Consider the DE which is linear with constant coefficients.

First we will work on solving the corresponding homogeneous equation. The auxiliary equation (using m as your variable) is $= 0$ which has root .

Because this is a repeated root, we don't have much choice but to use the exponential function corresponding to this root: to do reduction of order.
$y_2 = u e^{6x}$.
Then (using the prime notation for the derivatives)
$y_2^\prime =$
$y_2^{\prime\prime} =$
So, plugging $y_2$ into the left side of the differential equation, and reducing, we get
$y_2^{\prime\prime} - 12 y_2^\prime + 36 y_2 =$

So now our equation is $e^{6x} u^{\prime\prime} = x$. To solve for u we need only integrate $x e^{-6 x}$ twice, using a as our first constant of integration and b as the second we get
$u =$
Therefore $y_2 =$ , the general solution.

We knew from the beginning that $e^{6x}$ was a solution. We have worked out is that $x e^{6x}$ is another solution to the homogeneous equation, which is generally the case when we have multiple roots. Then $\frac{2+6x}{216}$ is the particular solution to the nonhomogeneous equation, and the general solution we derived is pieced together using superposition.

You can earn partial credit on this problem.