Differential Equation Problem: xy' + y = y^2, y(1)= -1

[thatguy47]

Find the solution of the differential equation that satisfies the given initial condition.
xy' + y = y^2, y(1)= -1

The answer is y = 1/(1-2x) but I get y^2 - y = 2 - x when I do it.

Here's my work:
x(dy/dx) = y^2 - y
(integral)(1/x)dx = (integral)(1/(y^2 - y))dy
lnx = ln(y^2 - y) + c
plug in y(1) = -1 to get ->> ln1 = ln(1+1) + c
0 = ln2 + c
c = -ln2
lnx = ln(y^2 - y) - ln2
ln(y^2 - y) = ln2 - lnx
e^(ln(y^2 - y)) = e^(ln2 - lnx)
y^2 - y = 2 - x

Does anyone know what I'm doing wrong?

 

[galactus]

\(\displaystyle \int \frac{1}{y^{2}-y}dy=ln(1-\frac{1}{y})+C\)