xdy/dx = y^2 - y
dy/(y^2 - y) = 1/x dx
dy/(y(y - 1)) = 1/x dx
∫[ ( 1/(y - 1) - 1/y)]dx = ∫(1/x)dx
ln(y - 1) - ln(y) = ln(x) + ln(C)
ln[ (y - 1)/y] = ln[xC]
(y - 1)/y = xC
y - 1 = Cxy
y(1 - Cx) = 1
y = 1/(1 + Cx)
Xy' + y = y^2
d(xy) / (xy)^2 = 1/x^2
-1/(xy) = -1/x + C
1/y = 1 -Cx
y = - 1/(Cx-1)
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Verified answer
xdy/dx = y^2 - y
dy/(y^2 - y) = 1/x dx
dy/(y(y - 1)) = 1/x dx
∫[ ( 1/(y - 1) - 1/y)]dx = ∫(1/x)dx
ln(y - 1) - ln(y) = ln(x) + ln(C)
ln[ (y - 1)/y] = ln[xC]
(y - 1)/y = xC
y - 1 = Cxy
y(1 - Cx) = 1
y = 1/(1 + Cx)
Xy' + y = y^2
d(xy) / (xy)^2 = 1/x^2
-1/(xy) = -1/x + C
1/y = 1 -Cx
y = - 1/(Cx-1)