Differential equation, How I can resolve this?, please

Question

Consider 1/2π< x <1/π for the differential equation

dy/dx = y^2 / x - y/x + 1/x^3

1- Find the particular solution of the form yp(x) = xα cot(1/x)

2-Find the general solution by doing y(x) = y_p + 1/z , where z is a function of x.

Answer 1

The particular solution is not of the form xα cot(1/x) .

A correct particular solution of the form y _p = (a / x) cot (1/x).

Substituting y _p = (a / x) cot (1/x). back to the equation

you will end up with a = 1.

Now with y _p = (1 / x) cot (1/x)

the substitution y = (1 / x) cot (1/x) +1/z

will produce the following Linear of First Order

z' + [ (2/x²)cot(1/x) - 1/x ] z = - 1/x.

Then God willing You will obtain the integrating factor μ (x) = csc²(1/x) - 1/x

The calculations here being from brutal to atrocious .....