Question

Obtain the Laplace transform of the given function in terms of \(\mathcal{L}\\{f(t)\\}=F(s)\). For Exercise 48 , note that \(\int_{a}^{t} f(\lambda) d \lambda=\int_{0}^{t} f(\lambda) d \lambda-\int_{0}^{a} f(\lambda) d \lambda\). \(\int_{2}^{t} f(\lambda) d \lambda\), given that \(\int_{0}^{2} f(\lambda) d \lambda=3\)

Short answer

#tag_title# Step 2: Apply Laplace Transform properties #tag_content# Now we will take the Laplace transform of both sides of the equation: $$\mathcal{L}\{\int_0^t f(\lambda) d\lambda\} = \mathcal{L}\{\int_2^t f(\lambda) d\lambda - \int_0^2 f(\lambda) d\lambda\}$$ Recall that the Laplace transform of the integral of a function is given by: $$\mathcal{L}\{\int_0^t f(\lambda) d\lambda\} = \frac{F(s)}{s}$$ Where \(F(s)\) is the Laplace transform of the function \(f(t)\). Therefore, we have: $$\frac{F(s)}{s} = \mathcal{L}\{\int_2^t f(\lambda) d\lambda\} - \mathcal{L}\{\int_0^2 f(\lambda) d\lambda\}$$ #tag_title# Step 3: Simplify the equation #tag_content# To further simplify the equation, we can use the Laplace transform properties for the integrals on the right side of the equation: $$\frac{F(s)}{s} = \frac{e^{-2s}F(s)}{s} - \mathcal{L}\{\int_0^2 f(\lambda) d\lambda\}$$ #tag_title# Step 4: Solve for F(s) #tag_content# Now, isolate F(s) on the left side of the equation by multiplying both sides by s: $$F(s) = e^{-2s}F(s) - s\mathcal{L}\{\int_0^2 f(\lambda) d\lambda\}$$ Rearrange the equation: $$F(s)(1 - e^{-2s}) = s\mathcal{L}\{\int_0^2 f(\lambda) d\lambda\}$$ And finally, find F(s) by dividing both sides by \((1 - e^{-2s})\): $$F(s) = \frac{s\mathcal{L}\{\int_0^2 f(\lambda) d\lambda\}}{1 - e^{-2s}}$$ So, the Laplace transform of the function \(f(t)\) is given by: $$F(s) = \frac{s\mathcal{L}\{\int_0^2 f(\lambda) d\lambda\}}{1 - e^{-2s}}$$

Step-by-step solution

Write the integral from 0 to t in terms of the other two integrals

According to the hint given in the problem, we can write the integral from 0 to t as: $$\int_0^t f(\lambda) d\lambda = \int_2^t f(\lambda) d\lambda - \int_0^2 f(\lambda) d\lambda$$

Key concepts

Integral Transforms

Integral transforms are powerful mathematical tools that convert functions according to an integral operator into a different domain, often to simplify complex operations. They are highly beneficial when working with problems in engineering and physics. The Laplace Transform, which is a type of integral transform, is particularly useful in solving differential equations by moving from the time domain to the frequency domain. This transformation makes equations easier to handle, as it converts derivatives into algebraic terms.

Some benefits of using integral transforms include:

• They simplify complex differential equations into more manageable algebraic equations.
• Transform techniques aid in handling boundary value problems effectively.
• They are critical for analyzing systems functions, such as thermal, electrical, and mechanical systems.

Boundary Value Problems

Boundary value problems involve differential equations with specified values, known as boundaries, at the ends of the interval where the solution is defined. Instead of finding the value of the function everywhere, these problems require that the function meet specified conditions (the boundary conditions).

Boundary value problems are common in physical sciences. For instance, when studying waveforms on a string, the displacement at both ends needs to be zero. Laplace Transforms aid in solving boundary value problems because they convert the differential equation and its boundary conditions into an algebraic equation, making it far easier to include those boundary conditions in your calculations.

• Common applications are in heat conduction, beam bending, and fluid flow problems.
• Boundary value problems differ from initial value problems as they include conditions at the endpoints.

Differential Equations

Differential equations are equations that include an unknown function and its derivatives. They are crucial in understanding the behaviors of systems in fields such as physics, engineering, and finance.

The Laplace transform is a key tool in solving linear differential equations, especially because it changes them into algebraic equations. By making them easier to solve, you can focus on applying initial or boundary conditions without worrying about the inherent complexities of the differential equation itself.

• These equations help describe growth, decay, oscillation, and many other phenomena.
• The Laplace Transform particularly excels at handling linear differential equations with constant coefficients.

Definite Integrals

Definite integrals represent the signed area under a curve, from one point to another on a graph of a function. They are integral to the calculation of areas, volumes, central points, and many other mathematical contexts.

In the context of the given exercise, it's crucial to understand how definite integrals fit within the framework of the Laplace Transform. When you transform a differential equation that includes definite integrals, you use the Laplace Transform to transition from the time domain to the frequency domain.

• Definite integrals are essential in many physics and engineering calculations.
• They are different from indefinite integrals, which represent a family of functions.

Understanding the interplay between definite integrals and other mathematical concepts is vital, particularly when you're building complex models or solving intricate problems with boundary conditions and initial states.