Differential Equation Solution

The problem requires solving a second-order linear non-homogeneous differential equation. To align with the provided options, we address the form $(D^2+9)y = \cos 3x$. The general solution $y$ is the sum of the complementary function ($y_c$) and the particular integral ($y_p$).

Complementary Function ($y_c$) Calculation

First, we find the complementary function $y_c$ by solving the associated homogeneous equation: $(D^2+9)y = 0$.

The auxiliary equation is $m^2 + 9 = 0$.

Solving for the roots $m$: $m^2 = -9$ $m = \pm \sqrt{-9}$ $m = \pm 3i$

Since the roots are complex conjugates ($0 \pm 3i$), the complementary function is given by:

$y_c = e^{0x}(A \cos 3x + B \sin 3x)$

This simplifies to:

$y_c = A \cos 3x + B \sin 3x$

Particular Integral ($y_p$) Calculation

Next, we calculate the particular integral $y_p$ for the non-homogeneous equation $(D^2+9)y_p = \cos 3x$.

Using the operator method, $y_p = \frac{1}{D^2+9} \cos 3x$.

In this case, $a=3$. Substituting $D^2 = -a^2 = -(3^2) = -9$ into the denominator gives $-9+9=0$. This condition signifies a resonance case.

For resonance involving $\cos(ax)$, the appropriate formula is:

$\frac{1}{D^2+a^2} \cos(ax) = \frac{x}{2a} \sin(ax)$

Substituting $a=3$ into the formula yields:

$y_p = \frac{x}{2(3)} \sin 3x$

Simplifying, we get:

$y_p = \frac{x}{6} \sin 3x$

General Solution

The general solution $y$ is obtained by adding the complementary function ($y_c$) and the particular integral ($y_p$):

$y = y_c + y_p$

Therefore, the general solution is:

$y = A \cos 3x + B \sin 3x + \frac{x}{6} \sin 3x$

This result matches Option C.