$\frac{dy}{dx} + 1 = 2y$, $y(1) = 1$

**My Work So Far**

Equation can be rearranged as:

$dy = (2y-1)dx \to$

$ \int dy = \int(2y-1)dx \to$

$ y = 2xy-x + C \to$

$y-2xy = C-x\to$

$y = \frac{C-x}{1-2x}$

For $y(1) = 1$:

$1 = \frac{C-1}{1-2(1)} \to$

$1 = \frac{C-1}{-1} \to$

$C = 1-1 \to C = 0$

Therefore, $y = \frac{-x}{1-2x}$

However, this is apparently wrong; the answer given is:

$y = \frac{1}{2}(1+e^{2x-2})$

Which is totally baffling to me; I've run through my solution over and over and can't find what I'm doing wrong, or what scenario would possibly lead to having to incorporate natural logs. Would anyone be able to point out where I'm going wrong in my solution?