Let y = y ( x ) be the solution of the differential equation d y d x = 1 + x e y - x ,   - 2 < x…

Given,

$\frac{dy}{dx}=1+x{e}^{y-x}$

$\Rightarrow \frac{dy-dx}{e^{y-x}}=xdx$

$\Rightarrow \int \frac{d\left(y-x\right)}{e^{y-x}}=\int xdx$

$\Rightarrow -e^{x-y}=\frac{x^2}{2}+c$

At $x=0,y=0\Rightarrow c=-1$

So, the particular solution is 

$e^{x-y}=\frac{2-x^2}{2}$

$\Rightarrow y=x-\ln \left(\frac{2-x^2}{2}\right)$

$\Rightarrow \frac{dy}{dx}=1+\frac{2x}{2-x^2}=\frac{2+2x-x^2}{2-x^2}$

$\Rightarrow \frac{dy}{dx}=\frac{x^2-2x-2}{x^2-2}$$$

$\Rightarrow \frac{dy}{dx}=\frac{x^2-2x-2}{\left(x+\sqrt{2}\right)\left(x-\sqrt{2}\right)}$

If $\frac{dy}{dx}=0\Rightarrow x^2-2x-2=0$

$\Rightarrow x=\frac{2\pm \sqrt{12}}{2}$

$\Rightarrow x=1\pm \sqrt{3}$

So minimum value occurs at $x=1-\sqrt{3}$

$y\left(1-\sqrt{3}\right)=\left(1-\sqrt{3}\right)-\ln \left(\frac{2-(4-2\sqrt{3})}{2}\right)$

$=\left(1-\sqrt{3}\right)-\ln \left(\sqrt{3}-1\right)$