Question

Let the solution curve $y=y\left(x\right)$ of the differential equation
$[\frac{x}{\sqrt{x^2-y^2}}+e^{\frac{y}{x}}]x\frac{dy}{dx}=x+[\frac{x}{\sqrt{x^2-y^2}}+e^{\frac{y}{x}}]y$ pass through the points $(1,0)$ and $(2\alpha ,\alpha ),\alpha >0$. Then $\alpha$ is equal to

Answer 1

$(\frac{1}{\sqrt{1-\frac{y^2}{x^2}}}+e^{\frac{y}{x}})\frac{dy}{dx}=1+(\frac{1}{\sqrt{1-\frac{y^2}{x^2}}}+e^{\frac{y}{x}})\frac{y}{x}$

Putting $y=tx$

$(\frac{1}{\sqrt{1-t^2}}+e^t)(t+x\frac{dt}{dx})=1+(\frac{1}{\sqrt{1-t^2}}+e^t)t$

$\Rightarrow x(\frac{1}{\sqrt{1-t^2}}+e^t)\frac{dt}{dx}=1$

$\Rightarrow {\sin }^{-1}t+e^t=\ln x+C$

$\Rightarrow {\sin }^{-1}\left(\frac{y}{x}\right)+e^{y/x}=\ln x+C$

at $x=1,y=0$

So, $0+e^0=0+C\Rightarrow C=1$

at $(2\alpha ,\alpha )$

${\sin }^{-1}\left(\frac{y}{x}\right)+e^{y/x}=\ln x+1$

$\Rightarrow \frac{\pi }{6}+e^{\frac{1}{2}}-1=\ln \left(2\alpha \right)$

$\Rightarrow \alpha =\frac{1}{2}{e}^{(\frac{\pi }{6}+e^{\frac{1}{2}}-1)}$