ORDINARY DIFFERENTIAL EQUATIONS 2y+12y+18y=0

Step 1
Consider the details provided
${\displaystyle 2y12y^{\prime }+18y=0}$
Further, simplify.
${\displaystyle y6y^{\prime }+9y=0}$
Immediately, enter the characteristic equation
${\displaystyle r^2+6r+9=0}$
Step 2
Now, find the factor of the equation.
${\displaystyle r^2+3r+3r+9=0}$
${\displaystyle r(r+3)+3(r+3)=0}$
${\displaystyle {(r+3)}^2=0}$
${\displaystyle r=-3}$
Now, write the solution of the equation, as both the root are same.
${\displaystyle y\left(x\right)=c_1{e}^{-3x}+c_{2}x{e}^{-3x}}$

Given: $2y^{″}+12y^{\prime }+18y=0$
Explanation:
$2y^{″}+12y^{\prime }+18y=0$
The equation can be written as
$[2D^2+12D+18]y=0whereD=\frac{d}{dx}$
The auxiliary equation of the differential equation is
$\begin{array}{}2m^2+12m+18=0\\ \Rightarrow 2(m^2+6m+9)=0\\ \Rightarrow m^2+6m+9=0\\ \Rightarrow (m+3{)}^2=0\\ \Rightarrow m=-3,-3\end{array}$
Here roots are real and same
Therefore, the solution of the differential equation is
$y=(c_1+c_{2}x)e^{-3x}$