Question

The roots of(x2 D2-xD +1)y=cos(logx)​

Answer 1

Answer:

We are given that the homogeneous linear differential equation,

Put x = e^z or z = logx

and x^2D^2= D°(D°-1) , xD = D°

Now, (D°(D°-1)-3D°+1)= sinx

The auxiliary equation is,

D°^2+2D°+1= 0

(D°+1)^2 = 0

D° = -1,-1

So, C.F. = (c°+c'z)e^-z

P.I. = { 1/(D°+1)^2}sinz

= {1/(D°^2+2D°+1)}sinz

= {1/(-1+2D°+1)}sinz

= {1/2D°}sinz

= (1/2)(-cosz)

=(-1/2)cosz

P.I. = (-1/2)cos(logx)

The required solution is, y = C.F. + P.I.

y = (c°+c'logx)x^-1 -(1/2)cos(logx)

where c° and c' are arbitrary constants.

Step-by-step explanation:

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