solve the initial value problem 9y'' - 12y' + 4y = 0, y(0) = 2, y'(0) = 3.

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.

Learn how to solve the differential equation 9y'' - 12y' + 4y = 0 with initial conditions y(0) = 2 and y'(0) = 3. This step-by-step guide explores the characteristic equation and repeated roots solution. Suitable for advanced high school or undergraduate students.

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