eozoischgc

2021-12-29

Auxiliary Equation:
Solve the following:
$\left(3{D}^{3}-22{D}^{2}+7D\right)y=0$

Foreckije

A.E. (Auxiliary equation)
$3{m}^{3}-22{m}^{2}+7m=0$
$m\left(3{m}^{2}-22m+7\right)=0$
$m=0,3{m}^{2}-21m-m+7=0$
$3m\left(m-7\right)-1\left(m-7\right)=0$
$\left(3m-1\right)\left(m-7\right)=0$
$m=7,m=\frac{1}{3}$
Then $m=0,m=7,m=\frac{1}{3}$
$y={c}_{1}{e}^{0}+{c}_{2}{e}^{7x}+{c}_{3}{e}^{\frac{x}{3}}$
$y={c}_{1}+{c}_{2}{e}^{7x}+{c}_{3}{e}^{\frac{x}{3}}$

Thomas Nickerson

To Solve: The above auxiliary equation.
Explanation- Rewrite the given expression,
$\left(3{D}^{3}-22{D}^{2}+7D\right)y=0$
Solving as follows by factorising the above equation , we get,
$D\left(3{D}^{2}-22D+7\right)=0$
$D\left(\left(3{D}^{2}-21D-D+7\right)\right)=0$
$D\left(3D\left(D-7\right)-1\left(D-7\right)\right)=0$
$D\left(3D-1\right)\left(D-7\right)=0$
Further solving, we get,

So, the complementry equation can be written as follows,
$C.E.={C}_{1}{e}^{0\cdot x}+{C}_{2}{e}^{\frac{1}{3}\cdot x}+{C}_{3}{e}^{7\cdot x}$
$={C}_{1}+{C}_{2}{e}^{\frac{1}{3}x}+{C}_{3}{e}^{7x}$
Answer: Hence, the solution of the auxiliary equation

karton