DE-BOOK Step 3 — Leaving the Laplace Domain

Section Step 3 — Leaving the Laplace Domain

Up to this point, we’ve been working inside the Laplace domain — a mathematical space where differential equations are replaced by algebraic ones. We used forward Laplace transforms to move from \(y(t)\) to \(Y(s)\text{,}\) solved for \(Y(s)\text{,}\) and simplified it to match known inverse forms.

🔗

Now, it’s time to take the final step: returning to the original domain where the solution is a function of time. To do this, we apply the inverse Laplace transform, denoted \(\laplacesym^{-1}\text{,}\) which undoes the effect of the forward transform.

🔗

Subsection What Is the Inverse Laplace Transform?

Symbolically, the inverse Laplace transform works much like the inverse of a standard function. For example, we can release \(y(t)\) from inside the cosine by applying the inverse cosine:

🔗

\begin{align*} \cos(y(t)) = t^2 \amp\quad\Rightarrow\quad \cos^{-1}(\cos(y(t))) = \cos^{-1}(t^2) \\ \amp\quad\Rightarrow\quad y(t) = \cos^{-1}(t^2) \end{align*}

Similarly, we can release \(y(t)\) from inside a Laplace transform by applying its inverse:

\begin{align*} \lap{y(t)} = {\DLBb\frac1{s}} \amp\quad\Rightarrow\quad \ilap{\lap{y(t)}} = \ilap{{\DLBb\frac1{s}}} \\ \amp\quad\Rightarrow\quad y(t) = \ilap{{\DLBb\frac1{s}}} \end{align*}

🔗

This is the idea behind the final step of the Laplace transform method as it takes us back into the original domain by releasing the \(y(t)\) contained inside \(Y = \lap{y(t)}\text{.}\) Visually, this looks like:

🔗

Step 3️⃣: Out of the Laplace Domain (Backward Transform).

\begin{equation*} \ul{\qquad\textbf{Original Domain}\qquad} \end{equation*}

🔗

\begin{equation*} \DLBa\textbf{Laplace Domain} \end{equation*}

🔗

\begin{gather*} \us{\large 🔻 🔻 🔻 🔻 🔻}{\text{Solution}}\\ y = e^{3t} + 2e^{-3t} - 2e^{2t} \end{gather*}

🔗

\begin{gather*} \text{Apply}\ \laplacesym^{-1}\\ \overleftarrow{\text{3️⃣ Backward}} \end{gather*}

🔗

\begin{gather*} \DLBa\dotsm\\ \DLBa Y = \frac{1}{s - 3} + \frac{2}{s + 3} - \frac{2}{s - 2} \end{gather*}

🔗

Checkpoint 219. 📖❓ What does the inverse transform do?

What is the purpose of the inverse Laplace transform \(\laplacesym^{-1}\) in solving a differential equation?

🔗

It recovers the original function \(y(t)\) from its Laplace-transformed version \(Y(s)\text{.}\)
Correct! The inverse transform brings us back to the time domain.
It rewrites a differential equation using algebra.
That’s what the forward transform does.
It simplifies partial fractions.
Partial fraction decomposition helps prepare \(Y(s)\) for the inverse, but isn’t the transform itself.
It evaluates the Laplace transform of \(y(t)\text{.}\)
This describes the forward transform, not the inverse.

🔗

Subsection Using the Table to Invert \(Y(s)\)

During the forward step, we used the table to map functions of \(t\) in the \(t\)-domain column, like \(e^{at}\) or \(\sin(bt)\text{,}\) into functions of \(s\) in the \(s\)-domain column. Now, we’ll do the opposite: match terms in \(Y(s)\) to the forms in the \(s\)-domain column and map them back to functions in the \(t\)-domain column.

🔗

For example, suppose we have the following \(Y(s)\) in the Laplace domain:

🔗

\begin{equation*} Y(s) = \frac{1}{s - 5} \end{equation*}

To return back this to the original domain we apply the inverse to both sides:

🔗

\begin{equation*} \ilap{Y(s)} = \ilap{\frac{1}{s - 5}} \quad\Rightarrow\quad y(t) = \ilap{\frac{1}{s - 5}} \end{equation*}

We have \(y(t)\text{,}\) but it needs to be a function of \(t\text{.}\) For that, we match \(\frac{1}{s - 5}\) to one of the forms in the \(s\)-domain column of the table.

