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Section 12.4 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, review the big picture, and 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.

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:
  1. How do we translate the differential equation into the Laplace domain?
  2. How do we isolate and tidy up \(Y(s)\) so it matches the inverse table?
  3. How do we return to the original time domain to reveal \(y(t)\text{?}\)

Laplace Method Roadmap.

\begin{equation*} \ul{\quad\ \textbf{Original Domain}\ \quad} \end{equation*}
\begin{equation*} \DLBa\textbf{2️⃣ Laplace Domain} \end{equation*}
\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*}
\begin{gather*} \underrightarrow{\text{1️⃣ Forward}}\\ \text{Apply}\ \laplacesym\\ \\ \\ \\ \\ \text{Apply}\ \laplacesym^{-1}\\ \overleftarrow{\text{3️⃣ Backward}} \end{gather*}
\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*}

✍🏻 Method 246. 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:
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).
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.
Step 3— Inverse Transform
Apply \(\laplacesym^{-1}\) term-by-term. The hidden original-domain solution \(y(t)\) is released, and the IVP is solved.

Checkpoint 247. Recap Checkpoint Questions.

Answer the following questions referring to the steps above.

(a) 📖❓ Laplace Method Snapshot.

Which option best represents the Laplace transform process?
  • Forward Transform → Find \(Y(s)\) → Prepare → Apply Inverse
  • Correct! The Laplace Transform Method involves applying the forward transform to obtain \(Y(s)\text{,}\) preparing for the backward transform, and applying the inverse transform to obtain \(y(t)\text{.}\)
  • Differentiate → Forward Transform → Find \(Y(s)\) → Verify Solution
  • Not quite. The Laplace Transform Method does not involve differentiating the equation; rather, it applies 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.

(b) 📖❓ Match Each Move.

(c) 📖❓ Algebra Tools Checklist.

Select every technique that may appear in Step 2.
  • Completing the square
  • Partial-fraction decomposition
  • Integration by parts
  • Chain Rule
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.

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:
  • Examples 1-2 ⟶ pure practice: constant-coefficient, homogeneous equations.
  • Examples 3-4 ⟶ common forcing terms: exponentials, polynomials, and sines/cosines.
  • Examples 5-6 ⟶ algebra heavyweights: repeated roots, mixed techniques, and higher order.

🌌 Example 248. Basic Second-Order, Homogeneous.

Solve the initial value problem using the Laplace transform:
\begin{equation*} y'' + 5y' + 6y = 0, \quad y(0) = 2,\quad y'(0) = -1 \end{equation*}
Solution.
\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*}
Laplace Domain Equation
First, isolate \(Y\) using algebra:
\begin{align*} (s^2 + 5s + 6)Y \amp = 2s + 9 \\ Y(s) \amp = \frac{2s + 9}{s^2 + 5s + 6} \end{align*}
Isolated \(Y\)
Factor the denominator:
\begin{gather*} Y(s) = \frac{2s + 9}{(s + 2)(s + 3)} \end{gather*}
Use partial fraction decomposition:
\begin{equation*} \frac{2s + 9}{(s + 2)(s + 3)} = \frac{A}{s + 2} + \frac{B}{s + 3} \end{equation*}
Multiply through by the denominator:
\begin{equation*} 2s + 9 = A(s + 3) + B(s + 2) \end{equation*}
Strategically selecting \(s\)-values gives \(A\) and \(B\text{:}\)
\(s=-2\text{:}\)
\(2(-2) + 9 = A(-2 + 3) + B(-2 + 2)\)
\(\implies\boxed{A = 5}\)
\(s=-3\text{:}\)
\(2(-3) + 9 = A(-3 + 3) + B(-3 + 2)\)
\(\implies\boxed{B = -3}\)
Updating the Laplace solution, we have
\begin{equation*} Y(s) = \frac{5}{s + 2} + \frac{-3}{s + 3} \end{equation*}
Prepared \(Y\)
Step 3— Inverse Transform
\begin{gather*} y(t) = \ilap{Y(s)} = 5\ilap{\frac{1}{s + 2}} - 3\ilap{\frac{1}{s + 3}} \end{gather*}
Using the table, the solution is:
\begin{equation*} y(t) = 5e^{-2t} - 3e^{-3t} \end{equation*}

🌌 Example 249. Nonhomogeneous Forcing Function.

Solve the initial value problem:
\begin{equation*} y'' - 2y' + y = e^{t}, \quad y(0) = 0,\quad y'(0) = 1 \end{equation*}
Solution.
\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*}
Laplace Domain Equation
Isolate \(Y\text{:}\)
\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*}
Isolated \(Y(s)\)
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*}
Take the inverse Laplace transform:
\begin{gather*} y(t) = \ilap{\frac{1}{(s-1)^2}} + \ilap{\frac{1}{(s-1)^3}} \end{gather*}
Therefore, the solution is:
\begin{equation*} y(t) = t e^{t} + \frac12 t^2 e^{t} \end{equation*}

Checkpoint 250. 📖❓ Order the Steps.

