DE-BOOK Summary & Worked Examples

Section Summary & Worked Examples

This closing section is where the entire Laplace transform method comes together. You have already practiced each of the three steps in isolation. Now we will zoom out, remind ourselves of the big picture, and then run walk through a variety of examples that illustrate how the method is applied from start to finish. These examples are meant to reinforce the process and highlight common patterns, strategies, and adjustments along the way.

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Subsection Laplace Transform Method— Quick Recap

Here is a visual of the full Laplace transform method. Each step corresponds to a move in the roadmap, and together they form a complete strategy for solving initial value problems. Notice that every arrow answers one of three guiding questions:

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How do we translate the differential equation into the Laplace domain?

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How do we isolate and tidy up \(Y(s)\) so it matches the inverse table?

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How do we return to the original time domain to reveal \(y(t)\text{?}\)

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Laplace Method Roadmap.

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

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\begin{equation*} \DLBa\textbf{2️⃣ Laplace Domain} \end{equation*}

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\begin{gather*} \os{\vphantom{m}}{y'' - 9y = 10e^{2t}}\\ y(0) = 1,\quad y'(0) = -7\\ \os{\large 🔺 🔺 🔺 🔺 🔺}{\text{Differential Equation}}\\ \\ \\ \us{\large 🔻 🔻 🔻 🔻 🔻}{\text{Solution}}\\ y = e^{3t} + 2e^{-3t} - 2e^{2t} \end{gather*}

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

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\begin{align*} \amp\DLBa s^2Y - 9Y - s + 7 = \frac{10}{s - 2}\\ \amp\qquad\qquad {\Big\downarrow}\quad\text{Solve for}\ Y\\ \amp\DLBa Y = \frac{1}{s^2 - 9}\left( \frac{10}{s - 2} + s - 7 \right)\\ \amp\qquad\qquad {\Big\downarrow}\ \ \text{Prepare for Inverse}\\ \amp\DLBa Y = \frac{1}{s - 3} + \frac{2}{s + 3} - \frac{2}{s - 2} \end{align*}

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✍🏻 Method 6. Laplace Transform Method (Three-Step Roadmap).

Given an \(n\)-th-order IVP

\begin{equation*} y^{(n)} + a_{n-1}y^{(n-1)} + \dots + a_1y' + a_0y = g(t), \end{equation*}

the particular solution is obtained through the following workflow:

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Step 1— Forward Transform 🔗: Apply \(\laplacesym\) term-by-term. Powers of \(s\) replace derivatives, and the initial conditions drop in automatically. The result is a single algebraic (Laplace-domain) equation.
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Step 2— Solve & Prepare \(Y(s)\) 🔗: Isolate \(Y(s)\text{,}\) then reshape it so every term matches an entry in the transform table. This may require tools like partial-fraction decomposition and completing the square.
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Step 3— Inverse Transform 🔗: Apply \(\laplacesym^{-1}\) term-by-term. The hidden original-domain solution \(y(t)\) is released, and the IVP is solved.
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Checkpoint 230. Recap Checkpoint Questions.

Answer the following questions referring to the steps above.

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(a) 📖❓ Laplace Method Snapshot.

Which option best represents the Laplace transform process?

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Forward Transform → Find \(Y(s)\) → Prepare → Apply Inverse
Correct! The Laplace Transform Method involves applying the forward transform, solving for \(Y(s) \text{,}\) preparing for the backward transform, and applying the inverse transform to find \(y(t) \text{.}\)
Differentiate → Forward Transform → Find \(Y(s)\) → Verify Solution
Not quite. The Laplace Transform Method does not involve differentiating the equation but rather applying the forward transform to convert it into an algebraic form.
Forward Transform → Integrate → Verify Solution
Incorrect. The Laplace Transform Method involves applying the forward transform to simplify the differential equation, not to verify the solution.
Forward Transform → Apply Inverse → Apply Initial Conditions
Incorrect. The Laplace Transform Method involves applying the forward transform to simplify the differential equation, not to check the solution.

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(b) 📖❓ Match Each Move.

Drag each task to the step it belongs to.

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Apply initial conditions
Match terms of \(y(t)\) the table
Step 1 — Forward Transform
Isolate \(Y(s)\)
Partial-fraction decomposition
Step 2 — Solve & Prepare \(Y(s)\)
Match terms of \(Y(s)\) the table
Step 3 — Inverse Transform

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(c) 📖❓ Algebra Tools Checklist.

