To solve this initial value problem using the Laplace transform, we’ll start by taking the Laplace transform of each term in the differential equation:

Given: $y'' - 8y' + 25y = 0, \quad y(0) = 0, \quad y'(0) = 5$

We apply the Laplace transform term by term. Let $\mathcal{L}\{y(t)\} = Y(s)$.

Step 1: Take the Laplace Transform of Each Term

1. Transform of $y''$: $\mathcal{L}\{y''\} = s^2 Y(s) - s y(0) - y'(0)$ Substituting $y(0) = 0$ and $y'(0) = 5$: $\mathcal{L}\{y''\} = s^2 Y(s) - 5$

2. Transform of $-8y'$: $\mathcal{L}\{-8y'\} = -8(s Y(s) - y(0))$ Since $y(0) = 0$: $\mathcal{L}\{-8y'\} = -8s Y(s)$

3. Transform of $25y$: $\mathcal{L}\{25y\} = 25Y(s)$

Step 2: Substitute and Form the Equation

Substituting each transformed term into the differential equation: $(s^2 Y(s) - 5) - 8s Y(s) + 25 Y(s) = 0$

Combine like terms: $(s^2 - 8s + 25) Y(s) = 5$

Thus, the equation we get by taking the Laplace transform of the differential equation is: $(s^2 - 8s + 25) Y(s) = 5$

Would you like to proceed with solving for $Y(s)$ or need further clarification on these steps?

Here are five related questions:

1. What is the next step to isolate $Y(s)$ in the equation?
2. How do we find the inverse Laplace transform of $Y(s)$ once it's isolated?
3. What role do initial conditions play in transforming the differential equation?
4. How does the characteristic equation help when solving with Laplace transforms?
5. What methods are used to simplify expressions in the Laplace domain?

Tip: When using Laplace transforms, always check if initial conditions can simplify the transformed expressions.