🔗

We see the best match is L\(_2\) with \(a = 5\text{,}\) so we get:

🔗

\begin{equation*} y(t) = e^{5t} \end{equation*}

🌌 Example 220. Inverses using the Laplace Transform Table.

Apply the inverse transform to reveal the solution, \(y(t)\text{,}\) in the original domain.

🔗

\begin{equation*} Y(s) = \frac{3}{s^2 + 9} \end{equation*}

🔗

Solution.

This matches L\(_4\) with \(b=3\text{.}\) So,

\begin{align*} \ilap{Y(s)} \amp = \ilap{\frac{3}{s^2 + 9}} \\ y(t) \amp = \sin(3t) \text{.} \end{align*}

🔗

\begin{equation*} Y(s) = \frac{s - 2}{(s - 2)^2 + 16} \end{equation*}

🔗

Solution.

This matches L\(_7\) with \(a=2\) and \(b=4\text{.}\) So,

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

🔗

\begin{equation*} Y(s) = \frac{24}{s^5} \end{equation*}

🔗

Solution.

Noting that \(24=4!\) and using L\(_3\) with \(n=4\text{,}\) we get

\begin{align*} \ilap{Y(s)} \amp = \ilap{\frac{4!}{s^5}} \\ y(t) \amp = t^4 \text{.} \end{align*}

🔗

Checkpoint 221.

(a) 📖❓ Use the Transform Table.

For each \(Y(s)\text{,}\) use the table of transforms to determine \(y(t)\text{.}\)

🔗

\begin{equation*} \ul{\hspace{50px} Y(S) \hspace{50px}} \end{equation*}

🔗

\begin{equation*} \ul{\hspace{170px} y(t) \hspace{170px}} \end{equation*}

🔗

\begin{equation*} \frac{1}{s+3} \end{equation*}

\(e^{-t}\)

\(e^{3t}\)

\(e^{-3t}\)

\(3t\)

\begin{equation*} \frac{3}{s^2 + 9} \end{equation*}

\(3\sin t\)

\(e^{3t}\)

\(\cos(3t)\)

\(\sin(3t)\)

\begin{equation*} \frac{24}{s^5} \end{equation*}

\(t^2\)

\(t^4\)

\(4!\ t^4\)

\(t^3\)

\begin{equation*} \frac{1}{(s+4)^2} \end{equation*}

\(e^{4t}\)

\(t^2e^{-4t}\)

\(e^{-4t}\)

\(te^{-4t}\)

(b) 📖❓ Table Matching Checkpoint.

Which of the following is the correct inverse Laplace transform of

\begin{equation*} Y(s) = \frac{6}{(s - 2)^2 + 36}\text{?} \end{equation*}

\(y(t) = e^{2t} \sin(6t)\)
Correct! This matches L6 with \(a = 2\text{,}\) \(b = 6\text{.}\)
\(y(t) = \sin(6t)\)
This is missing the exponential factor \(e^{2t}\text{.}\)
\(y(t) = e^{6t} \sin(2t)\)
This flips \(a\) and \(b\text{.}\) Be careful when matching the form.
\(y(t) = e^{-2t} \cos(6t)\)
This would require \(s + 2\) in the numerator and \(\cos\) instead of \(\sin\text{.}\)

🔗

So far, we’ve only dealt with expressions that match a single row in the transform table. But most Laplace domain solutions are a sum of multiple terms. Fortunately, we don’t need to invert the entire thing at once. We’ll use linearity to handle those in the next section.

🔗

Subsection Linearity of the Inverse Laplace Transform

Most Laplace-domain solutions are not single terms. Instead, they’re sums of several expressions. Fortunately, like the forward transform, the inverse Laplace transform is also linear. This means that inverse transforms of sums and constant multiples of functions can be broken down into smaller, more manageable steps.

🔗

Assuming \(f(s)\) and \(g(s)\) are functions in the Laplace domain, then we have:

🔗

\begin{gather*} \text{P}_{2}:\quad\ilap{af(s) \pm bg(s)} = a\ilap{f(s)} \pm b\ilap{g(s)}\quad (a, b\ \text{constants}) \end{gather*}

Here’s an example of this linearity property in action.

