DE-BOOK Matched Forms

Subsection 10.1.1 Matched Forms

The Laplace Transform Method begins with the forward Laplace transform of a differential equation with an unknown function \(y(t)\) into an algebraic equation involving the transformed function \(Y(s)\text{.}\) Once we have solved for \(Y(s)\text{,}\) the final step is to recover the original function, \(y(t)\text{,}\) by applying the backward (inverse) Laplace transform, denoted as \(\laplacesym^{-1}\text{.}\)

Much like how \(\cos^{-1}\) reverses the cosine function, applying \(\laplacesym^{-1}\) to both sides of a Laplace-transformed expression brings us back to the original function. For example, if

\begin{equation*} \lap{e^{-3t}} = \frac{1}{s+3}, \end{equation*}

then, applying the inverse Laplace transform gives:

\begin{equation*} \ilap{\frac{1}{s+3}} = e^{-3t}. \end{equation*}

To identify the appropriate inverse Laplace transform, we rely on the common transforms in the table below, where we match the function of \(s\) on the right and the inverse is the function of \(t\) on the left.

Table 16. Common Laplace Transforms. \(a, b\) are constant, \(n = 1, 2, \ldots\)

	\(t\)-functions	\(s\)-functions
	\(\downarrow\)	\(\downarrow\)
	\(f(t)\)	\(\lap{ f(t) }\)
\(L_1\)	\(1\)	\(\ds \frac{1}{s}\)	\(s \gt 0\)
\(L_2\)	\(e^{at}\)	\(\ds \frac{1}{s-a}\)	\(s \gt a\)
\(L_3\)	\(t^n\)	\(\ds \frac{n!}{s^{n+1}}\)	\(s \gt 0\)
\(L_4\)	\(\sin (bt)\)	\(\ds \frac{b}{s^2 + b^2}\)	\(s \gt 0\)
\(L_5\)	\(\cos(bt)\)	\(\ds \frac{s}{s^2 + b^2}\)	\(s \gt 0\)
\(L_6\)	\(t^n e^{at}\)	\(\ds \frac{n!}{(s-a)^{n+1}}\)	\(s \gt a\)
\(L_7\)	\(e^{at} \sin(bt)\)	\(\ds \frac{b}{(s-a)^2 + b^2}\)	\(s \gt a\)
\(L_8\)	\(e^{at} \cos(bt)\)	\(\ds \frac{s-a}{(s-a)^2 + b^2}\)	\(s \gt a\)

Many functions can be matched by simply observing the denominator. The following table summarizes common patterns for matching denominators with their associated inverse transforms:

Table 17. Matching Guide: \(s\)-function \(\rightarrow\) Inverse Transform

Denominator	\(s\) in Numerator?	Transform
\(s\)	-	\(L_1\)
\(s^{\text{power}}\)	-	\(L_3\)
\(s \pm \text{shift}\)	-	\(L_2\)
\((s \pm \text{shift})^{\text{power}}\)	-	\(L_6\)
\(s^2 + \text{number}\)	no yes	\(L_4\) \(L_5\)
\((s \pm \text{shift})^2 + \text{number}\)	no yes	\(L_7\) \(L_8 + L_7\)

Now, let’s look at some examples.

Example 18.

Find the inverse Laplace transforms of each function.

\(\ds\hspace{1.5em}\frac{3}{s^2 + 9}\)

Solution. Solution

This matches the \(L_4\) form with \(b=3\text{.}\) So,

\begin{equation*} \ilap{\frac{3}{s^2 + 9}} = \sin(3t). \end{equation*}

\(\ds\frac{s - 2}{(s - 2)^2 + 16}\)

Solution. Solution

This matches the \(L_7\) form with \(a=2\) and \(b=4\text{.}\) So,

\begin{equation*} \ilap{\frac{s - 2}{(s - 2)^2 + 16}} = e^{2t}\cos(4t)\text{.} \end{equation*}

\(\ds\hspace{2.4em}\frac{24}{s^5}\)

Solution. Solution

This matches the \(L_3\) with \(n=4\text{.}\)

\begin{equation*} \ilap{\frac{24}{s^5}} = \ilap{\frac{4!}{s^{4+1}}} = t^4. \end{equation*}

These examples demonstrate how to find the inverse Laplace Transform when the \(s\)-function matches a form on the right side of the common transforms table. However, not all functions will align perfectly with the table, and additional adjustments are needed to match the forms. In the next discussion, we will explore how to modify the \(s\)-function to match the common forms in the table.

Reading Questions Check-Point Questions

1. The table of common forward Laplace transforms can also be used to find inverse Laplace transforms.

The table of common forward Laplace transforms can also be used to find inverse Laplace transforms

True.
True. The same table used for forward transforms is used for backward transforms, but in reverse order.
False.
True. The same table used for forward transforms is used for backward transforms, but in reverse order.

2. The inverse Laplace transform undoes the effect of the forward Laplace transform, allowing us to recover the original function \(y(t)\) from \(Y(s)\).

