Question

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

• A. $\frac{1}{2}\exp \left(\frac{\pi }{6}+\sqrt{e}-1\right)$
• B. $\frac{1}{2}\exp \left(\frac{\pi }{3}+\sqrt{e}-1\right)$
• C. $\exp \left(\frac{\pi }{6}+\sqrt{e}+1\right)$
• D. $2\exp \left(\frac{\pi }{3}+\sqrt{e}-1\right)$

Answer 1

Answer: (A)

$\left(\frac{x}{\sqrt{x^2-y^2}}+e^{\frac{y}{x}}\right)x\frac{dy}{dx}=x+\left(\frac{x}{\sqrt{x^2{y}^2}}+e^{\frac{y}{x}}\right)y$
$\Rightarrow e^{\frac{y}{x}}(xdy-ydx)+\frac{x}{\sqrt{x^2-y^2}}(xdy-ydx)=xdx$
Dividing both side by $x^2$
$\Rightarrow e^{\frac{y}{x}}\left(\frac{xdy-ydx}{x^2}\right)+\frac{1}{\sqrt{1-{\left(\frac{y}{x}\right)}^2}}\left(\frac{xdy-ydx}{x^2}\right)=\frac{dx}{x}$
$\Rightarrow e^{\frac{y}{x}}\mid d\left(\frac{t}{x}\right)+\frac{1}{\sqrt{1-{\left(\frac{y}{x}\right)}^2}}d\left(\frac{y}{x}\right)=\frac{dy}{x}$
Integrate both side.
$\int e^{\frac{y}{x}}d\left(\frac{y}{x}\right)+\int \frac{1}{\sqrt{1-{\left(\frac{y}{x}\right)}^2}}d\left(\frac{y}{x}\right)=\int \frac{dx}{x}$
$\Rightarrow e^{\frac{y}{x}}+{\sin }^{-1}\left(\frac{y}{x}\right)=\ln x+c$
It passes through $(1,0)$
$1+0=0+c\Rightarrow c=1$
It passes through $(2\alpha ,\alpha )$
$e^{\frac{1}{2}}+\sin {}^{-1}\frac{1}{2}=\ln 2\alpha +1$
$\Rightarrow \ln 2\alpha =\sqrt{e}+\frac{\pi }{6}-1$
$\Rightarrow 2\alpha =e^{\left(\sqrt{e}+\frac{\pi }{6}-1\right)}$
$\Rightarrow \alpha =\frac{1}{2}{e}^{\left(\frac{\pi }{6}+\sqrt{e}-1\right)}$