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(A) Given the differential equation $\frac{dy}{dx} = 1 + xe^{y-x}$.Rearranging the terms: $\frac{dy}{dx} - 1 = xe^{y-x}$.Since $y-x = u$,then $\frac{d

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$(1-\sqrt{3})-\log_{e}(\sqrt{3}-1)$

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(A) Given the differential equation $\frac{dy}{dx} = 1 + xe^{y-x}$.Rearranging the terms: $\frac{dy}{dx} - 1 = xe^{y-x}$.Since $y-x = u$,then $\frac{du}{dx} = \frac{dy}{dx} - 1$. Substituting this,we get $\frac{du}{dx} = xe^u$.Separating variables: $e^{-u} du = x dx$.Integrating both sides: $\int e^{-u} du = \int x dx \Rightarrow -e^{-u} = \frac{x^2}{2} + C$.Substituting $u = y-x$: $-e^{-(y-x)} = \frac{x^2}{2} + C \Rightarrow -e^{x-y} = \frac{x^2}{2} + C$.Given $y(0)=0$,at $x=0, y=0$: $-e^0 = 0 + C \Rightarrow C = -1$.So,$-e^{x-y} = \frac{x^2}{2} - 1 \Rightarrow e^{x-y} = 1 - \frac{x^2}{2} = \frac{2-x^2}{2}$.Taking natural log: $x-y = \ln\left(\frac{2-x^2}{2}\right) \Rightarrow y = x - \ln\left(\frac{2-x^2}{2}\right)$.To find the minimum,set $\frac{dy}{dx} = 0$: $1 + xe^{y-x} = 0$. From the original equation,$\frac{dy}{dx} = 1 + x\left(\frac{2}{2-x^2}\right) = \frac{2-x^2+2x}{2-x^2} = 0$.Solving $-x^2+2x+2=0 \Rightarrow x^2-2x-2=0$. Roots are $x = \frac{2 \pm \sqrt{4+8}}{2} = 1 \pm \sqrt{3}$.Since $x \in(-\sqrt{2}, \sqrt{2})$,we take $x = 1-\sqrt{3}$.Substituting $x = 1-\sqrt{3}$ into $y(x)$: $y = (1-\sqrt{3}) - \ln\left(\frac{2-(1-\sqrt{3})^2}{2}\right) = (1-\sqrt{3}) - \ln\left(\frac{2-(1+3-2\sqrt{3})}{2}\right) = (1-\sqrt{3}) - \ln\left(\frac{2-4+2\sqrt{3}}{2}\right) = (1-\sqrt{3}) - \ln(\sqrt{3}-1)$.

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

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