Mastering Laplace Transform of Derivatives: Solved Problems for ODEs

📚 Introduction to Laplace Transforms and ODEs

The Laplace transform is a powerful mathematical tool used to solve linear ordinary differential equations (ODEs). It transforms a differential equation in the time domain into an algebraic equation in the complex frequency domain, which is often easier to solve. Once solved in the frequency domain, the inverse Laplace transform converts the solution back to the time domain.

• ⏱️ The Laplace transform simplifies solving ODEs, especially those with initial conditions.
• 📈 It finds wide applications in engineering, physics, and other scientific disciplines.
• 💡 Understanding the Laplace transform of derivatives is crucial for effectively applying this technique to ODEs.

📜 History and Background

The Laplace transform is named after Pierre-Simon Laplace, who introduced a similar transform in his work on probability theory. The modern form of the Laplace transform was developed in the 19th and 20th centuries by mathematicians and engineers seeking to solve differential equations arising in various fields.

• 🧑‍🏫 Pierre-Simon Laplace laid the groundwork for the transform.
• ⚙️ Oliver Heaviside developed operational calculus, a precursor to the modern Laplace transform.
• 🗓️ The transform gained widespread use in engineering during the mid-20th century.

🔑 Key Principles: Laplace Transform of Derivatives

The key to using Laplace transforms to solve ODEs lies in understanding how derivatives transform. Here are the essential formulas:

Let $L{f(t)} = F(s)$ denote the Laplace transform of the function $f(t)$. Then:

First Derivative: $L{f'(t)} = sF(s) - f(0)$

Second Derivative: $L{f''(t)} = s^2F(s) - sf(0) - f'(0)$

Third Derivative: $L{f'''(t)} = s^3F(s) - s^2f(0) - sf'(0) - f''(0)$

In general, for the $n^{th}$ derivative:

$L{f^{(n)}(t)} = s^nF(s) - s^{n-1}f(0) - s^{n-2}f'(0) - ... - f^{(n-1)}(0)$

• 🧮 Formula 1: $L{f'(t)} = sF(s) - f(0)$ describes the transform of the first derivative.
• 📈 Formula 2: $L{f''(t)} = s^2F(s) - sf(0) - f'(0)$ handles the second derivative.
• ✨ Generalization: The formula for $L{f^{(n)}(t)}$ extends this pattern to higher-order derivatives.

📝 Solved Problems

Let's illustrate these principles with solved problems. Remember to use partial fraction decomposition when finding the inverse Laplace transform!

Problem 1: Solve $y'' + y = 0$, with $y(0) = 1$ and $y'(0) = 0$.

Solution: Taking the Laplace transform of both sides, we get:

$s^2Y(s) - sy(0) - y'(0) + Y(s) = 0$

Substituting the initial conditions:

$s^2Y(s) - s + Y(s) = 0$

$Y(s)(s^2 + 1) = s$

$Y(s) = \frac{s}{s^2 + 1}$

Taking the inverse Laplace transform:

$y(t) = cos(t)$

• ✅ Step 1: Apply the Laplace transform to the entire differential equation.
• 🔢 Step 2: Substitute initial conditions into the transformed equation.
• ➗ Step 3: Solve for $Y(s)$, the Laplace transform of the solution.
• 🔙 Step 4: Find the inverse Laplace transform to obtain the solution $y(t)$.

Problem 2: Solve $y' + 2y = e^{-t}$, with $y(0) = 0$.

Solution: Taking the Laplace transform of both sides:

$sY(s) - y(0) + 2Y(s) = \frac{1}{s + 1}$

Substituting the initial condition:

$sY(s) + 2Y(s) = \frac{1}{s + 1}$

$Y(s)(s + 2) = \frac{1}{s + 1}$

$Y(s) = \frac{1}{(s + 1)(s + 2)}$

Using partial fraction decomposition: $Y(s) = \frac{A}{s + 1} + \frac{B}{s + 2}$

$1 = A(s + 2) + B(s + 1)$

Solving for $A$ and $B$, we find $A = 1$ and $B = -1$. Thus,

$Y(s) = \frac{1}{s + 1} - \frac{1}{s + 2}$

Taking the inverse Laplace transform:

$y(t) = e^{-t} - e^{-2t}$

• ➕ Step 1: Laplace transform both sides of the equation.
• ✔️ Step 2: Plug in the initial condition.
• ➗ Step 3: Solve for $Y(s)$.
• 🧩 Step 4: Use partial fractions if necessary to simplify $Y(s)$.
• ↩️ Step 5: Take the inverse Laplace transform to find $y(t)$.

Problem 3: Solve $y'' - 3y' + 2y = 0$, with $y(0) = 1$ and $y'(0) = 0$.

Solution: Applying the Laplace transform:

$s^2Y(s) - sy(0) - y'(0) - 3(sY(s) - y(0)) + 2Y(s) = 0$

Substitute initial conditions:

$s^2Y(s) - s - 3sY(s) + 3 + 2Y(s) = 0$

Rearrange:

$Y(s)(s^2 - 3s + 2) = s - 3$

$Y(s) = \frac{s - 3}{s^2 - 3s + 2} = \frac{s - 3}{(s - 1)(s - 2)}$

Partial fraction decomposition: $Y(s) = \frac{A}{s - 1} + \frac{B}{s - 2}$

$s - 3 = A(s - 2) + B(s - 1)$

Solving, we find $A = 2$ and $B = -1$.

$Y(s) = \frac{2}{s - 1} - \frac{1}{s - 2}$

Inverse Laplace transform:

$y(t) = 2e^{t} - e^{2t}$

• 🔄 Step 1: Transform the ODE using Laplace.
• 🌱 Step 2: Plug in initial values.
• 🧮 Step 3: Algebraically solve for $Y(s)$.
• 🔨 Step 4: Decompose using partial fractions.
• ↩️ Step 5: Invert back to time domain.

🌍 Real-World Applications

Laplace transforms are used extensively in:

• ⚡ Electrical Engineering: Circuit analysis, signal processing.
• ⚙️ Mechanical Engineering: Control systems, vibration analysis.
• 🌡️ Chemical Engineering: Process control, reaction kinetics.

💡 Conclusion

Mastering the Laplace transform of derivatives is essential for solving a wide range of ODEs. By understanding the key principles and practicing with solved problems, you can confidently apply this powerful technique in various fields. Remember to carefully apply initial conditions and use partial fraction decomposition when necessary. Keep practicing, and you'll become proficient in using Laplace transforms to solve ODEs!