Let $$y = y(x)$$ be the solution of the differential equation $$\frac{dy}{dx} = 1 + xe^{y-x}$$, $$-\sqrt{2} \lt x \lt \sqrt{2}$$, $$y(0) = 0$$, then the minimum value of $$y(x)$$, $$x \in (-\sqrt{2}, \sqrt{2})$$ is equal to:

The given differential equation is

$$\frac{dy}{dx}=1+xe^{\,y-x},\qquad -\sqrt{2}\lt x\lt\sqrt{2}.$$

Put $$u=y-x\;(\Longrightarrow\;y=u+x).$$ Then

$$\frac{dy}{dx}=\frac{du}{dx}+1,$$

and substituting in the equation gives

$$\frac{du}{dx}+1=1+xe^{u}\;\Longrightarrow\;\frac{du}{dx}=xe^{u}.$$

This is separable:

$$e^{-u}\,du=x\,dx.$$

Integrating,

$$\int e^{-u}\,du=\int x\,dx\;\Longrightarrow\;-e^{-u}=\frac{x^{2}}{2}+C.$$

Write the constant in the form $$e^{-u}=K-\frac{x^{2}}{2}.$$ Using the initial condition $$y(0)=0\;(u(0)=0)$$ gives

$$1=K-\frac{0}{2}\;\Longrightarrow\;K=1.$$

Thus

$$e^{-u}=1-\frac{x^{2}}{2}\quad\Bigl(|x|\lt\sqrt{2}\Bigr),\qquad u=-\ln\!\Bigl(1-\frac{x^{2}}{2}\Bigr).$$

Recalling $$y=u+x$$ we obtain the explicit solution

$$y(x)=x-\ln\!\Bigl(1-\frac{x^{2}}{2}\Bigr),\qquad -\sqrt{2}\lt x\lt\sqrt{2}.$$

To locate the extreme value of $$y(x)$$ differentiate:

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

Set this equal to zero:

$$1+\frac{x}{1-\dfrac{x^{2}}{2}}=0 \;\Longrightarrow\; 1-\frac{x^{2}}{2}+x=0 \;\Longrightarrow\; x^{2}-2x-2=0.$$

The roots are $$x=1\pm\sqrt{3}.$$ Only $$x_{0}=1-\sqrt{3}\;( \approx -0.732 )$$ lies in $$(-\sqrt{2},\sqrt{2})$$, so the minimum occurs there.

Evaluate $$y$$ at this point:

$$x_{0}=1-\sqrt{3},\qquad x_{0}^{2}=4-2\sqrt{3},\qquad \frac{x_{0}^{2}}{2}=2-\sqrt{3},$$

$$1-\frac{x_{0}^{2}}{2}=1-(2-\sqrt{3})=\sqrt{3}-1.$$

Therefore

$$y_{\min}=x_{0}-\ln\!\Bigl(1-\frac{x_{0}^{2}}{2}\Bigr) =(1-\sqrt{3})-\ln(\sqrt{3}-1).$$

This value matches Option D.

Hence, the correct answer is Option D.