Question

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

• A. $(2+\sqrt{3})+{\log }_{e}2$
• B. $(1+\sqrt{3})-{\log }_e(\sqrt{3}-1)$
• C. $(1-\sqrt{3})-{\log }_e(\sqrt{3}-1)$
• D. $(2-\sqrt{3})-{\log }_{e}2$

Answer 1

The correct option is C $(1-\sqrt{3})-{\log }_e(\sqrt{3}-1)$

$\frac{dy}{dx}=1+x{e}^{y-x}$
$\Rightarrow e^{-y}\frac{dy}{dx}=e^{-y}+x{e}^{-x}$
Let
$e^{-y}=t\Rightarrow -e^{-y}\frac{dy}{dx}=\frac{dt}{dx}\Rightarrow -\frac{dt}{dx}=t+x{e}^{-x}$
$\Rightarrow \frac{dt}{dx}+t=-x{e}^{-x}$
$\therefore$ Integrating function $e^{{\int }^{1dx}}=e^x$
$\therefore$ Solution is : $t\cdot e^x=\int -x{e}^{-x}.e^{x}dx$
$\Rightarrow t\cdot e^x=-\frac{x^2}{2}+c$
$\Rightarrow e^{x-y}=-\frac{x^2}{2}+c$
$\because y\left(0\right)=0\Rightarrow c=1$
$\therefore e^{x-y}=1-\frac{x^2}{2}\Rightarrow y\left(x\right)=x-\log (1-\frac{x^2}{2})$
Now $y^{\prime }\left(x\right)=1-\frac{1}{1-\frac{x^2}{2}}(0-x)$
$=\frac{2-x^2+2x}{2-x^2}$
$=\frac{(x-1{)}^2-3}{x^2-2}$
$\therefore y^{\prime }\left(x\right)=0\Rightarrow x=1\pm \sqrt{3}$

$\because x\in (-\sqrt{2},\sqrt{2})\therefore x=1-\sqrt{3}$ is point of local minima
$\therefore$ Minimum value of $y\left(x\right)=y(1-\sqrt{3})$
$=(1-\sqrt{3})-{\log }_e(1-\frac{(1-\sqrt{3}{)}^2}{2})$
$=(1-\sqrt{3})-{\log }_e(\sqrt{3}-1)$