Question

Solve `xy'+y=xy^2` ?

Answer 1

`y = 1/(x(C_1-log_e x))`

Explanation:

Making the substitution

`z=x y` we have

`z'=y+xy'` so

`xy'+y=xy^2->z'=x(z/x)^2`

this is a separable differential equation so

`(dz)/z^2= (dx)/x` and

`-1/z=log_ex+C_1` then

`y = 1/(x(C_1-log_e x))`