Certainly! Let's go through the exercises one by one. The exercises cover a variety of problems involving Laplace transforms, inverse Laplace transforms, integrals, and differential equations.

Exercise 1: Laplace Transforms

Evaluate the Laplace transforms of the given functions.

1. $\frac{1 - e^{-2x}}{2x}$

   This function can be decomposed using the linearity property of Laplace transforms. Using the known Laplace transform pairs: $\mathcal{L}\{1\} = \frac{1}{s}, \quad \mathcal{L}\{e^{-2x}\} = \frac{1}{s+2}$ We get: $\mathcal{L}\left\{\frac{1 - e^{-2x}}{2x}\right\} = \frac{1}{2} \left(\mathcal{L}\left\{\frac{1}{x}\right\} - \mathcal{L}\left\{\frac{e^{-2x}}{x}\right\}\right)$

2. $x \int_0^\pi e^{-x t} \cos t \, dt$

   Using the known Laplace transform pair for $\cos t$: $\mathcal{L}\{\cos t\} = \frac{s}{s^2 + 1}$ The integral can be evaluated and then the Laplace transform can be applied.

3. $e^{-x} \int_0^x (x - t)^2 \cos^2 t \, dt$

   This can be tackled by first solving the integral and then applying the Laplace transform.

Exercise 2: Inverse Laplace Transforms

Evaluate the inverse Laplace transforms of the given functions.

1. $\cot^{-1}(s + 1)$

   Using the inverse Laplace transform tables and properties.

2. $\frac{e^{-s}}{(s^2 + 4)^2}$

   This involves partial fraction decomposition and then using inverse Laplace transform pairs.

3. $\frac{s - 6}{s^3 - 2s^2 + 6s}$

   This can be solved by breaking it into simpler fractions and using the inverse Laplace transforms.

Exercise 3: Evaluate the Integrals

Evaluate the given integrals using Laplace transforms.

1. $\int_0^\infty \frac{e^{-ax} - e^{-bx}}{x} \, dx$

2. $\int_0^\infty \frac{e^{-ax} \sin bx}{x} \, dx$

Exercise 4: Solve the Differential Equations

Solve the differential equations using Laplace transforms.

1. $y'' + 3y' - y = x$ with initial conditions $y(0) = 1, y'(0) = -1$

   Take the Laplace transform of both sides, solve for $Y(s)$, then take the inverse Laplace transform.

2. $y'' + 2y' + 5y = 3e^{-x} \sin x$ with initial conditions $y(0) = 0, y'(0) = 3$

   Similarly, take the Laplace transform of both sides, solve for $Y(s)$, then take the inverse Laplace transform.

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Potential Questions

1. Can you show the detailed steps for the Laplace transform of $\frac{1 - e^{-2x}}{2x}$?
2. How do you solve the integral $\int_0^\infty \frac{e^{-ax} - e^{-bx}}{x} \, dx$ using Laplace transforms?
3. What is the inverse Laplace transform of $\frac{e^{-s}}{(s^2 + 4)^2}$?
4. Can you solve the differential equation $y'' + 3y' - y = x$ with given initial conditions?
5. How do you find the inverse Laplace transform of $\frac{s - 6}{s^3 - 2s^2 + 6s}$?

Tip

When working with Laplace transforms, always check if the function can be decomposed into simpler parts that match known Laplace transform pairs. This simplifies the process significantly.