Let y = y ( x ) be the solution of the differential equation...

Given,

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

$⇒ \dfrac{dy - dx}{{e}^{y - x}} = xdx$

$⇒ ∫ \dfrac{d\left(y - x\right)}{{e}^{y - x}} = ∫ xdx$

$⇒ - {e}^{x - y} = \dfrac{{x}^{2}}{2} + c$

At $x = 0, y = 0 ⇒ c = - 1$

So, the particular solution is

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

$⇒ y = x - \ln \left(\dfrac{2 - {x}^{2}}{2}\right)$

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

$⇒ \dfrac{dy}{dx} = \dfrac{{x}^{2} - 2x - 2}{{x}^{2} - 2}$

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

If $\dfrac{dy}{dx} = 0 ⇒ {x}^{2} - 2x - 2 = 0$

$⇒ x = \dfrac{2 \pm \sqrt{12}}{2}$

$⇒ x = 1 \pm \sqrt{3}$

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

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

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