Let the solution curve y = y x of the differential equation, x x 2 - y 2 + e y x x d y d x = x + x x 2 - y 2…

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

Taking $x$ common & cancelling them we get,

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

Let $y=vx$$\Rightarrow \frac{dy}{dx}=v+x\frac{dv}{dx}$

$\left(v+x\frac{dv}{dx}\right)\left(\frac{1}{\sqrt{1-v^2}}+e^v\right)=v\left(\frac{1}{\sqrt{1-v^2}}+e^v\right)+1$

$v+x\frac{dV}{dx}=v+\frac{1}{\left(\frac{1}{\sqrt{1-v^2}}+e^v\right)}$

$x\frac{dv}{dx}=\frac{1}{\left(\frac{1}{\sqrt{1-v^2}}+e^v\right)}$ $\Rightarrow \left(\frac{1}{\sqrt{1-v^2}}+e^v\right)dv=\frac{dx}{x}$

Integrating both side we get,

$\Rightarrow \int \left(\frac{1}{\sqrt{1-v^2}}+e^v\right)dv=\int \frac{dx}{x}$

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

Now $y\left(1\right)=0$ 

$\Rightarrow {\sin }^{-1}\left(\frac{0}{1}\right)+e^0=\ln 1+c$

$\Rightarrow c=1$

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

Now $y\left(2\alpha \right)=\alpha$ putting in equation (i) we get,

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

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