The initial value problem is given as:

$9y'' - 12y' + 4y = 0, \quad y(0) = 2, \quad y'(0) = 3.$

Step 1: Finding the Characteristic Equation

The differential equation is a linear homogeneous equation with constant coefficients. Let's start by finding the characteristic equation associated with it:

$9r^2 - 12r + 4 = 0.$

This can be simplified:

$(3r - 2)^2 = 0.$

Thus, the characteristic equation has a repeated root:

$r = \frac{2}{3}.$

Step 2: General Solution of the Differential Equation

Since the characteristic equation has a repeated root, the general solution for $y$ will be of the form:

$y(t) = (C_1 + C_2 t) e^{\frac{2}{3} t}.$

Step 3: Finding $C_1$ and $C_2$ Using Initial Conditions

We are given the initial conditions:

$y(0) = 2, \quad y'(0) = 3.$

1. Using $y(0) = 2$:

$y(0) = (C_1 + C_2 \cdot 0) e^{\frac{2}{3} \cdot 0} = C_1 \cdot 1 = C_1 = 2.$

2. Finding $y'(t)$:

$y'(t) = \left[ C_2 e^{\frac{2}{3} t} + \left( C_1 + C_2 t \right) \cdot \frac{2}{3} e^{\frac{2}{3} t} \right].$

Simplifying $y'(t)$:

$y'(t) = \left[ C_2 + \frac{2}{3}(C_1 + C_2 t) \right] e^{\frac{2}{3} t}.$

Substitute $t = 0$:

$y'(0) = \left[ C_2 + \frac{2}{3}(C_1) \right] e^{\frac{2}{3} \cdot 0} = C_2 + \frac{2}{3}(2) = C_2 + \frac{4}{3}.$

Using $y'(0) = 3$:

$C_2 + \frac{4}{3} = 3.$

Solving for $C_2$:

$C_2 = 3 - \frac{4}{3} = \frac{9}{3} - \frac{4}{3} = \frac{5}{3}.$

Step 4: Final Solution

The solution to the initial value problem is:

$y(t) = \left( 2 + \frac{5}{3} t \right) e^{\frac{2}{3} t}.$

Let me know if you need more details or have questions! Here are some related questions you might want to explore:

1. What is the behavior of the solution as $t$ increases?
2. How would the solution change if the roots of the characteristic equation were distinct?
3. What happens to the general solution if we have complex roots?
4. Can we use a different method, like the Laplace transform, to solve this problem?
5. How does the choice of initial conditions affect the constants in the solution?

Tip: For second-order homogeneous linear differential equations, the nature of the roots (real and distinct, real and repeated, or complex) determines the form of the solution.