Select every technique that may appear in Step 2.

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Completing the square
Partial-fraction decomposition
Integration by parts
Chain Rule

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The three-step outline provides the general strategy. If you ever feel lost, come back to this outline and roadmap. Regardless of the equation, the steps you take to navigate them will always follow one of these three lines.

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Subsection Worked Examples

The examples below walk through the full process, showing how to transform a differential equation, solve in the Laplace domain, and recover the final solution using the inverse transform. Each is annotated to show where each step begins and ends, so you can clearly see the roadmap in action. Here is a break down of what to expect:

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Examples 1–2 ⟶ pure practice: constant-coefficient, homogeneous equations.

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Examples 3–4 ⟶ common forcing terms: exponentials, polynomials, and sines/cosines.

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Examples 5–6 ⟶ algebra heavyweights: repeated roots, mixed techniques, and higher order.

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🌌 Example 231. Basic Second-Order, Homogeneous.

Solve the initial value problem using the Laplace transform:

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\begin{equation*} y'' + 5y' + 6y = 0, \quad y(0) = 2,\quad y'(0) = -1 \end{equation*}

Solution.

Specific-Roadmap.

\begin{equation*} \small\ul{\quad\ \textbf{Original Domain}\ \quad} \end{equation*}

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\begin{equation*} \small\DLBa\textbf{2️⃣ Laplace Domain} \end{equation*}

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\begin{gather*} \small \os{\vphantom{m}}{ y'' + 5y' + 6y = 0}\\ \small y(0) = 2,\ y'(0) = -1\\ \small \os{\large 🔺 🔺 🔺 🔺 🔺}{\text{Differential Equation}}\\ \\ \small \us{\large 🔻 🔻 🔻 🔻 🔻}{\text{Solution}}\\ \small y = e^{3t} + 2e^{-3t} - 2e^{2t} \end{gather*}

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

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\begin{align*} \amp\small\DLBa s^2Y - 2s + 1 + 5sY - 10 + 6Y = 0\\ \amp\small\qquad\qquad {\Big\downarrow}\quad\text{Solve for}\ Y\\ \amp\small\DLBa Y(s) = \frac{2s + 9}{s^2 + 5s + 6}\\ \amp\small\qquad\qquad {\Big\downarrow}\ \ \text{Prepare for Inverse}\\ \amp\small\DLBa Y(s) = \frac{1}{s + 2} + \frac{1}{s + 3} \end{align*}

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Step 1— Forward Transform

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\begin{align*} \lap{y'' + 5y' + 6y} = \lap{0}\\ s^2Y - sy(0) - y'(0) + 5(sY - y(0)) + 6Y \amp = 0\\ s^2Y - 2s + 1 + 5sY - 10 + 6Y \amp = 0 \end{align*}

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Laplace Domain Equation

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Step 2— Solve & Prepare \(Y(s)\)

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First, isolate \(Y\) using algebra:

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\begin{align*} (s^2 + 5s + 6)Y \amp = 2s + 9 \\ Y(s) \amp = \frac{2s + 9}{s^2 + 5s + 6} \end{align*}

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Isolated \(Y\)

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Factor the denominator:

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

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Use partial fraction decomposition:

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

Multiply through by the denominator:

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\begin{equation*} 2s + 9 = A(s + 3) + B(s + 2) \end{equation*}

Strategically selecting \(s\)-values gives \(A\) and \(B\text{:}\)

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\(s=-2\text{:}\)

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\(2(-2) + 9 = A(-2 + 3) + B(-2 + 2)\)

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\(\implies\boxed{A = 5}\)

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\(s=-3\text{:}\)

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\(2(-3) + 9 = A(-3 + 3) + B(-3 + 2)\)

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\(\implies\boxed{B = -3}\)

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Updating the Laplace solution, we have

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\begin{equation*} Y(s) = \frac{1}{s + 2} + \frac{1}{s + 3} \end{equation*}

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Prepared \(Y\)

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Step 3— Inverse Transform

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\begin{gather*} y(t) = \ilap{Y(s)} = \ilap{\frac{1}{s + 2}} + \ilap{\frac{1}{s + 3}} \end{gather*}

Using the table, the solution is:

\begin{equation*} y(t) = e^{-2t} + e^{-3t} \end{equation*}

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🌌 Example 232. Nonhomogeneous Forcing Function.

