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

Gregory Emery

Gregory Emery

Answered question

2021-12-27

Identify the characteristic equation, solve for the characteristic roots, and solve the 2nd order differential equations.
(4D44D323D2+12D+36)y=0

Answer & Explanation

Jack Maxson

Jack Maxson

Beginner2021-12-28Added 25 answers

To find: The characteristic equation and the solution.
Solution:
Let y=Aemx(A0) be the solution of the given differential equation is
(4D44D323D2+12D+36)y=0
Then the characteristic equation is
4m44m323m2+12m+36=0
(x24x+4)(4x2+12x+9)=0
(x2)2(2x+3)2=0
x=2,2,x=32,32
Then the solution is y(x)=e2x(4+c2x)+e32x(c3+c3x)
Where c1,c2,c3,c4 are arbitrary constant.
Natalie Yamamoto

Natalie Yamamoto

Beginner2021-12-29Added 22 answers

A.E. is 4m44m323m2+12m+36=0
If m=2,643292+24+36=0
m=2 is a root of inspection. By synthetic division,
i.e.,4m3+4m215m18=0
If m=2
32+163018=0
Again m=2 is a root.
By synthetic division.
4m2+12m+9=0
(2m+3)2=0
m=32,32
The roots of the A.E.
are 2,2,32,32
Thus, y=(C1+C2x)e2x+(C3+C4x)e3x2.
Vasquez

Vasquez

Expert2022-01-09Added 669 answers

Solution:
(4D44D323D2+12D+36)y=0
the auxilary equation, 4m44m3+23m2+12m+36=0
Use long division we get, m=32 of order 2 and m=2 of order 2
The general solution,
y(x)=Ae32x+Bxe32x+Ce2x+Dxe2x
y(x)=(A+Bx)e32x+(C+Dx)e2x

Do you have a similar question?

Recalculate according to your conditions!

New Questions in Differential Equations

Ask your question.
Get an expert answer.

Let our experts help you. Answer in as fast as 15 minutes.

Didn't find what you were looking for?