Laplace Transform Double Derivative: Solved!

Unlocking the Laplace Transform of the Double Derivative

The "laplace transform double derivative" is a fundamental concept within Laplace transform theory, providing a powerful tool for solving differential equations, particularly those encountered in electrical engineering, mechanical systems, and control theory. This article provides a comprehensive explanation of how to derive and apply this transform.

Understanding the Laplace Transform

Before diving into the double derivative, a quick recap of the Laplace transform itself is helpful.

Defining the Laplace Transform

The Laplace transform is an integral transform that converts a function of time, f(t), into a function of a complex variable s, denoted as F(s). Mathematically, it’s defined as:

F(s) = ∫₀^∞ e^-st f(t) dt

Where:

• f(t) is the time-domain function.
• F(s) is the frequency-domain representation (Laplace transform) of f(t).
• s is a complex variable (s = σ + jω, where σ is the real part and ω is the imaginary part).
• The integral is evaluated from 0 to infinity.

Importance of the Laplace Transform

The Laplace transform is valuable because it converts differential equations into algebraic equations. This makes the solution process significantly simpler. By solving for F(s) algebraically and then performing the inverse Laplace transform, we can obtain the solution f(t) to the original differential equation.

Deriving the Laplace Transform of the Double Derivative

The key goal is to find the Laplace transform of f”(t), the second derivative of f(t). We’ll achieve this using integration by parts, relying on our knowledge of the Laplace transform of the first derivative.

Laplace Transform of the First Derivative

First, recall the Laplace transform of the first derivative, f'(t):

L{f'(t)} = sF(s) – f(0)

Where f(0) is the initial value of the function f(t) at t = 0. This formula is crucial for deriving the double derivative.

Applying Integration by Parts Twice

To find the Laplace transform of f”(t), we apply integration by parts twice. This involves splitting the integral into two parts and applying the formula ∫ u dv = uv – ∫ v du.

1. First Application: Let’s write the Laplace Transform integral for f”(t):

   L{f”(t)} = ∫₀^∞ e^-st f”(t) dt

   Let u = e^-st and dv = f”(t) dt. Then du = -se^-st dt and v = f'(t). Applying integration by parts:

   ∫₀^∞ e^-st f”(t) dt = [e^-st f'(t)]₀^∞ – ∫₀^∞ f'(t) (-se^-st) dt

   Assuming that e^-stf'(t) approaches 0 as t approaches infinity (a common condition for the Laplace transform to converge), the first term becomes -f'(0). The integral becomes:

   -f'(0) + s∫₀^∞ e^-st f'(t) dt

2. Second Application: Now, we need to evaluate ∫₀^∞ e^-st f'(t) dt, which is the Laplace transform of f'(t), which we know from the previous section is equal to sF(s) – f(0). Substituting, we get:

   -f'(0) + s[sF(s) – f(0)]

3. Result: Simplifying the expression, we obtain the Laplace transform of the double derivative:

   L{f”(t)} = s²F(s) – sf(0) – f'(0)

Formula Summary

The final formula for the Laplace transform of the double derivative is:

*L{f''(t)} = s<sup>2</sup>F(s) - sf(0) - f'(0)*

Where:

• F(s) is the Laplace transform of f(t).
• f(0) is the initial value of f(t) at t = 0.
• f'(0) is the initial value of f'(t) at t = 0.

Applying the Laplace Transform Double Derivative: Example

Consider the second-order differential equation:

y”(t) + 3y'(t) + 2y(t) = 0

With initial conditions:

y(0) = 1
y'(0) = 0

Transforming the Equation

Apply the Laplace transform to each term in the differential equation:

• L{y”(t)} = s²Y(s) – sy(0) – y'(0) = s²Y(s) – s
• L{3y'(t)} = 3[sY(s) – y(0)] = 3sY(s) – 3
• L{2y(t)} = 2Y(s)
• L{0} = 0

The transformed equation becomes:

s²Y(s) – s + 3sY(s) – 3 + 2Y(s) = 0

Solving for Y(s)

Rearrange and solve for Y(s):

Y(s)(s² + 3s + 2) = s + 3

Y(s) = (s + 3) / (s² + 3s + 2)

Y(s) = (s + 3) / ((s + 1)(s + 2))

Inverse Laplace Transform

Perform a partial fraction decomposition to find the inverse Laplace transform:

Y(s) = A / (s + 1) + B / (s + 2)

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

Therefore:

Y(s) = 2 / (s + 1) – 1 / (s + 2)

Applying the inverse Laplace transform:

y(t) = 2e^-t – e^-2t

This is the solution to the given differential equation. The Laplace transform of the double derivative played a crucial role in converting the differential equation into an algebraic equation, simplifying the solution process.

And there you have it! Hopefully, this deep dive into the laplace transform double derivative helped clear things up. Now go forth and conquer those tricky differential equations!