Solve the initial value problem:

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\begin{equation*} y'' - 2y' + y = e^{t}, \quad y(0) = 0,\quad y'(0) = 1 \end{equation*}

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Solution.

Specific-Roadmap.

\begin{equation*} \small\ul{\quad\ \textbf{Original Domain}\ \quad} \end{equation*}

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\begin{equation*} \small\DLBa\textbf{2️⃣ Laplace Domain} \end{equation*}

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\begin{gather*} \small \os{\vphantom{m}}{ y'' - 2y' + y = e^{t}}\\ \small y(0) = 0,\ y'(0) = 1\\ \small \os{\large 🔺 🔺 🔺 🔺 🔺}{\text{Differential Equation}}\\ \\ \\ \small \us{\large 🔻 🔻 🔻 🔻 🔻}{\text{Solution}}\\ \small y(t) = t e^{t} + \frac12 t^2 e^{t} \end{gather*}

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

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\begin{align*} \amp\small\DLBa s^2Y - sy(0) - y'(0)\\ \amp\small\DLBa \qquad\qquad - 2(sY - y(0)) + Y = \frac{1}{s-1}\\ \amp\small\qquad\qquad {\Big\downarrow}\quad\text{Solve for}\ Y\\ \amp\small\DLBa Y(s) = \frac{1}{(s-1)^3} + \frac{1}{(s-1)^2}\\ \amp\small\qquad\qquad {\Big\downarrow}\ \ \text{Prepare for Inverse}\\ \amp\small\DLBa Y(s)\ \text{already matches} \end{align*}

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Step 1— Forward Transform

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\begin{align*} \lap{y'' - 2y' + y} = \lap{e^{t}}\\ s^2Y - sy(0) - y'(0) - 2(sY - y(0)) + Y \amp = \frac{1}{s-1}\\ s^2Y - 1 - 2sY + Y \amp = \frac{1}{s-1} \end{align*}

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Laplace Domain Equation

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Step 2— Solve & Prepare \(Y(s)\)

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Isolate \(Y\text{:}\)

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\begin{align*} (s^2 - 2s + 1)Y \amp = \frac{1}{s-1} + 1\\ (s-1)^2 Y \amp = \frac{1}{s-1} + 1\\ Y(s) \amp = \frac{1}{(s-1)^3} + \frac{1}{(s-1)^2} \end{align*}

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Isolated \(Y(s)\)

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Step 3— Inverse Transform

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Notice each term of \(Y(s)\) is already in the transform table form:

\begin{equation*} \frac{1}{(s-1)^2} \quad\text{and}\quad \frac{1}{(s-1)^3}. \end{equation*}

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Take the inverse Laplace transform:

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\begin{gather*} y(t) = \ilap{\frac{1}{(s-1)^2}} + \ilap{\frac{1}{(s-1)^3}} \end{gather*}

Therefore, the solution is:

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\begin{equation*} y(t) = t e^{t} + \frac12 t^2 e^{t} \end{equation*}

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Checkpoint 233. 📖❓ Order the Steps.

Put these phrases in the order they occur when you solved Example 2.

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Rewrite \(Y(s)\) as table terms
3
Isolate \(Y(s)\)
2
Transform the differential equation
1
Apply \(\laplacesym^{-1}\)
4

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🌌 Example 234. Forcing with a Polynomial.

Solve the initial value problem:

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\begin{equation*} y'' + y = t^2, \quad y(0) = 1,\quad y'(0) = 0 \end{equation*}

Solution.

