JEE Main 2021 — Mathematics Calculus

Given,

$$\frac{dy}{dx}=1+x{e}^{y-x}$$dxdy​=1+xey−x

$$\Rightarrow \frac{dy-dx}{{e}^{y-x}}=xdx$$⇒ey−xdy−dx​=xdx

$$\Rightarrow \int \frac{d(y-x)}{{e}^{y-x}}=\int xdx$$⇒∫ey−xd(y−x)​=∫xdx

$$\Rightarrow -{e}^{x-y}=\frac{{x}^{2}}{2}+c$$⇒−ex−y=2x2​+c

At $x=0,y=0\Rightarrow c=-1$x=0,y=0⇒c=−1

So, the particular solution is

$${e}^{x-y}=\frac{2-{x}^{2}}{2}$$ex−y=22−x2​

$$\Rightarrow y=x-\mathrm{ln}(\frac{2-{x}^{2}}{2})$$⇒y=x−ln(22−x2​)

$$\Rightarrow \frac{dy}{dx}=1+\frac{2x}{2-{x}^{2}}=\frac{2+2x-{x}^{2}}{2-{x}^{2}}$$⇒dxdy​=1+2−x22x​=2−x22+2x−x2​

$\Rightarrow \frac{dy}{dx}=\frac{{x}^{2}-2x-2}{{x}^{2}-2}$⇒dxdy​=x2−2x2−2x−2​\Rightarrow dydx=x2-2x-2x+2x-2

If $\frac{dy}{dx}=0\Rightarrow {x}^{2}-2x-2=0$dxdy​=0⇒x2−2x−2=0

$$\Rightarrow x=\frac{2\pm \sqrt{12}}{2}$$⇒x=22±12​​

$$\Rightarrow x=1\pm \sqrt{3}$$⇒x=1±3​

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

$$y(1-\sqrt{3})=(1-\sqrt{3})-\mathrm{ln}(\frac{2-(4-2\sqrt{3})}{2})$$y(1−3​)=(1−3​)−ln(22−(4−23​)​)

$$=(1-\sqrt{3})-\mathrm{ln}(\sqrt{3}-1)$$=(1−3​)−ln(3​−1)