🔗

🌌 Example 222. Inverse Transform of a Sum.

Find \(y(t)\) if \(Y(s) = \dfrac{3}{s} + \dfrac{5}{s^2 + 16}\text{.}\)

🔗

Solution.

We break this into two parts and use the transform table to invert each one.

🔗

\begin{equation*} \ilap{Y(s)} = \ilap{\frac{3}{s}} + \ilap{\frac{5}{s^2 + 16}} \end{equation*}

The first term matches L\(_1\) and gives \(3\text{.}\) The second term matches L\(_4\) with \(b = 4\text{,}\) giving \(5\sin(4t)\text{.}\) So the solution is:

🔗

\begin{equation*} y(t) = 3 + 5\sin(4t). \end{equation*}

🔗

Checkpoint 223. 📖❓ Apply Linearity.

Suppose

\begin{equation*} Y(s) = \frac{6}{s - 2} + \frac{4}{s^2 + 9}\text{.} \end{equation*}

What is \(y(t)\text{?}\)

\(6e^{2t} + 4\sin(3t)\)
Perfect! The first term matches \(\frac{1}{s - a}\) and the second is a sine with \(b = 3\text{.}\)
\(6e^{-2t} + 4\sin(3t)\)
Double check the sign in \(\frac{1}{s - 2}\text{.}\) It corresponds to \(e^{2t}\text{,}\) not \(e^{-2t}\text{.}\)
\(6e^{2t} + 4\cos(3t)\)
The second term matches a sine form, not cosine.
\(6e^{2t} + 4t^3\)
\(\frac{4}{s^2 + 9}\) corresponds to \(\sin(3t)\text{,}\) not a polynomial term.

🔗

Linearity makes inverting multi-term expressions much more manageable. Sometimes you’ll run into expressions that don’t quite match anything in the table. They’re close—but not perfect. In the next section, we’ll look at how to make small algebraic adjustments that bring those expressions into a form we can invert.

🔗

Subsection Minor Adjustments for a Perfect Match

Sometimes, the Laplace-domain function \(Y(s)\) is close to a known table form—but not quite there. In these cases, a few algebraic adjustments will help reveal the match. These are not major rewrites, just small nudges: splitting a numerator, factoring out constants, or rewriting terms to expose a recognizable pattern.

🔗

Separate Fractions to Match Multiple Forms.

Suppose your function has a sum in the numerator, like

\begin{equation*} Y(s) = \frac{2s + 5}{s^2 + 4}\text{.} \end{equation*}

The denominator matches both sine and cosine forms in the table, but the numerator doesn’t match any of the entries. The fix? Separate the addition in the numerator:

🔗

\begin{equation*} Y(s) = \frac{2s}{s^2 + 4} + \frac{5}{s^2 + 4}\text{.} \end{equation*}

Then factor constants and attempt to match with the table again:

🔗

\begin{equation*} Y(s) = 2\ \us{\ds\knowl{./knowl/xref/lt-L5-table.html}{\text{L}_5}\text{✔️}}{\ul{\frac{s}{s^2 + 4}}} + 5\ \us{\ds\text{almost}\ \knowl{./knowl/xref/lt-L4-table.html}{\text{L}_4}}{\ul{\frac{1}{s^2 + 4}}}\!\text{.} \end{equation*}

The first term perfectly matches, the second term needs \(b=2\) in the numerator. Missing numbers in the numerator is common with inverse transforms, but there is an easy fix. We multiply numerator and denominator by the missing number:

🔗

📝: 🚩 Multiplying by \(1\).

\begin{equation*} Y(s) = 2\frac{s}{s^2 + 4} + 5\frac{1}{s^2 + 4} \cdot \us{\ds 🚩}{\boxed{\frac{2}{2}}} = 2\ \us{\ds\knowl{./knowl/xref/lt-L5-table.html}{\text{L}_5}\text{✔️}}{\ul{\frac{s}{s^2 + 4}}} + \frac{5}{2}\ \us{\ds\knowl{./knowl/xref/lt-L4-table.html}{\text{L}_4}\text{✔️}}{\ul{\frac{2}{s^2 + 4}}}\text{.} \end{equation*}

With these minor adjustments, the inverse transform becomes straightforward:

\begin{equation*} y(t) = 2\cos(2t) + \frac{5}{2}\sin(2t)\text{.} \end{equation*}

🔗

Checkpoint 224. 📖❓ Splitting to Match Table Forms.