Specific-Roadmap.

\begin{equation*} \small\ul{\quad\ \textbf{Original Domain}\ \quad} \end{equation*}

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\begin{equation*} \small\DLBa\textbf{2️⃣ Laplace Domain} \end{equation*}

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\begin{gather*} \small \os{\vphantom{m}}{ y'' + y = t^2 }\\ \small y(0) = 1,\ y'(0) = 0\\ \small \os{\large 🔺 🔺 🔺 🔺 🔺}{\text{Differential Equation}}\\ \\ \small \us{\large 🔻 🔻 🔻 🔻 🔻}{\text{Solution}}\\ \small y(t) = 1 + \cos(t) + t\sin(t) \end{gather*}

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\begin{align*} \amp\small\DLBa s^2Y - sy(0) - y'(0) + Y = \frac{2}{s^3}\\ \amp\small\qquad\qquad {\Big\downarrow}\quad\text{Solve for}\ Y\\ \amp\small\DLBa Y(s) = \frac{2 + s}{s(s^2 + 1)}\\ \amp\small\qquad\qquad {\Big\downarrow}\ \ \text{Prepare for Inverse}\\ \amp\small\DLBa Y(s) = \frac{1}{s} + \frac{s}{s^2 + 1} + \frac{1}{(s^2 + 1)^2} \end{align*}

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Step 1— Forward Transform

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\begin{gather*} \lap{y'' + y} = \lap{t^2}\\ s^2Y - sy(0) - y'(0) + Y = \frac{2}{s^3}\\ s^2Y - s + Y = \frac{2}{s^3} \end{gather*}

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Laplace Domain Equation

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Step 2— Solve & Prepare \(Y(s)\)

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Isolate \(Y(s)\text{:}\)

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\begin{align*} (s^2 + 1) Y \amp = \frac{2}{s^3} + s\\ Y(s) \amp = \frac{2 + s^4}{s^3 (s^2 + 1)} = \frac{s(s^3 + 2)}{s^3 (s^2 + 1)} = \frac{2 + s}{s(s^2 + 1)} \end{align*}

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Isolated \(Y(s)\)

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Step 2— Solve & Prepare \(Y(s)\)

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Break \(Y(s)\) into recognizable pieces:

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

Each term now matches an entry in the Laplace transform table.

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Step 3— Inverse Transform

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Take the inverse Laplace transform term by term:

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\begin{gather*} y(t) = \ilap{\frac{1}{s}} + \ilap{\frac{s}{s^2 + 1}} + \ilap{\frac{1}{(s^2 + 1)^2}} \end{gather*}

Therefore, the solution is:

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\begin{equation*} y(t) = 1 + \cos(t) + t\sin(t) \end{equation*}

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🌌 Example 235. Repeated Root in the Characteristic Equation.

Solve the initial value problem using the Laplace transform:

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\begin{equation*} y'' + 6y' + 9y = 0, \quad y(0) = 3,\quad y'(0) = -3 \end{equation*}

Solution.

Specific-Roadmap.

\begin{equation*} \small\ul{\quad\ \textbf{Original Domain}\ \quad} \end{equation*}

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\begin{equation*} \small\DLBa\textbf{2️⃣ Laplace Domain} \end{equation*}

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\begin{gather*} \small \os{\vphantom{m}}{ y'' + 6y' + 9y = 0 }\\ \small y(0) = 3,\ y'(0) = -3\\ \small \os{\large 🔺 🔺 🔺 🔺 🔺}{\text{Differential Equation}}\\ \\ \small \us{\large 🔻 🔻 🔻 🔻 🔻}{\text{Solution}}\\ \small y(t) = (3 - 24t)e^{-3t} \end{gather*}

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\begin{align*} \amp\small\DLBa s^2Y - 3s + 3 + 6sY - 18 + 9Y = 0\\ \amp\small\qquad\qquad {\Big\downarrow}\quad\text{Solve for}\ Y\\ \amp\small\DLBa Y(s) = \frac{3s - 15}{(s + 3)^2}\\ \amp\small\qquad\qquad {\Big\downarrow}\ \ \text{Prepare for Inverse}\\ \amp\small\DLBa Y(s) = \frac{3}{s + 3} - \frac{24}{(s + 3)^2} \end{align*}

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Step 1— Forward Transform

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\begin{align*} \lap{y'' + 6y' + 9y} = \lap{0}\\ s^2Y - sy(0) - y'(0) + 6(sY - y(0)) + 9Y \amp = 0\\ s^2Y - 3s + 3 + 6sY - 18 + 9Y \amp = 0 \end{align*}

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Laplace Domain Equation

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Step 2— Solve & Prepare \(Y(s)\)

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Isolate \(Y(s)\text{:}\)

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\begin{align*} (s^2 + 6s + 9)Y \amp = 3s - 15\\ Y(s) \amp = \frac{3s - 15}{(s + 3)^2} \end{align*}

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Isolated \(Y(s)\)

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Step 2— Solve & Prepare \(Y(s)\)

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Separate the numerator:

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\begin{equation*} Y(s) = \frac{3(s + 3) - 24}{(s + 3)^2} = \frac{3}{s + 3} - \frac{24}{(s + 3)^2}. \end{equation*}

Now each term matches an entry in the Laplace table.

