Найти общее решение дифференциального уравнения y^2*dx=(2-x)*dy

y^2*dx=(2-x)*dy,
[y^(-2)]*dy = [(2-x)^(-1)]*dx,

1) [y^(-2)]*dy = - [y^(-1)] + C, C = Const,

2) (2-x) = t,
-dx = dt,

[(2-x)^(-1)]*dx =
= - [t^(-1)]dt =
= - ln(t) + B =
= - ln(2-x) + B, B = Const,

3) - [y^(-1)] + C = - ln(2-x) + B,
[y^(-1)] = ln(2-x) + P, P = Const,

y = 1/[ln(2-x) + P].

ОТВЕТ: y = 1/[ln(2-x) + P],
P = Const,