Let the solution curve y=y(x) of the differential equation, [ x √x^2-y^2+e^ y x]x dy dx=x+[ x √x^2-y^2+e^ y x]y pass through the points (1,0) and (2α…

Given $x(\frac{x}{\sqrt{{x}^{2}-{y}^{2}}}+{e}^{\frac{y}{x}})\frac{dy}{dx}=y(\frac{x}{\sqrt{{x}^{2}-{y}^{2}}}+{e}^{\frac{y}{x}})+x$x(x2−y2​x​+exy​)dxdy​=y(x2−y2​x​+exy​)+x

Taking $x$x common & cancelling them we get,

$$\frac{dy}{dx}\times (\frac{1}{\sqrt{1-{(\frac{y}{x})}^{2}}}+{e}^{\frac{y}{x}})=\frac{y}{x}(\frac{1}{\sqrt{1-{(\frac{y}{x})}^{2}}}+{e}^{\frac{y}{x}})+1$$dxdy​×(1−(xy​)2​1​+exy​)=xy​(1−(xy​)2​1​+exy​)+1

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

$$(v+x\frac{dv}{dx})(\frac{1}{\sqrt{1-{v}^{2}}}+{e}^{v})=v(\frac{1}{\sqrt{1-{v}^{2}}}+{e}^{v})+1$$(v+xdxdv​)(1−v2​1​+ev)=v(1−v2​1​+ev)+1

$$v+x\frac{dV}{dx}=v+\frac{1}{(\frac{1}{\sqrt{1-{v}^{2}}}+{e}^{v})}$$v+xdxdV​=v+(1−v2​1​+ev)1​

$x\frac{dv}{dx}=\frac{1}{(\frac{1}{\sqrt{1-{v}^{2}}}+{e}^{v})}$xdxdv​=(1−v2​1​+ev)1​ $\Rightarrow (\frac{1}{\sqrt{1-{v}^{2}}}+{e}^{v})dv=\frac{dx}{x}$⇒(1−v2​1​+ev)dv=xdx​

Integrating both side we get,

$$\Rightarrow \int (\frac{1}{\sqrt{1-{v}^{2}}}+{e}^{v})dv=\int \frac{dx}{x}$$⇒∫(1−v2​1​+ev)dv=∫xdx​

$\Rightarrow {\mathrm{sin}}^{-1}v+{e}^{v}=\mathrm{ln}x+c$⇒sin−1v+ev=lnx+c $\Rightarrow {\mathrm{sin}}^{-1}(\frac{y}{x})+{e}^{\frac{y}{x}}=\mathrm{ln}x+c$⇒sin−1(xy​)+exy​=lnx+c

Now $y(1)=0$y(1)=0

$$\Rightarrow {\mathrm{sin}}^{-1}(\frac{0}{1})+{e}^{0}=\mathrm{ln}1+c$$⇒sin−1(10​)+e0=ln1+c

$$\Rightarrow c=1$$⇒c=1

$\Rightarrow {\mathrm{sin}}^{-1}(\frac{y}{x})+{e}^{\frac{y}{x}}=\mathrm{ln}x+1$⇒sin−1(xy​)+exy​=lnx+1 ........(i)

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

$$\Rightarrow {\mathrm{sin}}^{-1}(\frac{\alpha }{2\alpha })+{e}^{\frac{\alpha }{2\alpha }}=\mathrm{ln}2\alpha +1$$⇒sin−1(2αα​)+e2αα​=ln2α+1

$\Rightarrow \frac{\pi }{6}+{e}^{\frac{1}{2}}=\mathrm{ln}2\alpha +1$⇒6π​+e21​=ln2α+1 $\Rightarrow \alpha =\frac{1}{2}{e}^{\frac{\pi }{6}+\sqrt{e}-1}$⇒α=21​e6π​+e​−1