Find the solution of the following Second Order Differential Equations.4y4y′+y=0

Calculation:
Convert ${\displaystyle 4y4y^{\prime }+y=0}$ into characteristics equation.
${\displaystyle 4D^2+4D+1=0}$
${\displaystyle {\left(2D\right)}^2+2\cdot 2D\cdot 1+1^2=0}$
${\displaystyle {(2D+1)}^2=0}$
${\displaystyle D=-\frac{1}{2}}$
Thus, the solution of differential equation ${\displaystyle 4y4y+y=0\text{is}y=A{e}^{-\frac{x}{2}}+Bx{e}^{-\frac{x}{2}}}$

The auxiliary equation of this equation is:
${\displaystyle m^2+4m+m=0}$
On comparing this equation with the general quadratic equation,
${\displaystyle a{x}^2+bx+c=0}$
Then, ${\displaystyle a=1,b=4,c=1}$
Now, use the quadratic form to find the value of m.
${\displaystyle x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}}$
${\displaystyle m=\frac{-4\pm \sqrt{4^2-4\left(1\right)\left(4\right)}}{2\left(4\right)}}$
${\displaystyle m=\frac{-4\pm \sqrt{0}}{8}}$
${\displaystyle m=-\frac{4}{8}-0,-\frac{4}{8}+0}$
${\displaystyle m=-\frac{1}{2},-\frac{1}{2}}$
Since values of m are equal, then the solution of the given differential equation is:
${\displaystyle y=(c_1+c_{2}x)e^{-\frac{1}{2}x}}$
Where, ${\displaystyle C_1,C_2}$ are arbitrary constant.

\(\begin{array}{} 4y+4y+y=0