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 already matches \(L_6\) and does not require splitting.

🔗

Adding and Subtracting to Match Shifts.

Some forms contain a shift, like \((s - a)\text{.}\) For example, the form

\begin{equation*} Y(s) = \frac{s + 1}{(s - 2)^2 + 9} \end{equation*}

looks similar to L\(_8\), but the numerator should be \(s-2\text{.}\) This can be fixed by subtracting \(2\) and adding \(2\text{,}\) like so:

🔗

📝: 🚩 Adding \(0\).

\begin{equation*} Y(s) = \frac{s - \os{\ds🚩}{\overline{2 + 2}} + 1}{(s - 2)^2 + 9} = \frac{s - 2 + 3}{(s - 2)^2 + 9} \end{equation*}

Now you can separate the fraction as before:

🔗

\begin{equation*} Y(s) = \us{\ds\knowl{./knowl/xref/lt-L8-table.html}{\text{L}_8}\text{✔️}}{\ul{\frac{s - 2}{(s - 2)^2 + 9}}} + \us{\ds\knowl{./knowl/xref/lt-L7-table.html}{\text{L}_7}\text{✔️}}{\ul{\frac{3}{(s - 2)^2 + 9}}} \end{equation*}

These can be a little tricky so here’s a few additional examples to prepare you for what you may encounter.

🔗

🌌 Example 225. Handling Coefficients on \(s\).

Find \(y(t)\) if \(\ds\quad Y(s) = \frac{3s + 9}{(s-2)^2 + 16}\)

🔗

Solution.

The first term needs \(s-2\) in the numerator, but before adding and subtracting \(2\text{,}\) let’s get \(3\) out of the way by factoring it out:

🔗

\begin{align*} Y(s) = 3\frac{s + 3}{(s-2)^2 + 16} \amp = 3\left(\frac{s - 2 + 2 + 3}{(s-2)^2 + 16}\right) \text{.} \end{align*}

Now, splitting the fraction inside the parentheses and distributing the \(3\) gives us

🔗

\begin{align*} Y(s) \amp = 3\left(\frac{s - 2}{(s-2)^2 + 16} + \frac{5}{(s-2)^2 + 16}\right)\\ \amp = 3\frac{s - 2}{(s-2)^2 + 16} + 15\frac{1}{(s-2)^2 + 16}\\ \amp = 3\ub{\frac{s - 2}{(s-2)^2 + 16}}_{\large\text{matches } \knowl{./knowl/xref/lt-L8.html}{\text{L\(_8\)}}} + \frac{15}{4}\ub{\frac{4}{(s-2)^2 + 16}}_{\large\text{matches } \knowl{./knowl/xref/lt-L7.html}{\text{L\(_7\)}}}\text{.} \end{align*}

Finally, the backward transform is given by

🔗

\begin{align*} y(t) \amp = 3e^{2t}\cos(4t) + \frac{15}{4}e^{2t}\sin(4t) \text{.} \end{align*}

🔗

Checkpoint 226. 📖❓ Give the Numerator of the Split Fractions.

🔗

🌌 Example 227. Complete the Square First.

Find \(y(t)\) if \(\ds\quad Y(s) = \frac{s+3}{s^2 + 8s + 19}\)

🔗

Solution.

