Chapter 10.10, Problem 1AP

To determine

The Laplace transform $L\left[f\right]$ of the given function $f$.

Answer to Problem 1AP

Solution:

The Laplace transform $L\left[f\right]$ of the given function $f$ is $-\frac{4}{s}+\frac{3}{s^2}$, where $s>0$.

Explanation of Solution

Given:

A function $f$ is defined on an interval $\left[0,\infty \right)$ as,

$f\left(t\right)=3t-4$

Approach:

The function $F\left(s\right)$ is Laplace transform of a function $f$ defined on an interval $\left[0,\infty \right)$

$F\left(s\right)={\displaystyle {\int }_0^{\infty }{e}^{-st}f\left(t\right)dt}$ $\left(1\right)$

It is provided that the improper integral converges as,

$\begin{array}{rll}F\left(s\right)& =& {\displaystyle {\int }_0^{\infty }{e}^{-st}f\left(t\right)dt}\\ & =& \underset{N\to \infty }{\lim }{\displaystyle {\int }_0^N{e}^{-st}f\left(t\right)dt}\end{array}$

If and only if the limit in the above expression exists and is finite, the improper integral converges.

Calculation:

The Laplace transform is written as,

$\begin{array}{rll}L\left[f\right]& =& {\displaystyle {\int }_0^{\infty }{e}^{-st}f\left(t\right)dt}\\ L\left[3t-4\right]& =& {\displaystyle {\int }_0^{\infty }{e}^{-st}\left(3t-4\right)dt}\\ & =& \underset{N\to \infty }{\lim }{\left[-\frac{3t{e}^{-st}}{s}+\frac{4e^{-st}}{s}\right]}_0^N+{\displaystyle {\int }_0^{\infty }\frac{3e^{-st}}{s}dt}\\ & =& \underset{N\to \infty }{\lim }{\left[-\frac{3t{e}^{-st}}{s}+\frac{4e^{-st}}{s}-\frac{3e^{-st}}{s^2}\right]}_0^N\end{array}$

Further simplify the above expression.

$\begin{array}{rll}\underset{N\to \infty }{\lim }N{e}^{-sN}& =& 0,\\ s>0\end{array}$

$\begin{array}{rll}L\left[3t-4\right]& =& \underset{N\to \infty }{\lim }\left[-\frac{3N{e}^{-sN}}{s}+\frac{4e^{-sN}}{s}-\frac{3e^{-sN}}{s^2}-\frac{4}{s}+\frac{3}{s^2}\right]\\ & =& -\frac{4}{s}+\frac{3}{s^2}\\ s>0\end{array}$

Therefore, the Laplace transform $L\left[f\right]$ of the given function $f$ is $-\frac{4}{s}+\frac{3}{s^2}$, where $s>0$.

Conclusion:

Hence, the Laplace transform $L\left[f\right]$ of the given function $f$ is $-\frac{4}{s}+\frac{3}{s^2}$, where $s>0$.