Let y=y(x) be the solution of the differential


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 $x=1, y=0 \Rightarrow c=0$

$y=x \sin (\ln x)$

$A=\int_{1}^{e^{x}} x \sin (\ln x) d x$

$a=e^{t}, d x=e^{t} d t \Rightarrow \int_{0}^{x} e^{2 t} \sin (t) d t=A$


$a=\frac{1}{5}, \beta=\frac{1}{5}$ so $10(\alpha+\beta)=4$

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