# Identify the characteristic equation, solve for the characteristic roots, and

Identify the characteristic equation, solve for the characteristic roots, and solve the 2nd order differential equations.
$\left(18{D}^{3}-33{D}^{2}+20D-4\right)y=0$
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Robert Pina
Finding characteristic equation by putting r at the place of D then solve r to find characteristic root
$\left(18{D}^{3}-33{D}^{2}+20D-4\right)y=0$
Characteristic equation
$18{r}^{3}-33{r}^{2}+20r-4=0$
Now solving the above characteristic equation to find characteristic root.
$18{r}^{3}-33{r}^{2}+20r-4=0$
put $r=\frac{1}{2}$
$18×\frac{1}{8}-33×\frac{1}{4}+20×\frac{1}{2}-4$
$\frac{9}{4}-\frac{33}{4}+10-4$
$-6+6=0$
So $x=\frac{1}{2}$ is a characteristic root
Now $18{r}^{3}-33{r}^{2}+20r-4=0$
$18{r}^{2}\left(r-\frac{1}{2}\right)-24r\left(r-\frac{1}{2}\right)+8\left(r-\frac{1}{2}\right)=0$
$\left(r-\frac{1}{2}\right)\left[18{r}^{2}-24r+8\right]=0$
Now $18{r}^{2}-24r+8=0$
$9{r}^{2}-12r+4=0$
$9{r}^{2}-\left(6+6\right)r+4=0$
$9{r}^{2}-6r-6r+4=0$
$3r\left(3r-2\right)-2\left(3r-2\right)=0$
$\left(3r-2\right)\left(3r-2\right)=0$
$r=\frac{2}{3}$ (repeated)
Now characteristic roots are
(repeated)
So solution for defferential equation will be
$y\left(x\right)={c}_{1}{e}^{\frac{1}{2}x}+\left({c}_{2}+{c}_{3}x\right){e}^{\frac{2}{3}x}$
here ${c}_{1},{c}_{2},{c}_{3}$ are arbitrary constant

Donald Cheek
Therefore, $18{m}^{3}-33{m}^{2}+20m-4=0$
Where ${m}^{3}={D}^{3}y,{m}^{2}=Dy,m=Dy$, as y
$⇒\left(2m-1\right){\left(3m-2\right)}^{2}=0$

Hence the roots are $\frac{1}{2},\frac{2}{3},\frac{2}{3}$
So, the general solution of the given equation is
$y={c}_{1}{e}^{\frac{x}{2}}+\left({c}_{2}+{c}_{3}x\right){e}^{2\frac{x}{3}}$

Vasquez