1. Given differential equation

$\frac{dy}{dx}=1+x e^{y-x}, \qquad y(0)=0, \qquad -\sqrt2<x<\sqrt2.$

We need the minimum value of $y(x)$ on $\left(-\sqrt2,\sqrt2\right)$.

2. Simplify the differential equation

Let

$u=y-x.$

Then

$\frac{du}{dx}=\frac{dy}{dx}-1.$

From the given equation,

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

So,

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

This is separable:

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

Integrating,

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

$-e^{-u}=\frac{x^2}{2}+C.$

Hence,

$e^{-u}=C_1-\frac{x^2}{2}.$

Using $y(0)=0$, we get $u(0)=y(0)-0=0$. Therefore,

$e^{-u(0)}=1=C_1.$

So,

$e^{-u}=1-\frac{x^2}{2}.$

Thus,

$u=-\ln\left(1-\frac{x^2}{2}\right).$

Since $u=y-x$,

$y=x-\ln\left(1-\frac{x^2}{2}\right).$

3. Find critical points of $y(x)$

Differentiate:

$y'(x)=1+\frac{x}{1-x^2/2}.$

Combining,

$y'(x)=\frac{1-x^2/2+x}{1-x^2/2}.$

Since $1-\frac{x^2}{2}>0$ on $(-\sqrt2,\sqrt2)$, critical points come from

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

Multiply by $2$:

$2+2x-x^2=0$

$x^2-2x-2=0.$

So,

$x=1\pm\sqrt3.$

Now,

• $1+\sqrt3>\sqrt2$, so it is outside the interval.
• $1-\sqrt3\in(-\sqrt2,\sqrt2)$, so this is the only critical point.

Thus the minimum occurs at

$x=1-\sqrt3.$

4. Evaluate $y$ at the critical point

We compute

$y(1-\sqrt3)=(1-\sqrt3)-\ln\left(1-\frac{(1-\sqrt3)^2}{2}\right).$

First,

$(1-\sqrt3)^2=1+3-2\sqrt3=4-2\sqrt3.$

Hence,

$\frac{(1-\sqrt3)^2}{2}=2-\sqrt3.$

Therefore,

$1-\frac{(1-\sqrt3)^2}{2}=1-(2-\sqrt3)=\sqrt3-1.$

So,

$y_{\min}=(1-\sqrt3)-\ln(\sqrt3-1).$

5. Match with the options

This is exactly

$\boxed{(1-\sqrt3)-\log_e(\sqrt3-1)}.$

So the correct option is D.

6. Comparison with stored answer

Stored correct answer: D

Our derived answer: D

They agree.