For the differential equation
(xlogx)dy=(logx−y)dx:
(A) Degree of the given differential equation is 1.
(B) It is a homogeneous differential equation.
(C) Solution is
2ylogx+A=(logx)2, where
A is an arbitrary constant.
(D) Solution is \(2y \log x + A = \llog(\ln x)\), where
A is an arbitrary constant.