right)={\int }_0^{\infty }{e}^{-pt}{t}^k{e}^{-at}dt$

$$\begin{gathered} ={\int }_0^{\infty }{e}^{-t(a+p)}{t}^{k}dt \\ =\frac{\Gamma (k+1)}{{(p+a)}^{k+1}} \end{gathered}$$

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