Find the general solution of the differential equation y'''-y''+y'-y=0

ankarskogC 2021-10-04 Answered
Find the general solution of the differential equation
yy+yy=0
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Expert Answer

broliY
Answered 2021-10-05 Author has 97 answers

We have the differential equation
yy+yy=0
We immediately see that it is second-order homogeneous linear DE with constant coefficients. The usual technique for this type of D.E is to first suppose that there is a solution of the form y=emx.
Differentiate this supposition with respect to x, then we have
y=memx
and,
y=m2emx
and y=m3emx
Substitute into the original differential equation then we obtain
m3emxm2emx+memxemx=0
emx(m3m2+m1)=0
Since emx can not be equal 0, then we have
m3m2+m1=0
Then, (m1)(m2+1)=0
Then the roots are
m1=1
m2,3=±i
m2=+i,, m3=i
which are one real, two conjurate complex and distinct roots.
Then using theorem and theorem, the general solution of the differential equation y+6y+13y=0 is
y=k1ex+c1e(i)x+c2e(i)x
y=k1ex+c1exi+c2eξ
where k1,c1 and c2 are an arbitrary constants.
Which simply is
y=k1ex+k2cos(x)+k3sin(x)
where
eξ=cos(x)+sin(x)
and eξ=cos(x)sin(x)
and k2=(c1+c2),k3=(c1c2) are a new arbitrary constants

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