To solve this initial value problem using the Laplace transform, let’s go through each part step-by-step:

Step 1: Take the Laplace Transform of Both Sides of the Differential Equation

Given: $y' + 4y = 32t, \quad y(0) = 2$

We’ll apply the Laplace transform to each term. Denote the Laplace transform of $y(t)$ by $Y(s)$.

1. Laplace transform of $y'$: $\mathcal{L}\{y'\} = sY(s) - y(0) = sY(s) - 2$

2. Laplace transform of $4y$: $\mathcal{L}\{4y\} = 4Y(s)$

3. Laplace transform of $32t$: $\mathcal{L}\{32t\} = \frac{32}{s^2}$

So, applying the Laplace transform to the entire differential equation gives: $sY(s) - 2 + 4Y(s) = \frac{32}{s^2}$

Step 2: Solve for $Y(s)$

Combine terms involving $Y(s)$: $(s + 4)Y(s) = \frac{32}{s^2} + 2$

Now, solve for $Y(s)$: $Y(s) = \frac{32}{s^2(s + 4)} + \frac{2}{s + 4}$

Step 3: Simplify and Take the Inverse Laplace Transform to Find $y(t)$

Now that we have $Y(s)$, let’s take the inverse Laplace transform to find $y(t)$.

Rewrite $Y(s)$ by breaking down the terms: $Y(s) = \frac{32}{s^2(s + 4)} + \frac{2}{s + 4}$

Each term in this expression can now be individually inverted to find $y(t)$.

Would you like further details on the partial fraction decomposition or the inverse Laplace transform process?

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Follow-up Questions

1. What is the significance of using the Laplace transform in solving differential equations?
2. How does the initial condition $y(0) = 2$ influence the solution in Laplace transforms?
3. What are the steps to perform partial fraction decomposition for complex rational expressions?
4. How can we verify the solution obtained from the Laplace transform by direct substitution?
5. Why is the Laplace transform particularly useful for handling non-homogeneous differential equations?

Tip:

The Laplace transform is powerful for converting differential equations into algebraic ones, making complex problems more manageable by working in the $s$-domain before converting back to the time domain.