Mary Hammonds

2021-12-30

ORDINARY DIFFERENTIAL EQUATIONS

$2y+12y+18y=0$

yotaniwc

Beginner2021-12-31Added 34 answers

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.

${r}^{2}+3r+3r+9=0$

$r(r+3)+3(r+3)=0$

${(r+3)}^{2}=0$

$r=-3$

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

Beginner2022-01-01Added 37 answers

The differential equation is given as

$2y12{y}^{\prime}+18y=0$

Corresponding auxiliary equation is given as

$2{m}^{2}+12m+18=0$

$2{m}^{2}+6m+6m+18=0$

$2m(m+3)+6(m+3)=0$

$(2m+6)(m+3)=0$

$2(m+3)(m+3)=0$

Roots of auxiliary equations are

$m=-3,-3$

Thus the general solution of the ordinary differential equation is

$y=({C}_{1}+{C}_{2}x){e}^{-3x}$

Corresponding auxiliary equation is given as

Roots of auxiliary equations are

Thus the general solution of the ordinary differential equation is

Vasquez

Expert2022-01-09Added 669 answers

Given:

Explanation:

The equation can be written as

The auxiliary equation of the differential equation is

Here roots are real and same

Therefore, the solution of the differential equation is

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I Completed the square on the bottom but what do you do now?

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inverse laplace transform - with symbolic variables:

$F(s)=\frac{2{s}^{2}+(a-6b)s+{a}^{2}-4ab}{({s}^{2}-{a}^{2})(s-2b)}$

My steps:

$F(s)=\frac{2{s}^{2}+(a-6b)s+{a}^{2}-4ab}{(s+a)(s-a)(s-2b)}$

$=\frac{A}{s+a}+\frac{B}{s-a}+\frac{C}{s-2b}+K$

$K=0$

$A=F(s)\ast (s+a)$