JEE Main · 2021 · 18 Mar 2021 · Shift 2 · Differential Equations

Writing the equation as $xdy=\left(y+\sqrt{x^2-y^2}\right)dx$ and putting $y=x\sin \theta$ gives $dy=x\cos \theta d\theta +\sin \theta dx$ and $\sqrt{x^2-y^2}=x\cos \theta$, so

Hence $\theta =\ln x+C$, and $y(1)=0$ forces $C=0$, giving $y=x\sin (\ln x)$. For $1\le x\le e^{\pi }$ the quantity $\ln x$ lies in $[0,\pi ]$, so $y\ge 0$ throughout and the bounded area is

after the substitution $u=\ln x$. Using $\int e^{2u}\sin udu=\frac{e^{2u}\left(2\sin u-\cos u\right)}{5}$,

Comparing with $\alpha e^{2\pi }+\beta$ gives $\alpha =\beta =\frac{1}{5}$, so $10(\alpha +\beta )=4$.