Let $y=y(x)$ be the solution of the differential
equation $x d y=y d x=\sqrt{\left(x^{2}-y^{2}\right)} d x, x \geq 1$, with
$y(1)=0 .$ If the area bounded by the line $x=1, x=e^{\pi}, y=0$ and $y=y(x)$ is
$\alpha \mathrm{e}^{2 \pi}+\beta$, then the value of $10(\alpha+\beta)$ is equal to_________
$x d y-y d x=\sqrt{x^{2}-y^{2}} d x$
$\Rightarrow \frac{x d y-y d x}{x^{2}}=\frac{1}{x} \sqrt{1-\frac{y^{2}}{x^{2}}} d x$
$\Rightarrow \int \frac{d\left(\frac{y}{x}\right)}{\sqrt{1-\left(\frac{y}{x}\right)^{2}}}=\int \frac{d x}{x}$
$\Rightarrow \sin ^{-1}\left(\frac{y}{x}\right)=\ell n|x|+c$
at $x=1, y=0 \Rightarrow c=0$
$y=x \sin (\ln x)$
$A=\int_{1}^{e^{x}} x \sin (\ln x) d x$
$a=e^{t}, d x=e^{t} d t \Rightarrow \int_{0}^{x} e^{2 t} \sin (t) d t=A$
$\Rightarrow$
$a=\frac{1}{5}, \beta=\frac{1}{5}$ so $10(\alpha+\beta)=4$
Click here to get exam-ready with eSaral
For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.