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

veksetz

veksetz

Answered question

2021-12-31

Identify the characteristic equation, solve for the characteristic roots, and solve the 2nd order differential equations.
(18D333D2+20D4)y=0

Answer & Explanation

Robert Pina

Robert Pina

Beginner2022-01-01Added 42 answers

Finding characteristic equation by putting r at the place of D then solve r to find characteristic root
(18D333D2+20D4)y=0
Characteristic equation
18r333r2+20r4=0
Now solving the above characteristic equation to find characteristic root.
18r333r2+20r4=0
put r=12
18×1833×14+20×124
94334+104
6+6=0
So x=12 is a characteristic root
Now 18r333r2+20r4=0
18r2(r12)24r(r12)+8(r12)=0
(r12)[18r224r+8]=0
Now 18r224r+8=0
9r212r+4=0
9r2(6+6)r+4=0
9r26r6r+4=0
3r(3r2)2(3r2)=0
(3r2)(3r2)=0
r=23 (repeated)
Now characteristic roots are
r=12 and r=23 (repeated)
So solution for defferential equation will be
y(x)=c1e12x+(c2+c3x)e23x
here c1,c2,c3 are arbitrary constant
Donald Cheek

Donald Cheek

Beginner2022-01-02Added 41 answers

Therefore, 18m333m2+20m4=0
Where m3=D3y,m2=Dy,m=Dy, as y
(2m1)(3m2)2=0
2n1=0 on (3m2)2=0
Hence the roots are 12,23,23
So, the general solution of the given equation is
y=c1ex2+(c2+c3x)e2x3
Vasquez

Vasquez

Expert2022-01-09Added 669 answers

18m333m2+20m4=0m3=D3y,m2=Dy,m=Dy(2m1)(3m2)2=02n1=0 on (3m2)2=012,23,23y=c1ex/2+(c2+c3x)e2x/3

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