The inverse Laplace transform undoes the effect of the forward Laplace transform, allowing us to recover the original function \(y(t)\) from \(Y(s)\)

True.
True. The inverse Laplace transform reverses the forward transformation, converting \(Y(s)\) back into \(y(t)\text{.}\)
False.
True. The inverse Laplace transform reverses the forward transformation, converting \(Y(s)\) back into \(y(t)\text{.}\)

3. In the Laplace Transform Method, the backward transform \(\fillinmath{XXXXX}\).

In the Laplace Transform Method, the backward transform \(\fillinmath{XXXXX}\)

converts a differential equation into an algebraic equation.
No, this describes the forward Laplace transform.
solves the algebraic equation for \(Y(s)\text{.}\)
No, solving for \(Y(s)\) happens after applying the forward transform.
recovers the original function \(y(t)\) from \(Y(s)\text{.}\)
Correct! The inverse Laplace transform brings us back to the original function \(y(t)\text{.}\)
eliminates initial conditions from the equation.
No, initial conditions are incorporated into the transformed equation, not eliminated.

4. \(\ds\quad \ilap{\frac{3}{s^2 + 9}} = \fillinmath{XXXXXX}\).

\(\ds\quad \ilap{\frac{3}{s^2 + 9}} = \fillinmath{XXXXXX}\)

\(\sin(3t)\)
Correct! The inverse Laplace transform of \(\ds\frac{3}{s^2 + 9}\) is \(\sin(3t)\text{.}\)
\(\cos(3t)\)
No, the correct transform for \(\ds\frac{3}{s^2 + 9}\) is \(\sin(3t)\text{,}\) not \(\cos(3t)\text{.}\)
\(e^{3t}\)
No, this is not the correct inverse transform for the given expression.
\(\ds\frac{3}{s-3}\)
No, this is not an inverse transform expression.

5. \(\ds\quad \ilap{\frac{1}{s+3}} = \fillinmath{XXXXXX}\).

\(\ds\quad \ilap{\frac{1}{s+3}} = \fillinmath{XXXXXX}\)

\(e^{-3t}\)
Correct! The inverse Laplace transform of \(\ds\frac{1}{s+3}\) is indeed \(e^{-3t}\text{.}\)
\(e^{3t}\)
No, the correct answer is \(e^{-3t}\text{,}\) not \(e^{3t}\text{.}\)
\(e^{-t}\)
No, the exponent should be \(-3t\text{,}\) not \(-t\text{.}\)
\(\ds\frac{1}{s-3}\)
No, this is not the correct inverse Laplace transform.

6. \(\ds\quad \ilap{\frac{24}{s^5}} = \fillinmath{XXXXXX}\).

\(\ds\quad \ilap{\frac{24}{s^5}} = \fillinmath{XXXXXX}\)

\(t^3\)
No, this is incorrect. The correct answer is \(t^4\text{.}\)
\(t^4\)
Correct! The inverse Laplace transform of \(\ds\frac{24}{s^5}\) is \(t^4\text{.}\)
\(t^2\)
No, the correct answer is \(t^4\text{,}\) not \(t^2\text{.}\)
\(\ds\frac{1}{s^5}\)
No, this is the original function in the \(s\)-domain, not its inverse transform.

7. \(\ds\quad \ilap{\frac{2}{s^2 + 4}} = \fillinmath{XXXXXX}\).

\(\ds\quad \ilap{\frac{2}{s^2 + 4}} = \fillinmath{XXXXXX}\)

\(\sin(2t)\)
Correct! The inverse Laplace transform of \(\ds\frac{2}{s^2 + 4}\) is \(\sin(2t)\text{.}\)
\(\cos(2t)\)
No, the correct inverse Laplace transform is \(\sin(2t)\text{,}\) not \(\cos(2t)\text{.}\)
\(e^{2t}\)
No, the correct inverse transform is \(\sin(2t)\text{,}\) not \(e^{2t}\text{.}\)
\(t^2\)
No, the correct answer is \(\sin(2t)\text{,}\) not \(t^2\text{.}\)

8. \(\ds\quad \ilap{\frac{1}{s - 5}} = \fillinmath{XXXXXX}\).

\(\ds\quad \ilap{\frac{1}{s - 5}} = \fillinmath{XXXXXX}\)

\(e^{5t}\)
Correct! The inverse Laplace transform of \(\ds\frac{1}{s - 5}\) is \(e^{5t}\text{.}\)
\(e^{-5t}\)
No, the correct answer is \(e^{5t}\text{,}\) not \(e^{-5t}\text{.}\)
\(\cos(5t)\)
No, this is not the correct inverse transform for the given expression.
\(\sin(5t)\)
No, the correct inverse transform for \(\ds\frac{1}{s - 5}\) is \(e^{5t}\text{.}\)

9. \(\ds\quad \ilap{\frac{1}{(s+4)^2}} = \fillinmath{XXXXXX}\).

\(\ds\quad \ilap{\frac{1}{s - 5}} = \fillinmath{XXXXXX}\)

\(te^{-4t}\)
Correct! The inverse Laplace transform of \(\ds\frac{1}{(s+4)^2}\) is \(te^{-4t}\text{.}\)
\(e^{4t}\)
No, the correct answer is \(te^{-4t}\text{,}\) not \(e^{4t}\text{.}\)
\(t^2e^{-4t}\)
No, the correct inverse transform is \(te^{-4t}\text{,}\) not \(t^2e^{-4t}\text{.}\)
\(e^{-4t}\)
No, the correct answer is \(te^{-4t}\text{,}\) not \(e^{-4t}\text{.}\)

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