Solve $y' = (1+\frac{y-1}{2x})^2$ Solve $y' = (1+\frac{y-1}{2x})^2$
My first thought was to expand to see if I can get a linear form:
$$y' = 1 + \frac{y-1}{x} + (\frac{y-1}{2x})^2 = 1 + \frac{y-1}{x} + \frac{y^2-2y+1}{4x^2}$$
$$y'  = 1 + \frac{y}{x} -\frac{1}{x} + \frac{y^2}{4x^2} +  \frac{-2y}{4x^2} +  \frac{1}{4x^2} \rightarrow y'+y(\frac{2}{4x^2}-\frac{1}{x}) = \frac{(2x-1)^2+y^2}{4x^2}$$
However, this is not even a bernoulli equation. How should I approach this problem?

$$y' = \left(1+\frac{y-1}{2x}\right)^2$$
Substitute :
$$z=1+\frac{y-1}{2x}$$
$$y'=2(z-1+xz')$$
Then  the DE is separable.
$$2xz'=(z-1)^2+1$$
