We are given the second-order linear homogeneous differential equation:

$9y'' + 6y' + y = 0$

with the initial conditions $y(1) = 1$ and $y'(1) = 3$.

Step 1: Solve the characteristic equation

To solve this, we first solve the associated characteristic equation by assuming $y = e^{rt}$. Substituting this into the differential equation yields:

$9r^2 e^{rt} + 6r e^{rt} + e^{rt} = 0$

Since $e^{rt}$ is never zero, we can divide through by $e^{rt}$:

$9r^2 + 6r + 1 = 0$

This is a quadratic equation, and we can solve it using the quadratic formula:

$r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

For the equation $9r^2 + 6r + 1 = 0$, we have $a = 9$, $b = 6$, and $c = 1$. Substituting these into the quadratic formula gives:

$r = \frac{-6 \pm \sqrt{6^2 - 4(9)(1)}}{2(9)} = \frac{-6 \pm \sqrt{36 - 36}}{18} = \frac{-6 \pm 0}{18} = \frac{-6}{18} = -\frac{1}{3}$

Thus, the characteristic equation has a repeated root $r = -\frac{1}{3}$.

Step 2: General solution

Since we have a repeated root, the general solution to the differential equation is:

$y(t) = (C_1 + C_2 t)e^{-\frac{1}{3}t}$

Step 3: Apply the initial conditions

We now use the given initial conditions $y(1) = 1$ and $y'(1) = 3$ to determine the constants $C_1$ and $C_2$.

Condition 1: $y(1) = 1$

Substitute $t = 1$ into the general solution:

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

This simplifies to:

$(C_1 + C_2)e^{-\frac{1}{3}} = 1$

or