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.