Mary Hammonds

2021-12-30

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

yotaniwc

Step 1
Consider the details provided
$2y12{y}^{\prime }+18y=0$
Further, simplify.
$y6{y}^{\prime }+9y=0$
Immediately, enter the characteristic equation
${r}^{2}+6r+9=0$
Step 2
Now, find the factor of the equation.
${r}^{2}+3r+3r+9=0$
$r\left(r+3\right)+3\left(r+3\right)=0$
${\left(r+3\right)}^{2}=0$
$r=-3$
Now, write the solution of the equation, as both the root are same.
$y\left(x\right)={c}_{1}{e}^{-3x}+{c}_{2}x{e}^{-3x}$

alkaholikd9

The differential equation is given as
$2y12{y}^{\prime }+18y=0$
Corresponding auxiliary equation is given as
$2{m}^{2}+12m+18=0$
$2{m}^{2}+6m+6m+18=0$
$2m\left(m+3\right)+6\left(m+3\right)=0$
$\left(2m+6\right)\left(m+3\right)=0$
$2\left(m+3\right)\left(m+3\right)=0$
Roots of auxiliary equations are
$m=-3,-3$
Thus the general solution of the ordinary differential equation is
$y=\left({C}_{1}+{C}_{2}x\right){e}^{-3x}$

Vasquez

Given: $2{y}^{″}+12{y}^{\prime }+18y=0$
Explanation:
$2{y}^{″}+12{y}^{\prime }+18y=0$
The equation can be written as

The auxiliary equation of the differential equation is
$\begin{array}{}2{m}^{2}+12m+18=0\\ ⇒2\left({m}^{2}+6m+9\right)=0\\ ⇒{m}^{2}+6m+9=0\\ ⇒\left(m+3{\right)}^{2}=0\\ ⇒m=-3,-3\end{array}$
Here roots are real and same
Therefore, the solution of the differential equation is
$y=\left({c}_{1}+{c}_{2}x\right){e}^{-3x}$

Do you have a similar question?