Subsection 10.2.1 Splitting Fractions
One useful algebraic manipulation in preparing for the inverse Laplace transform is splitting a rational function that has a sum or difference in the numerator. This technique simplifies the expression into recognizable forms, making it easier to apply known Laplace transform pairs. Let’s explore two examples where splitting fractions proves useful.
Example 20. Find the Inverse Laplce Transform.
Solution 1. \(\ds \ilap{\frac{2s - 3}{s^2 + 16}}\)
To do this, we split the fraction into two simpler terms:
\begin{equation*}
\ds \frac{2s - 3}{s^2 + 16} = \frac{2s}{s^2 + 16} - \frac{3}{s^2 + 16}.
\end{equation*}
Now, we can apply the inverse Laplace transform to each term separately. The first term matches with
\(L_5\), and the second term corresponds to
\(L_4\):
\begin{align*}
\amp\ \ilap{\frac{2s - 3}{s^2 + 16}} \\
\amp\ = 2\ \ilap{\frac{s}{s^2 + 16}} - 3\ \ilap{\frac{1}{s^2 + 16}} \\
\amp\ = 2 \cos(4t) - \frac{3}{4} \sin(4t).
\end{align*}
Solution 2. \(\ds \ilap{\frac{s^3 - 7}{s^7}}\)
We can split the fraction into the terms,
\begin{equation*}
\ds \frac{s^3 - 7}{s^7} = \frac{s^3}{s^7} - \frac{7}{s^7} = \frac{1}{s^4} - \frac{7}{s^7}.
\end{equation*}
and apply the inverse Laplace transform,
\(L_3\), to each
\begin{align*}
\ilap{\frac{s^3 - 7}{s^7}} =\amp\ \ilap{\frac{1}{s^4} - \frac{7}{s^7}}\\
=\amp\ \ilap{\frac{1}{s^4}} - 7\ilap{\frac{1}{s^7}}\\
=\amp\ \frac{1}{3!}\ilap{\frac{3!}{s^4}} - \frac{7}{6!}\ilap{\frac{6!}{s^7}}\\
=\amp\ \frac{1}{6}t^3 - \frac{7}{720}t^6
\end{align*}
This section demonstrates how relatively simple modifications, such as splitting fractions, can make it easier to find the inverse Laplace transform. However, some functions of \(s\) may require more advanced techniques. In the next section, we will explore how completing the square can help match the forms needed for a backward transform when the \(s\)-variable is shifted.
Reading Questions Check-Point Questions
1. The function \(\ds \frac{s - 4}{s^2 - 16}\) can be split into separate terms that match the forms for sine and cosine inverse Laplace transforms.
The function \(\ds \frac{s - 4}{s^2 - 16}\) can be split into separate terms that match the forms for sine and cosine inverse Laplace transforms
- True
Incorrect. To match with the forms for sine and cosine inverse Laplace transforms, the denominator should have the form \(\ul{s^2+\text{number}}\text{.}\)
- False
Correct! The denominator should have the form \(\ul{s^2+\text{number}}\text{,}\) not \(\ul{s^2-\text{number}}\text{.}\)
2. Which of the following functions can be split into separate terms that match known inverse Laplace transforms?
Which of the following functions can be split into separate terms that match known inverse Laplace transforms?
- \(\ds\frac{3}{s^2 + 4}\)
Incorrect. This function already matches a known sine form and doesn’t require splitting.
- \(\ds\frac{s + 2}{s^2 + 9}\)
Correct! This function can be split as \(\ds\frac{s}{s^2 + 9} + \frac{2}{s^2 + 9}\text{,}\) matching cosine and sine forms.
- \(\ds\frac{5}{(s - 3)^4}\)
Incorrect. This function matches with the \(L_6\) and doesn’t require splitting.
3. \(\quad\ds \frac{s - 6}{s^2 - 36} = \) .
\(\quad\ds \frac{s - 6}{s^2 - 36} = \)
- \(\ds\frac{s}{s^2 - 36} - \frac{6}{s^2 - 36}\)
Correct! Splitting the function into these two terms allows you to use cosine and sine inverse Laplace transforms on each.
- \(\ds\frac{s}{s^2 - 36} + \frac{6}{s^2 - 36}\)
Incorrect. The second term should be subtracted, not added.
- \(\ds\frac{1}{s - 6}\)
Incorrect.
- \(\ds\frac{1}{s + 6}\)
Incorrect.
4. \(\quad\ds \ilap{\frac{s + 2}{s^2 + 1}} = \) .
\(\quad\ds \ilap{\frac{s + 2}{s^2 + 1}} = \)
- \(\cos(t) + 2\sin(t)\)
Correct! The function can be split as \(\ds\frac{s}{s^2 + 1} + 2\frac{1}{s^2 + 1}\text{,}\) which corresponds to \(\cos(t)\) and \(\sin(t)\text{.}\)
- \(\cos(2t) + \sin(2t)\)
Incorrect.
- \(\cos(t) + 2\sin(2t)\)
Incorrect.
- \(\cos(t) + \sin(2t)\)
Incorrect.
