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Evaluate the Laplace transform of the given function. $$ f(t)=e^{2 t} u_{0}(t-2) $$

Short Answer

Expert verified
\( \mathcal{L}\{ e^{2t} u_{0}(t-2) \} = \frac{e^{-4s}}{s-2} \), for \( \Re(s) > 2 \).

Step by step solution

01

- Understand the Laplace Transform of an Exponential Multiplied by a Step Function

The Laplace transform of a function multiplied by a Heaviside step function, such as \( u_c(t) \), can be found by shifting the original function by the amount \( c \) in the time domain. This is expressed by the formula \( \mathcal{L} \{ e^{at} u_{0}(t-c) \} = \frac{e^{-cs}}{s-a} \) for \( \Re(s) > a \).
02

- Apply the Shifting Property

For the function \( f(t) = e^{2t} u_{0}(t-2) \), the step function shifts the exponential function by 2 units. Using the shifting property, we can write the Laplace transform as \( \mathcal{L}\{e^{2(t-2)}\}e^{-2s} \), which simplifies to \( \mathcal{L}\{e^{2t}\}e^{-4s} \) since we have translated the function by 2 units to the right.
03

- Calculate the Laplace Transform of the Exponential Function

From the Laplace transform table, we know that \( \mathcal{L}\{e^{at}\} = \frac{1}{s-a} \) for \( \Re(s) > a \). For the exponential function \( e^{2t} \), this gives us \( \mathcal{L}\{e^{2t}\} = \frac{1}{s-2} \), assuming that \( \Re(s) > 2 \).
04

- Combine the Results

Multiply the result from Step 3 by \( e^{-4s} \) to incorporate the effect of the step function. This gives us the final Laplace transform of the given function: \[ \mathcal{L}\{ e^{2t} u_{0}(t-2) \} = \frac{1}{s-2}e^{-4s} \] for \( \Re(s) > 2 \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Heaviside Step Function
The Heaviside step function, denoted as \( u_c(t) \), plays a critical role in Laplace transforms, particularly when dealing with piecewise functions. This function is defined to be zero for any time \( t \) less than some value \( c \), and one for any time \( t \) greater than or equal to \( c \). It effectively 'turns on' a function at a specific point in time.

In the context of Laplace transforms, the Heaviside step function allows us to handle functions that start or stop abruptly. For instance, if we have a function that is zero until time \( c \) and then follows a certain behavior for \( t \geq c \), we can represent this as the product of the function with a Heaviside step function. This technique is particularly useful in solving boundary value problems where conditions change at a certain point in time.

The visualization of this function is quite simple: imagine a light switch that remains off (zero) and suddenly turns on (one) at the desired time, shedding light on how the function behaves only after a certain point.
Exponential Functions
Exponential functions are an essential class of functions characterized by an equation of the form \( e^{at} \), where \( e \) is the base of the natural logarithm and \( a \) is a constant. These functions exhibit growth or decay at a constant rate and are ubiquitous in mathematical models for natural phenomena, including population growth, radioactive decay, and interest calculation.

In the realm of Laplace transforms, exponential functions are transformed into algebraic expressions which are easier to manipulate and solve in comparison to their original exponential form. By applying the Laplace transform, differential equations involving exponential terms can be converted into polynomial equations, enabling easier solutions for complex systems.
Shifting Property
The shifting property of the Laplace transform is an essential tool for dealing with functions that are delayed in time. Specifically, when you have a function that is multiplied by a Heaviside step function, such as \( f(t) = e^{at} u_c(t) \), the shifting property can be used to find its Laplace transform.

This property states that if a function \( f(t) \) is delayed by \( c \) units in time, then its Laplace transform is given by \( e^{-cs} \) times the Laplace transform of the original, undelayed function \( f(t) \). Applying it effectively helps in solving equations that involve time shifts and makes handling boundary conditions more straightforward in various applications, including control systems and signal processing.
Boundary Value Problems
Boundary value problems are a type of differential equation problem where the solution must satisfy certain conditions at the boundaries of the domain. These problems often arise in physical scenarios such as the temperature distribution along a rod at steady state, or the vibrations of a drumhead.

When solving boundary value problems using Laplace transforms, we use the transform to turn differential equations into algebraic ones, which are simpler to solve. Additionally, the Heaviside step function may come into play when the boundary conditions or sources change value at a particular point in time. By understanding and applying the Laplace transform effectively, with its shifting property, these typically challenging problems become more tractable, leading to solutions that encapsulate the physics of the scenario precisely.

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Most popular questions from this chapter

Evaluate the Laplace transform of the given function using appropriate theorems and examples from this section. $$ f(t)=t \sinh 3 t $$

Evaluate the inverse Laplace transform of the given function. $$ F(s)=\frac{2 s-1}{s^{2}-s-2} $$

Solution of Bessel's equation via the Laplace transform. (s) Show that after applying the Laplace transform to Bessel's equation of order 0, we obtain $$-\frac{d}{d s}\left[y^{2} Y-s y(0)-y^{\prime}(0)\right]+s Y-y(0)-Y^{\prime}=0$$ (b) Simplify the equation in \(Y\) to obtain \(\left(s^{2}+1\right) Y^{\prime \prime}=-8 Y\). (c) Solve the first order ordinary differential equation in \(Y\), and obtain that \(Y=\) \(\frac{K}{\sqrt{8^{2}+1}}\), where \(K\) is a constant. (d) Use the result of Exercise 50 to identify your solution as \(K J_{0}(t)\). You may wonder why the method did not yield a second solution, as expected from a second order differeatial equation. The reason is that by applying the Laplace transform. we are assuming that the (unknown) solution and its first and second derivatives have a Laplace transform. Since the second solution of Bessel's equation, \(Y_{0}\), is known to behave like Int for \(t\) near zero (Section 4.7, Exercise 28), the term ty" behaves like a constant multiple of \(\frac{1}{t}\) and does not have a Laplace transform, which explains why the method did not capture this second solution.

Use the Laplace transform to solve the given boundary volue problem, Give your answer in the form of an integral and simplify as much as possible. Whenever possible, wse the examples from this section without repeating the derivations. $$ \begin{aligned} &\frac{\partial u}{\partial t}=\frac{\partial^{2} u}{\partial x^{2}}, 00 \\ &u(0, t)=70, t>0 \\ &u(x, 0)=0,0

Solve the given initial value problem with the Laplace transform. $$ y^{\prime \prime}+y^{\prime}+4 y=0, \quad y(0)=1, y^{\prime}(0)=1 $$

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