To solve each first-order initial value problem using the Laplace transform, let's go through them step-by-step.

Problem (b): $y' + 4y = 0$, $y(0) = 2$

1. Apply the Laplace transform: Taking the Laplace transform of both sides of the equation $y' + 4y = 0$: $\mathcal{L}\{y'\} + 4\mathcal{L}\{y\} = \mathcal{L}\{0\}$

2. Use Laplace transform properties: The Laplace transform of $y'$ is $sY(s) - y(0)$, and $\mathcal{L}\{y\} = Y(s)$, where $Y(s)$ is the Laplace transform of $y(t)$. Substituting these into the equation: $(sY(s) - 2) + 4Y(s) = 0$

3. Solve for $Y(s)$: $sY(s) - 2 + 4Y(s) = 0$ $Y(s)(s + 4) = 2$ $Y(s) = \frac{2}{s + 4}$

4. Find $y(t)$ using the inverse Laplace transform: The inverse Laplace transform of $\frac{2}{s + 4}$ is $2e^{-4t}$.

   So, the solution is: $y(t) = 2e^{-4t}$

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Problem (c): $y' - y = 2t$, $y(0) = 0$

1. Apply the Laplace transform: Taking the Laplace transform of both sides of the equation $y' - y = 2t$: $\mathcal{L}\{y'\} - \mathcal{L}\{y\} = \mathcal{L}\{2t\}$

2. Use Laplace transform properties: Substituting $\mathcal{L}\{y'\} = sY(s) - y(0)$ and $\mathcal{L}\{y\} = Y(s)$: $(sY(s) - 0) - Y(s) = \frac{2}{s^2}$ $Y(s)(s - 1) = \frac{2}{s^2}$ $Y(s) = \frac{2}{s^2(s - 1)}$

3. Perform partial fraction decomposition: To find $Y(s)$ in a simpler form, decompose $\frac{2}{s^2(s - 1)}$: $\frac{2}{s^2(s - 1)} = \frac{A}{s} + \frac{B}{s^2} + \frac{C}{s - 1}$

   Solving for $A$, $B$, and $C$ (omitted here for brevity), we find: $Y(s) = \frac{2}{s^2} + \frac{2}{s - 1}$

4. Find $y(t)$ using the inverse Laplace transform: $y(t) = 2t + 2e^t$

So, the solution is: $y(t) = 2t + 2e^t$

Let me know if you want further details or if you have any questions!

Here are five related questions that might expand your understanding:

1. How does the Laplace transform handle different initial conditions?
2. What is the importance of partial fraction decomposition in finding the inverse Laplace?
3. How would the solutions change if the constants in each equation were different?
4. Can the Laplace transform solve higher-order differential equations?
5. What are the specific properties of the Laplace transform that make it suitable for differential equations?

Tip: When solving initial value problems using the Laplace transform, keep the transform properties (linearity, differentiation, shifting) in mind to simplify complex expressions.