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The given initial value problem is:

\[
9y'' + 6y' + y = 0, \quad y(1) = 1, \quad y'(1) = 3
\]

This is a second-order linear homogeneous differential equation. Let's solve it step by step:

### Step 1: Solve the characteristic equation
Assume the solution is of the form \(y = e^{rt}\), where \(r\) is a constant. Substituting \(y = e^{rt}\) into the differential equation:

\[
9r^2 e^{rt} + 6r e^{rt} + e^{rt} = 0
\]

Factor out \(e^{rt}\):

\[
e^{rt} (9r^2 + 6r + 1) = 0
\]

Since \(e^{rt} \neq 0\), we solve the characteristic equation:

\[
9r^2 + 6r + 1 = 0
\]

Using the quadratic formula \(r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), with \(a = 9\), \(b = 6\), and \(c = 1\):

\[
r = \frac{-6 \pm \sqrt{6^2 - 4(9)(1)}}{2(9)} = \frac{-6 \pm \sqrt{36 - 36}}{18} = \frac{-6 \pm 0}{18}
\]

Thus, \(r = \frac{-6}{18} = -\frac{1}{3}\).

This is a repeated root, so the general solution is:

\[
y(t) = (C_1 + C_2 t) e^{-t/3}
\]

### Step 2: Apply the initial conditions
We are given two initial conditions: \(y(1) = 1\) and \(y'(1) = 3\).

1. Apply \(y(1) = 1\):

\[
y(1) = (C_1 + C_2 \cdot 1) e^{-1/3} = 1
\]

This simplifies to:

\[
(C_1 + C_2) e^{-1/3} = 1
\]

So,

\[
C_1 + C_2 = e^{1/3}
\]

2. Apply \(y'(1) = 3\):
First, differentiate \(y(t) = (C_1 + C_2 t) e^{-t/3}\) using the product rule:

\[
y'(t) = \left(C_2 e^{-t/3} + (C_1 + C_2 t) \left(-\frac{1}{3}\right) e^{-t/3}\right)
\]

Simplifying:

\[
y'(t) = e^{-t/3} \left(C_2 - \frac{1}{3} (C_1 + C_2 t)\right)
\]

At \(t = 1\):

\[
y'(1) = e^{-1/3} \left(C_2 - \frac{1}{3}(C_1 + C_2 \cdot 1)\right) = 3
\]

Simplify:

\[
e^{-1/3} \left(C_2 - \frac{1}{3}(C_1 + C_2)\right) = 3
\]

\[
C_2 - \frac{1}{3}(C_1 + C_2) = 3e^{1/3}
\]

Multiply through by 3:

\[
3C_2 - (C_1 + C_2) = 9e^{1/3}
\]

\[
2C_2 - C_1 = 9e^{1/3}
\]

### Step 3: Solve for \(C_1\) and \(C_2\)
We have the system of equations:

1. \(C_1 + C_2 = e^{1/3}\)
2. \(-C_1 + 2C_2 = 9e^{1/3}\)

Add these two equations:

\[
(C_1 + C_2) + (-C_1 + 2C_2) = e^{1/3} + 9e^{1/3}
\]

\[
3C_2 = 10e^{1/3}
\]

\[
C_2 = \frac{10}{3} e^{1/3}
\]

Substitute \(C_2\) into \(C_1 + C_2 = e^{1/3}\):

\[
C_1 + \frac{10}{3} e^{1/3} = e^{1/3}
\]

\[
C_1 = e^{1/3} - \frac{10}{3} e^{1/3} = -\frac{7}{3} e^{1/3}
\]

### Final Solution:
Substituting the values of \(C_1\) and \(C_2\) into the general solution:

\[
y(t) = \left(-\frac{7}{3} e^{1/3} + \frac{10}{3} e^{1/3} t\right) e^{-t/3}
\]

Thus, the solution is:

\[
y(t) = e^{1/3} \left(-\frac{7}{3} + \frac{10}{3} t\right) e^{-t/3}
\]

Do you have any questions or would you like further details on any part?

Here are some related questions:
1. What happens when the characteristic equation has complex roots?
2. How does the method of undetermined coefficients work for non-homogeneous equations?
3. Can repeated roots occur in higher-order differential equations as well?
4. What is the significance of the Wronskian in determining the linear independence of solutions?
5. How would you apply variation of parameters to solve a non-homogeneous version of this equation?

**Tip:** Always check your solution by substituting it back into the original differential equation to ensure it satisfies both the equation and the initial conditions.

Solve the initial value problem: 9y'' + 6y' + y = 0, y(1) = 1, y'(1) = 3. Use t as the independent variable.

Learn how to solve the second-order homogeneous differential equation 9y'' + 6y' + y = 0 with the initial conditions y(1) = 1 and y'(1) = 3. This problem involves solving the characteristic equation and applying initial values for a complete solution.

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