Question Number 101841 by bemath last updated on 05/Jul/20

xy′ + y = y^2

Answered by bobhans last updated on 05/Jul/20

xy′ = y^2 −y ⇒ (dy/(y(y−1))) = (dx/x) ∫ (dy/(y−1)) − ∫ (dy/y) = (dx/x) ln(((y−1)/y)) = ln∣Cx∣ ⇒ 1−(1/y) = ∣Cx∣ (1/y) = 1−∣Cx∣ ⇔ y = (1/(1−∣Cx∣)) (Bob − )

Commented bybemath last updated on 05/Jul/20

macho......^o^