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Step 3— Inverse Transform

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Take the inverse Laplace transform term by term:

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\begin{gather*} y(t) = \ilap{\frac{3}{s + 3}} - \ilap{\frac{24}{(s + 3)^2}} \end{gather*}

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\begin{equation*} y(t) = (3 - 24t)e^{-3t} \end{equation*}

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🌌 Example 236. Third-Order Equation.

Solve the initial value problem using the Laplace transform:

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\begin{equation*} y''' + y'' - y' - y = 0,\quad y(0) = -1,\quad y'(0) = 0,\quad y''(0) = -1 \end{equation*}

Solution.

Specific-Roadmap.

\begin{equation*} \small\ul{\quad\ \textbf{Original Domain}\ \quad} \end{equation*}

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\begin{equation*} \small\DLBa\textbf{2️⃣ Laplace Domain} \end{equation*}

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\begin{gather*} \small \os{\vphantom{m}}{ y''' + y'' - y' - y = 0 }\\ \small y(0) = 1,\ y'(0) = 0,\ y''(0) = -1\\ \small \os{\large 🔺 🔺 🔺 🔺 🔺}{\text{Differential Equation}}\\ \\ \small \us{\large 🔻 🔻 🔻 🔻 🔻}{\text{Solution}}\\ \small y(t) = -\frac12e^{t} - \frac12e^{-t} \end{gather*}

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\begin{align*} \amp\small\DLBa s^3Y + s^2 + 1 + s^2Y + s\\ \amp\small\DLBa \qquad\qquad - sY - 1 - Y = 0\\ \amp\small\qquad\qquad {\Big\downarrow}\quad\text{Solve for}\ Y\\ \amp\small\DLBa Y(s) = \frac{-s^2 - s}{s^3 + s^2 - s - 1}\\ \amp\small\qquad\qquad {\Big\downarrow}\ \ \text{Prepare for Inverse}\\ \amp\small\DLBa Y(s) = -\frac12\frac{1}{s - 1} - \frac12\frac{1}{s + 1} \end{align*}

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Step 1— Forward Transform

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\begin{gather*} \lap{y''' + y'' - y' - y} = \lap{0}\\ s^3Y - s^2y(0) - sy'(0) - y''(0) + s^2Y - sy(0) - y'(0) - (sY - y(0)) - Y = 0\\ s^3Y - s^2(-1) - (-1) + s^2Y - s(-1) - (sY - (-1)) - Y = 0\\ s^3Y + s^2 + 1 + s^2Y + s - sY - 1 - Y = 0 \end{gather*}

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Step 2— Solve & Prepare \(Y(s)\)

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Isolate \(Y(s)\text{:}\)

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\begin{align*} (s^3 + s^2 - s - 1)Y \amp = -s^2 - s\\ Y(s) \amp = \frac{-s^2 - s}{s^3 + s^2 - s - 1} \end{align*}

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Step 2— Solve & Prepare \(Y(s)\)

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This denominator factors by grouping:

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\begin{align*} s^3 + s^2 - s - 1 \amp = s^2(s + 1) - (s + 1)\\ \amp = (s^2 - 1)(s + 1) = (s - 1)(s + 1)^2 \end{align*}

Before jumping to PFD, simplify \(Y\) as much as possible first:

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

Now PFD is a little easier. Here is the general form:

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\begin{equation*} Y(s) = \frac{-s}{(s - 1)(s + 1)} = \frac{A}{s - 1} + \frac{B}{s + 1} \end{equation*}

Applying the standard process gives: \(A = -\sfrac12\text{,}\) \(B = -\sfrac12\)

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So the prepared \(Y\) is:

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\begin{equation*} Y(s) = \frac{-\sfrac12}{s - 1} + \frac{-\sfrac12}{s + 1} \end{equation*}

Step 3— Inverse Transform

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Take the inverse Laplace transform term by term:

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\begin{gather*} y(t) = -\frac12\ilap{\frac{1}{s - 1}} - \frac12\ilap{\frac{1}{s + 1}} \end{gather*}

giving us the solution:

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\begin{equation*} y(t) = -\frac12e^{t} - \frac12e^{-t} \end{equation*}

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🌌 Example 237. Sinusoidal Forcing with Exponential Growth.

