To solve this initial value problem using the Laplace transform, let's go through the steps.

The problem is: $y'' + 9y = f(t)$ where $f(t) = 3t$ for $0 \leq t \leq 6$ and $f(t) = 18$ for $t > 6$. We are also given the initial conditions $y(0) = 0$ and $y'(0) = 0$.

Step 1: Define the Laplace Transform of $y$

Let $Y(s)$ denote the Laplace transform of $y(t)$, so $\mathcal{L}\{y(t)\} = Y(s)$.

Step 2: Laplace Transform of the Differential Equation

Taking the Laplace transform of both sides of the equation $y'' + 9y = f(t)$, we use the property: $\mathcal{L}\{y''\} = s^2 Y(s) - s y(0) - y'(0)$ Given $y(0) = 0$ and $y'(0) = 0$, this simplifies to: $\mathcal{L}\{y''\} = s^2 Y(s)$ The Laplace transform of $9y$ is $9 Y(s)$.

Thus, the Laplace-transformed equation becomes: $s^2 Y(s) + 9 Y(s) = \mathcal{L}\{f(t)\}$

Step 3: Laplace Transform of $f(t)$

The function $f(t)$ is defined piecewise, so we express it using the Heaviside step function: $f(t) = 3t \, u(t) + (18 - 3t) u(t - 6)$ where $u(t)$ is the unit step function.

Now, we find the Laplace transform of $f(t)$: $\mathcal{L}\{f(t)\} = \mathcal{L}\{3t\} + \mathcal{L}\{(18 - 3t) u(t - 6)\}$

1. Transform of $3t$: $\mathcal{L}\{3t\} = 3 \cdot \frac{1}{s^2} = \frac{3}{s^2}$

2. Transform of $(18 - 3t) u(t - 6)$: Using the shifting property $\mathcal{L}\{g(t - a) u(t - a)\} = e^{-as} \mathcal{L}\{g(t)\}$, where $g(t) = 18 - 3t$: $\mathcal{L}\{(18 - 3t) u(t - 6)\} = e^{-6s} \mathcal{L}\{18 - 3t\}$ Now, $\mathcal{L}\{18 - 3t\} = \frac{18}{s} - \frac{3}{s^2}$ So, $\mathcal{L}\{(18 - 3t) u(t - 6)\} = e^{-6s} \left( \frac{18}{s} - \frac{3}{s^2} \right)$

Combining these, we have: $\mathcal{L}\{f(t)\} = \frac{3}{s^2} + e^{-6s} \left( \frac{18}{s} - \frac{3}{s^2} \right)$

Step 4: Substitute and Solve for $Y(s)$

The Laplace-transformed equation is: $s^2 Y(s) + 9 Y(s) = \frac{3}{s^2} + e^{-6s} \left( \frac{18}{s} - \frac{3}{s^2} \right)$ Factor out $Y(s)$ on the left side: $Y(s) (s^2 + 9) = \frac{3}{s^2} + e^{-6s} \left( \frac{18}{s} - \frac{3}{s^2} \right)$ Solving for $Y(s)$, we get: $Y(s) = \frac{\frac{3}{s^2} + e^{-6s} \left( \frac{18}{s} - \frac{3}{s^2} \right)}{s^2 + 9}$

This expression for $Y(s)$ is the solution in the Laplace domain.

Would you like to proceed with the inverse Laplace transform to find $y(t)$ explicitly in the time domain, or do you have questions about this setup?

Here are some related questions to consider:

1. How does the Laplace transform handle piecewise functions like $f(t)$?
2. What role do initial conditions play in solving differential equations with Laplace transforms?
3. How would the solution change if $y(0)$ or $y'(0)$ had non-zero values?
4. What is the significance of the term $e^{-6s}$ in the Laplace-transformed equation?
5. How does using the Heaviside function simplify working with piecewise-defined functions?

Tip: The Heaviside step function is an effective way to handle piecewise functions when applying the Laplace transform, as it allows you to break up the function into manageable parts.