<s>[INST] 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?
 [/INST] The first thing I would try is $v = y - 1$
$$
v' = \left(1+\frac{v}{2x}\right)^2
$$
Let $v = wx$
$$
w + xw' = \left(1+\frac{w}{2}\right)^2
$$
which is basically
$$
xw' = g(w)
$$
now see if you can solve this ODE.
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