Completing the square for the denominator of \(Y(s)\) gives:

\begin{equation*} Y(s) = \frac{s+3}{s^2 + 8s + 19} = \frac{s+3}{(s + 4)^2 + 3}\text{.} \end{equation*}

🔗

The form of the denominator and the \(s\) in the numerator indicates that we are trying to match L\(_8\) (\(a=-4\text{,}\) \(b=\sqrt{3}\)), but we need \(s+4\) in the numerator. So we add and subtract \(4\text{,}\) then split the fraction:

🔗

\begin{equation*} Y(s) = \frac{s + 4 - 4 + 2}{(s + 4)^2 + 3} = \us{\ds\knowl{./knowl/xref/lt-L8-table.html}{\text{L}_8}\text{✔️}}{\ul{\frac{s+4}{(s + 4)^2 + 3}}} + \us{\ds\text{almost}\ \knowl{./knowl/xref/lt-L7-table.html}{\text{L}_7}}{\ul{\frac{-2}{(s + 4)^2 + 3}}}\text{.} \end{equation*}

Since the second term needs a \(\sqrt{3}\) in the numerator, we multiply it by \(\sfrac{\sqrt{3}}{\sqrt{3}}\text{:}\)

🔗

\begin{equation*} Y(s) = \frac{s + 4 - 4 + 2}{(s + 4)^2 + 3} = \us{\ds\knowl{./knowl/xref/lt-L8-table.html}{\text{L}_8}\text{✔️}}{\ul{\frac{s+4}{(s + 4)^2 + 3}}} + \frac{-2}{\sqrt{3}}\us{\ds\knowl{./knowl/xref/lt-L7-table.html}{\text{L}_7}\text{✔️}}{\ul{\frac{\sqrt{3}}{(s + 4)^2 + 3}}}\text{.} \end{equation*}

From here, the inverse transform comes from the table:

🔗

\begin{equation*} y(t) = e^{-4t}\cos(\sqrt{3}t) - \frac{2}{\sqrt{3}}e^{-4t}\sin(\sqrt{3}t)\text{.} \end{equation*}

🔗

Checkpoint 228. 📖❓ Give the Next Step.

What is the next step toward finding the inverse of \(Y(s)\text{?}\)

\begin{equation*} Y(s) = \frac{s-1}{s^2 - 2s + 5} \end{equation*}

🔗

Factor the denominator
Incorrect. The quadratic cannot be factored; try something else.
Complete the square in the denominator
Correct! Completing the square helps you match the form needed to invert with L\(_8\).
Cancel out the \(s\) in the numerator and denominator.
Canceling parts of rational expressions like this isn’t valid.
Differentiate the entire function
Nope—this isn’t a derivative problem.

🔗

With practice, these adjustments become intuitive, greatly simplifying the process of leaving the Laplace domain to recover the solution to the original differential equation.

🔗

Step 3 — Leaving the Laplace Domain Summary.

Apply the inverse transform \(\laplacesym^{-1}\) term-by-term to \(Y(s)\text{,}\) converting the algebraic answer back into the time-domain solution \(y(t)\text{.}\)

🔗

Subsection 📤 Wrap-Up

🗝️ \(\textbf{Key Takeaways...}\)

The inverse Laplace transform \(\ilap{\phantom{f}}\) recovers the solution to the differential equation you are trying to solve. \(y(t)\text{.}\)

🔗
Once the terms of \(Y(s)\) match the forms in the transform table, the inverse is the function of \(t\) in the same row.

🔗
If \(Y(s)\) is a sum of terms, use linearity to invert each part separately.

🔗
If \(Y(s)\) is close but doesn’t match a table entry, try splitting the numerator, factoring constants, or completing the square.

🔗
Always look at the denominator when matching forms in the table, they provide the best way to identify the correct form to match.

🔗

🔗

Check Your Understanding.

Checkpoint 229. 🤔💭 Separation of Variables Reading Questions.

(a) 🤔💭 Adjusting the Numerator.

Select the number that belongs in both blanks to directly apply the inverse.

\begin{equation*} \ilap{\frac{5}{s^4}} = \frac{5}{\fillinmath{X}}\ilap{\frac{\fillinmath{X}}{s^4}} \end{equation*}

🔗

\(6\)
Correct!
\(9\)
Incorrect.
\(3\)
Incorrect.
\(24\)
Incorrect.

🔗

(b) 🤔💭 Find the Inverse Transform.

\begin{equation*} Y(s) = \frac{s + 2}{s^2 + 1} \quad\Rightarrow\quad y(t) = \fillinmath{XXXXXXXXXXXXXXX} \end{equation*}

\(\cos(t) + 2\sin(t)\)
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.

🔗

You have attempted of activities on this page.

🔗

Prev Top Next