1.1 - Laplace Transform of Derivatives
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Introduction to Laplace Transform of Derivatives
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Welcome everyone! Today, we’re going to delve into the Laplace Transform of derivatives. Can anyone tell me why the Laplace Transform is an important tool in differential equations?
I think it converts differential equations into algebraic equations, which are easier to solve.
Exactly! It simplifies the problem-solving process. Now, let's start with the first derivative. The formula is L{f'(t)} = sF(s) - f(0). What does this mean?
It means we can express the Laplace Transform of the first derivative in terms of the function's value at zero.
Well done! Remember this: 'First Derivative gives F(s)'—it's a good mnemonic to remember the relationship.
Deriving the First Derivative Formula
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Let’s dive into how we prove L{f'(t)} = sF(s) - f(0). First, we use integration by parts. Can someone outline what integration by parts entails?
It involves splitting up a function so we can integrate it more simply, right?
Correct! Here, we set u = f(t) and dv = e^{-st}dt. Who can help with the next step of the proof?
We find d = -se^{-st} and then apply the integration limit from 0 to infinity.
Great! And what happens as t approaches infinity?
The entire term goes to zero because f(t) is of exponential order.
Exactly! So we conclude with L{f'(t)} = sF(s) - f(0). This formulation is vital as we proceed to higher derivatives.
Laplace Transform of Higher Derivatives
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Now, let's tackle the second derivative. What does L{f''(t)} equal?
It’s s^2F(s) - sf(0) - f'(0)!
Exactly! It builds on the first derivative. What about the n-th derivative? Can anyone summarize that?
L{f(n)(t)} = s^nF(s) - (sum of values at zero)... uh, I’m not sure about the summation part.
No problem! The general formula is L{f(n)(t)} = s^nF(s) - sum from k=0 to n-1 of s^{n-1-k} f(k)(0). This encapsulates all derivatives and their respective initial conditions.
Application to Initial Value Problems
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Let's see the practical applications. For instance, how would we approach an IVP like y'' + 5y' + 6y = 0 with given initial conditions?
We would apply the Laplace Transform to each term, right?
Correct! What does that yield?
We get (s^2Y(s) - sy(0) - y'(0)) + 5(sY(s) - y(0)) + 6Y(s) = 0.
Right! Substitute the initial conditions to simplify it further. Let’s discuss the next steps in solving for Y(s).
Introduction & Overview
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Quick Overview
Standard
This section focuses on the Laplace Transform of derivatives, illustrating how to apply this tool to first and higher-order derivatives for more effective solutions of ordinary differential equations (ODEs). Key formulas and proofs are provided to highlight the transformation's significance in engineering and physical sciences.
Detailed
Detailed Summary
The Laplace Transform is a fundamental tool in the fields of engineering and physical sciences, particularly when dealing with ordinary differential equations (ODEs). By transforming differential equations into algebraic equations, the Laplace Transform simplifies the problem-solving process.
This section provides a thorough examination of the Laplace Transform of derivatives. Firstly, we present the Laplace Transform of the first derivative which states:
$$L\{f'(t)\} = sF(s) - f(0)$$
This formula allows us to express the first derivative of a function in terms of its Laplace Transform. The proof involves integration by parts.
Next, we extend the concept to the second derivative and higher-order derivatives, leading to the general formula:
$$L\{f^{(n)}(t)\} = s^nF(s) - s^{n-1}f(0) - s^{n-2}f'(0) - ... - f^{(n-1)}(0)$$
This allows us to handle complex ODEs more easily. Additionally, examples of solving Initial Value Problems (IVPs) demonstrate practical applications of these transformations. The section concludes with emphasis on the Laplace Transform's utility in real-world applications like control systems and circuit analysis.
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Introduction to Laplace Transform of Derivatives
Chapter 1 of 6
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Chapter Content
The Laplace Transform is a powerful tool for solving differential equations, especially those arising in engineering and physical sciences. One of the key applications of the Laplace Transform is in converting differential equations into algebraic equations, which are easier to manipulate and solve. In this topic, we focus on the Laplace Transform of derivatives, which allows us to systematically handle ordinary differential equations (ODEs) involving first and higher-order derivatives.
Detailed Explanation
The Laplace Transform converts functions of time (like those found in differential equations) into functions of a complex variable (s). This transformation simplifies complex calculations, especially in the realm of engineering and physics. By turning differential equations into algebraic form, we make them easier to solve. This section specifically deals with how derivatives of functions can be transformed using the Laplace method, which is essential for tackling ordinary differential equations (ODEs).
Examples & Analogies
Imagine trying to predict the motion of a vehicle. The equations of motion can be quite complex if we directly deal with speed and acceleration. However, if we transform these equations into a different form, it becomes much easier to apply them and solve for the vehicle's path.
Laplace Transform of the First Derivative
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Chapter Content
Let f(t) be a function such that f′(t) exists and is continuous on [0, ∞). Then:
L{f′(t)}=sF(s)−f(0)
Detailed Explanation
This formula shows how the Laplace Transform of the first derivative of a function relates to the transform of the function itself (denoted as F(s)). Essentially, it states that the Laplace Transform of the derivative involves multiplying the original transform F(s) by the variable s and subtracting the initial value of the function at t = 0. This formula highlights the relationship between derivatives and the Laplace Transform, making it easier to solve differential equations that involve derivatives.