🌌 Example 251. Forcing with a Polynomial.

Solve the initial value problem:
\begin{equation*} y'' + y = t^2, \quad y(0) = 1,\quad y'(0) = 0 \end{equation*}
Solution.
\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*}
Laplace Domain Equation
Isolate \(Y(s)\text{:}\)
\begin{align*} (s^2 + 1) Y \amp = \frac{2}{s^3} + s\\ Y(s) \amp = \frac{s}{s^2 + 1} + \frac{2}{s^3(s^2 + 1)} \end{align*}
Isolated \(Y(s)\)
Break \(Y(s)\) into recognizable pieces:
\begin{equation*} Y(s) = \frac{s}{s^2 + 1} + \frac{2}{s^3(s^2 + 1)} = \frac{3s}{s^2 + 1} - \frac{2}{s} + \frac{2}{s^3}. \end{equation*}
Each term now matches an entry in the Laplace transform table.
Take the inverse Laplace transform term by term:
\begin{gather*} y(t) = \ilap{\frac{3s}{s^2 + 1}} - \ilap{\frac{2}{s}} + \ilap{\frac{2}{s^3}} \end{gather*}
Therefore, the solution is:
\begin{equation*} y(t) = 3\cos(t) + t^2 - 2 \end{equation*}

🌌 Example 252. Repeated Root in the Characteristic Equation.

Solve the initial value problem using the Laplace transform:
\begin{equation*} y'' + 6y' + 9y = 0, \quad y(0) = 3,\quad y'(0) = -3 \end{equation*}
Solution.
\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*}
Laplace Domain Equation
Isolate \(Y(s)\text{:}\)
\begin{align*} (s^2 + 6s + 9)Y \amp = 3s - 15\\ Y(s) \amp = \frac{3s - 15}{(s + 3)^2} \end{align*}
Isolated \(Y(s)\)
Separate the numerator:
\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.
Take the inverse Laplace transform term by term:
\begin{gather*} y(t) = \ilap{\frac{3}{s + 3}} - \ilap{\frac{24}{(s + 3)^2}} \end{gather*}
So
\begin{equation*} y(t) = (3 - 24t)e^{-3t} \end{equation*}

🌌 Example 253. Third-Order Equation.

Solve the initial value problem using the Laplace transform:
\begin{equation*} y''' + y'' - y' - y = 0,\quad y(0) = -1,\quad y'(0) = 0,\quad y''(0) = -1 \end{equation*}
Solution.
\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*}
Isolate \(Y(s)\text{:}\)
\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*}
This denominator factors by grouping:
\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:
\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:
\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: \(A = -\sfrac12\text{,}\) \(B = -\sfrac12\)
So the prepared \(Y\) is:
\begin{equation*} Y(s) = \frac{-\sfrac12}{s - 1} + \frac{-\sfrac12}{s + 1} \end{equation*}
Take the inverse Laplace transform term by term:
\begin{gather*} y(t) = -\frac12\ilap{\frac{1}{s - 1}} - \frac12\ilap{\frac{1}{s + 1}} \end{gather*}
giving us the solution:
\begin{equation*} y(t) = -\frac12e^{t} - \frac12e^{-t} \end{equation*}

🌌 Example 254. Sinusoidal Forcing with Exponential Growth.

Solve the initial value problem using the Laplace transform:
\begin{equation*} y'' + 2y = e^t \cos(3t), \quad y(0) = 0, \quad y'(0) = 0\text{.} \end{equation*}
Solution.
\begin{gather*} \lap{y'' + 2y} = \lap{e^{t} \cos(3t)}\\ s^2 Y + 2Y = \frac{s - 1}{(s - 1)^2 + 9} \end{gather*}
Laplace Domain Equation
Isolate \(Y(s)\text{:}\)
\begin{align*} Y(s) \amp = \frac{s - 1}{(s^2 + 2)[(s - 1)^2 + 9]} \end{align*}
Isolated \(Y(s)\)
Rewrite \(Y(s)\) into pieces that match table entries:
\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.
Take the inverse Laplace transform of each term:
\begin{gather*} y(t) = \ilap{\frac{1}{(s - 1)^2 + 9}} - \ilap{\frac{s}{s^2 + 2}} \end{gather*}
Therefore, the solution is:
\begin{equation*} y(t) = e^{t}\cos(3t) - \cos(\sqrt{2}t) \end{equation*}
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.

Checkpoint 255. 🤔💭 Reading Questions.

(a) 📖❓ Who Does What?

(b) 📖❓ Where Do ICs Appear?

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