Solve the initial value problem using the Laplace transform:

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\begin{equation*} y'' + 2y = e^t \cos(3t), \quad y(0) = 0, \quad y'(0) = 0 \end{equation*}

Solution.

Specific-Roadmap.

\begin{equation*} \small\ul{\quad\ \textbf{Original Domain}\ \quad} \end{equation*}

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\begin{equation*} \small\DLBa\textbf{2️⃣ Laplace Domain} \end{equation*}

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\begin{gather*} \small \os{\vphantom{m}}{ y'' + 2y = e^{t} \cos(3t) }\\ \small y(0) = 0,\ y'(0) = 0\\ \small \os{\large 🔺 🔺 🔺 🔺 🔺}{\text{Differential Equation}}\\ \\ \small \us{\large 🔻 🔻 🔻 🔻 🔻}{\text{Solution}}\\ \small y(t) = e^{t}\cos(3t) - \cos(\sqrt{2}t) \end{gather*}

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\begin{align*} \amp\small\DLBa s^2 Y + 2Y = \frac{s - 1}{(s - 1)^2 + 9}\\ \amp\small\qquad\qquad {\Big\downarrow}\quad\text{Solve for}\ Y\\ \amp\small\DLBa Y(s) = \frac{s - 1}{(s^2 + 2)[(s - 1)^2 + 9]}\\ \amp\small\qquad\qquad {\Big\downarrow}\ \ \text{Prepare for Inverse}\\ \amp\small\DLBa Y(s) = \frac{1}{(s - 1)^2 + 9} - \frac{s}{s^2 + 2} \end{align*}

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Step 1— Forward Transform

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\begin{gather*} \lap{y'' + 2y} = \lap{e^{t} \cos(3t)}\\ s^2 Y + 2Y = \frac{s - 1}{(s - 1)^2 + 9} \end{gather*}

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Laplace Domain Equation

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Step 2— Solve & Prepare \(Y(s)\)

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Isolate \(Y(s)\text{:}\)

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\begin{align*} Y(s) \amp = \frac{s - 1}{(s^2 + 2)[(s - 1)^2 + 9]} \end{align*}

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Isolated \(Y(s)\)

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Step 2— Solve & Prepare \(Y(s)\)

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Rewrite \(Y(s)\) into pieces that match table entries:

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

Each piece is now recognizable for inverse transforming.

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Step 3— Inverse Transform

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Take the inverse Laplace transform of each term:

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\begin{gather*} y(t) = \ilap{\frac{1}{(s - 1)^2 + 9}} - \ilap{\frac{s}{s^2 + 2}} \end{gather*}

Therefore, the solution is:

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\begin{equation*} y(t) = e^{t}\cos(3t) - \cos(\sqrt{2}t) \end{equation*}

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These examples highlight how the Laplace Transform Method simplifies solving differential equations by converting them into algebraic equations and then back into the time domain. By mastering these steps, you can tackle a wide range of initial-value problems with ease and precision.

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Subsection 📤 Wrap-Up

🗝️ \(\textbf{Key Takeaways...}\)

The three-step Laplace roadmap never changes; the variety comes from the algebra you need in Step 2.

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Forward transform ⟶ algebra-only world ⟶ isolate \(Y(s)\) ⟶ prepare ⟶ inverse transform.

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Check Your Understanding.

Checkpoint 238. 🤔💭 Separation of Variables Reading Questions.

(a) 🤔💭 Who Does What?

Match each step with its primary role.

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Turns the DE into an algebra problem
Step 1 — Forward Transform
Makes every term inverse-friendly
Step 2 — Solve & Prepare \(Y(s)\)
Reveals \(y(t)\)
Step 3 — Inverse Transform

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(b) 🤔💭 Where Do ICs Appear?

Initial conditions influence only Step 1.

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True
Right — after Step 1 they are baked into \(Y(s)\text{.}\)
False

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