Examples & Analogies
Think of this in terms of a runner's speed. The Laplace Transform of speed (which is the derivative of distance) gives us the overall behavior of the runner's journey. By knowing the starting point, we can understand how speed changes over time.
Laplace Transform of the Second Derivative
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Chapter Content
L{f″(t)}=s²F(s)−sf(0)−f′(0)
Detailed Explanation
This formula extends the idea of the Laplace Transform to a second derivative. It expresses the Laplace Transform of the second derivative as a function of the transform of the original function and its initial values. More specifically, it involves the square of s, the transform of the function, and the initial values, which enables the systematic treatment of second derivatives in differential equations.
Examples & Analogies
Consider a car accelerating from a stop. The first derivative represents the car's speed, while the second derivative represents how quickly that speed is changing (acceleration). Using this transform allows us to analyze and predict the car's behavior over time based on initial conditions.
Laplace Transform of the n-th Derivative
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Chapter Content
For n∈N, the Laplace Transform of the n-th derivative is:
L{f(n)(t)}=s^nF(s)−s^(n−1)f(0)−s^(n−2)f′(0)−⋯−f(n−1)(0)
Detailed Explanation
This formula generalizes the concept of the Laplace Transform for any n-th derivative of a function. It states that the transform can be expressed in terms of the n-th power of s, the original function's transform, and the initial values of the function and its derivatives. This allows us to manage higher-order differential equations efficiently.
Examples & Analogies
Imagine a mechanical system like a suspension bridge. Each derivative represents different motions (position, speed, acceleration, etc.). The Laplace Transform helps us analyze the entire system's response to loads and stresses starting from initial conditions.
Applications of the Laplace Transform of Derivatives
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Chapter Content
Laplace Transform is widely used to solve Initial Value Problems (IVPs).
Example:
y″ + 5y′ + 6y = 0, y(0) = 2, y′(0) = 1
Apply Laplace on both sides:
L{y″} + 5L{y′} + 6L{y} = 0
(s²Y(s)−sy(0)−y′(0)) + 5(sY(s)−y(0)) + 6Y(s) = 0
Substitute initial conditions:
(s²Y(s)−2s−1) + 5(sY(s)−2) + 6Y(s) = 0 ⇒ Y(s)(s² + 5s + 6) = 2s + 1 + 10 = 2s + 11
Solve for Y(s), then use partial fractions and inverse Laplace.
Detailed Explanation
This chunk demonstrates a practical application of the Laplace Transform in solving Initial Value Problems (IVPs). By applying the Laplace Transform to each term of a differential equation, we can convert it into an algebraic equation that is simpler to manage. After substituting initial conditions, we can solve for the transformed function Y(s) and eventually return to the time domain using the inverse Laplace Transform.
Examples & Analogies
Think of it like solving a complex puzzle. Instead of dealing with pieced-together parts directly, using the Laplace Transform is like translating the puzzle on paper, making it easier to visualize and solve. Once we have the solution, we can translate it back to its original form.
Summary of Key Formulas
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Chapter Content
• Laplace Transform of Derivatives converts differentiation into algebraic terms involving s.
• This transformation simplifies solving differential equations.
• Formulas:
o L{f′(t)} = sF(s) - f(0)
o L{f″(t)} = s²F(s) - sf(0) - f′(0)
o L{f(n)(t)} = s^nF(s) - ∑(s^(n-1-k)f(k)(0)), for k=0 to n-1
Detailed Explanation
This summary encapsulates the fundamental concepts regarding the Laplace Transform of derivatives. It highlights how these transforms are instrumental in simplifying the solution of differential equations by converting differentiation into algebraic operations involving the variable s. Remembering these core formulas will aid in efficiently solving differential equations in various applications.
Examples & Analogies
Just like knowing the key steps in a recipe makes cooking easier, understanding these formulas makes solving differential equations straightforward. When you see a derivative in a problem, you can use these transformations to navigate through complexities with ease.
Key Concepts
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Laplace Transform of the First Derivative: L{f'(t)} = sF(s) - f(0)
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Laplace Transform of the Second Derivative: L{f''(t)} = s^2F(s) - sf(0) - f'(0)
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Laplace Transform of the n-th Derivative: L{f(n)(t)} = s^nF(s) - sum from k=0 to n-1 of s^{n-1-k} f(k)(0)
Examples & Applications
Example 1: For L{f'(t)} where f(t) = t, we find that L{f'(t)} = L{1} = 1/s^2.
Example 2: Apply the Laplace Transform to the equation y'' + 5y' + 6y = 0 using y(0) = 2 and y'(0) = 1.
Memory Aids
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Rhymes
If you want to solve with ease, just take the Laplace and seize.
Stories
Once in a land of equations, lived a function who wanted to find its path easily. It discovered a magical transform that turned complex derivatives into simple algebra, and everyone celebrated!
Memory Tools
For derivatives: L = S^F - I, where 'F' is Function, 'I' is Initial value.
Acronyms
D F (for Derivative Formula)
= F(s) - f(0) or S^nF(s) - Σ terms!
Flash Cards
Glossary
- Laplace Transform
An integral transform that converts a function of time into a function of a complex variable.
- Derivative
A measure of the rate at which a quantity changes.
- Initial Value Problem (IVP)
A problem that seeks to find a function satisfying a differential equation and specific values at a given point.
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