Question

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 $\alpha e^{2 \pi}+\beta$ then the value of $10(\alpha+\beta)$ is equal to ______.

Answer 1

$x d y-y d x=\sqrt{x^{2}-y^{2}} d x$
$\Rightarrow \frac{x d y-y d x}{x^{2}}=\frac{1}{x} \sqrt{1-\frac{y^{2}}{x^{2}}} d x$
$\Rightarrow \int \frac{d\left(\frac{y}{x}\right)}{\sqrt{1-\left(\frac{y}{x}\right)^{2}}}=\int \frac{d x}{x}$
$\Rightarrow \sin ^{-1}\left(\frac{y}{x}\right)=\ln |x|+c$
at $x=1, y=0 \Rightarrow c=0$
$y=x \sin (\ell n x)$
$A=\int\limits_{1}^{e^{\pi}} x \sin (\ell n x) d x$
$x=e^{t}, d x=e^{t} d t $
$\Rightarrow \int\limits_{0}^{\pi} e^{2 t} \sin (t) d t=A$
$\alpha e^{2 \pi}+\beta=\left(\frac{e^{2 t}}{5}(2 \sin t-\cos t)\right)_{0}^{\pi}=\frac{1+e^{2 \pi}}{5}$
$\alpha=\frac{1}{5}, \beta=\frac{1}{5}$ so $10(\alpha+\beta)=4$