Solve the differential equation:

Question

dy/dx = y^2 + y

Answer 1

dy/(y^2+y) = dx
but 1/(y^2+y) = 1/(y-1) - 1/y, so
(1/(y-1) - 1/y)dy = dx
ln(y-1) - ln y = x+k
ln ((y-1)/y) = x+k
(y-1)/y = e^(x+k) = e^k * e^x = ce^x
y-1 = cye^x
y(1-ce^x) = 1
y = 1/(1-ce^x)

Hmmm. Wolframalpha says

y = ce^x/(e^cx-1)

but they are the same, since

u/(u-1) = 1 + 1/(u-1) so there's just a different constant c. Still, better double-check my math.