Question

Let $y=y\left(x\right)$ be the solution of the differential equation $xdy–ydx=\sqrt{(x^2-y^2)}dx,x\ge 1,$ with $y\left(1\right)=0.$ If the area bounded by the line $x=1,x=e^{\pi },y=0$ and $y=y\left(x\right)$ is $\alpha e^{2\pi }+b$, then the value of $10(\alpha +\beta )$ is equal to

Answer 1

$xdy–ydx=\sqrt{(x^2-y^2)}dx\Rightarrow x\frac{dy}{dx}-y=\sqrt{(x^2-y^2)}\Rightarrow \frac{dy}{dx}-\frac{y}{x}=\sqrt{1-{\left(\frac{y^2}{x^2}\right)}^2}$
Now, $y=vx\Rightarrow \frac{dy}{dx}=v+x\frac{dv}{dx}$
$\Rightarrow v+x\frac{dv}{dx}-v=\sqrt{1-v^2}\Rightarrow \frac{dv}{\sqrt{1-v^2}}=\frac{dx}{x}\Rightarrow {\sin }^{-1}v=\ln |x|+C\Rightarrow {\sin }^{-1}\frac{y}{x}=\ln |x|+C$
We know $y\left(1\right)=0$
$\Rightarrow 0=C$
As $x\ge 1$, so
$y=x\sin \left(\ln x\right)$
Now, the required area
$A=\underset{1}{\overset{e^{\pi }}{\int }}x\sin \left(\ln x\right)dx$
Let $x=e^t\Rightarrow dx=e^{t}dt$
$\Rightarrow A=\underset{0}{\overset{\pi }{\int }}{e}^{2t}\sin tdt$
Let
$I=\int e^{2t}\sin tdt\Rightarrow I=\frac{e^{2t}\sin t}{2}-\frac{1}{2}\int e^{2t}\cos tdt\Rightarrow I=\frac{e^{2t}\sin t}{2}-\frac{1}{2}[\frac{e^{2t}\cos t}{2}+\frac{1}{2}\int e^{2t}\sin tdt]\Rightarrow I=\frac{e^{2t}\sin t}{2}-\frac{e^{2t}\cos t}{4}-\frac{I}{4}\Rightarrow I=\frac{e^{2t}}{5}(2\sin t-\cos t)+C$
Where $C$ is arbitrary constant
Therefore,
$A={\left[\frac{e^{2t}}{5}(2\sin t-\cos t)\right]}_0^{\pi }\Rightarrow A=\frac{e^{2\pi }+1}{5}\Rightarrow \alpha =\frac{1}{5},\beta =\frac{1}{5}\therefore 10(\alpha +\beta )=4$