Question

Let $y=y(x)$ be the solution of the differential equation $xdy-ydx=\sqrt{\left(x^2-y^2\right)}dx,x\ge 1$, 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

Answer: 4

$xdy-ydx=\sqrt{x^2-y^2}dx$
$\Rightarrow \frac{xdy-ydx}{x^2}=\frac{1}{x}\sqrt{1-\frac{y^2}{x^2}}dx$
$\Rightarrow \int \frac{d\left(\frac{y}{x}\right)}{\sqrt{1-{\left(\frac{y}{x}\right)}^2}}=\int \frac{dx}{x}$
$\Rightarrow {\sin }^{-1}\left(\frac{y}{x}\right)=\ln |x|+c$
at $x=1,y=0\Rightarrow c=0$
$y=x\sin (\ell nx)$
$A=\underset{1}{\overset{e^{\pi }}{\int }}x\sin (\ell nx)dx$
$x=e^t,dx=e^{t}dt$
$\Rightarrow \underset{0}{\overset{\pi }{\int }}{e}^{2t}\sin (t)dt=A$
$\alpha e^{2\pi }+\beta ={\left(\frac{e^{2t}}{5}(2\sin t-\cos t)\right)}_0^{\pi }=\frac{1+e^{2\pi }}{5}$
$\alpha =\frac{1}{5},\beta =\frac{1}{5}$ so $10(\alpha +\beta )=4$