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The general solution of the differential equation $\frac{\mathrm{d} y}{\mathrm{~d} x}=1-x+y-x y$ is (where $C$ is a constant of integration)
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$\log (1+y)=x-\frac{x^2}{2}+C$
$\begin{aligned} & \frac{\mathrm{d} y}{\mathrm{~d} x}=1-x+y-x y=(1-x)(1+y) \\ & \Rightarrow \int \frac{\mathrm{d} y}{1+y}=\int(1-x) \mathrm{d} x \\ & \Rightarrow \log (1+y)=x-\frac{x^2}{2}+C\end{aligned}$
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