Find the general/particular solution of the following homogeneous linear differential

Joanna Benson 2021-12-28 Answered
Find the general/particular solution of the following homogeneous linear differential equations with constant coefficients.
(D3+D221D45)y=0
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Expert Answer

Daniel Cormack
Answered 2021-12-29 Author has 34 answers
Given differential equation
(D3+D221D45)y=0
Auxillary equation of D.E is
r3+r221r45=0
r35r2+6r230r+9r45=0
r2(r5)+6r(r5)+9(r5)=0
(r5)(r2+6r+9)
(r5)(r2+2(r)(3)+(3)2)
(r5)(r+3)2=0
r5=0 (r+3)2=0
r=5 r+3=0,r+3=0
r=33
general solution y=c1e5x+(c2+c3x)e3x

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puhnut1m
Answered 2021-12-30 Author has 33 answers
Step 1
Given differential equation is
(D3+D221D45)y=0
Step 2
Its auxilary equation is
f(m)=m3+m221m45=0
m3+m221m45=0 (1)
put m=5
LHS=53+5221×545=125+2510545
=150150=0=RHS
(m5) is a factor of f(m)
m2(m5)+(m5)+9(m5)=0
(m5)(m2+6m+9)=0
(m5)(m+3)(m+3)=0
Therefore m=5,3,3
C.F=c1e5x+c2e3x+c3xe3x
Solution is
y(x)=c1e5x+(c2+c3x)e3x

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Vasquez
Answered 2022-01-09 Author has 460 answers

(D3+D221D45)y=0r3+r221r45=0r35r2+6r230r+9r45=0r2(r5)+6r(r5)+9(r5)=0(r5)(r2+6r+9)(r5)(r2+2(r)(3)+(3)2)(r5)(r+3)2=0r5=0 (r+3)2=0r=5 r+3=0,r+3=0r=33y=c1e5x+(c2+c3x)e3x

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