y=cotx den    n_C1 y_n-1(0)  -- n_C3 y_n--3(0) +........... =?ans kyaaaaaaaaaaaaaaaa hei

if   Cos^-1(y/b) =n.log(x/n)  den x²y_n+2 +(2n+1)x.y_n+1 +2n²y_n =?

plz urgenteez  jaldi bolo

2)y/b=cos(nlnx/n)y'=-bsin(nlnx/n)(n/x)

xy'=-nbsin(nlnx/n)xy"+y'=-nbcos(nlnx/n)(n/x)x²y"+xy'+n²y=0

Applying leibnitz theorem and taking nth derivative on both sides,

x²y_n+2+nc₁(2x)y_n+1+nc₂(2)y_n+xy_n+1+nc₁y_n+n²y_n=0

x²y_n+2+(2n+1)xy_n+1+2n²y_n=0

plz tell me

I think the answer to the first qn is cos(   npi/2 )

as cos ( x ) differentiated n -times is cos ( x + npi/2 ) .

It s simply ( y sin x ) differentiated  n times and expanded by Leibneitz theorem with x = 0 , sin ( x ) = 0 , cosx = 1

Gud ans sir.I kept on differntiating cotx n times but couldnot succeed.I never got the idea of taking sinx to the other side.Thank you.