Step $1$

Recall the method for solving an $n$ th$-$order homogenous differential equation by finding roots of an auxiliary equation.

$a_n{y}^{(n)}+a_{n-1}{y}^{(n-1)}+$cdots$+a_1{y}^{\text{'}}+a_{0}y=0$

Rewrite this as a polynomial in $m,$ replacing the $j$ th derivative with $m^j.$

$a_n{m}^n+a_{n-1}{m}^{n-1}+$cdots$+a_{1}m+a_0=0$

Then for every root $m_i$ of multiplicity $k$ of this polynomial, the general solution for the equation must contain the following linear combination.

$c_1{e}^{m_{i}x}+c_{2}x{e}^{m_{i}x}+$cdots$+c_k{x}^{k-1}{e}^{m_{i}x}$

We are given the following second$-$order homogeneous differential equation.

$y^{\text{'}\text{'}}+12y^{\text{'}}+36y=0$

Therefore, the auxiliary equation is a second$-$degree polynomial.

$m^2+12m+36=0$

Solving for $m,$ the root of the auxiliary equation is $m_1=$ with multiplicity $$