1. Tardigrade
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  4. Let the solution curve y=y(x) of the differential equation, [(x/√x2-y2)+e(y/x)] x (d y/d x)=x+[(x/√x2-y2)+e(y/x)] y pass through the points (1,0) and (2 α, α), α >0. Then α is equal to

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

JEE MainJEE Main 2022Differential Equations

Solution:

$\left(\frac{x}{\sqrt{x^{2}-y^{2}}}+e^{\frac{y}{x}}\right) x \frac{d y}{d x}=x+\left(\frac{x}{\sqrt{x^{2} y^{2}}}+e^{\frac{y}{x}}\right) y$
$\Rightarrow e^{\frac{y}{x}}(x d y-y d x)+\frac{x}{\sqrt{x^{2}-y^{2}}}(x d y-y d x)=x d x$
Dividing both side by $x ^{2}$
$\Rightarrow e^{\frac{y}{x}}\left(\frac{x d y-y d x}{x^{2}}\right)+\frac{1}{\sqrt{1-\left(\frac{y}{x}\right)^{2}}}\left(\frac{x d y-y d x}{x^{2}}\right)=\frac{d x}{x}$
$\Rightarrow e^{\frac{y}{x}} \mid d\left(\frac{t}{x}\right)+\frac{1}{\sqrt{1-\left(\frac{y}{x}\right)^{2}}} d\left(\frac{y}{x}\right)=\frac{d y}{x}$
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 $(2 \alpha, \alpha)$
$e ^{\frac{1}{2}}+\sin { }^{-1} \frac{1}{2}=\ln 2 \alpha+1$
$\Rightarrow \ln 2 \alpha=\sqrt{ e }+\frac{\pi}{6}-1$
$\Rightarrow 2 \alpha= e ^{\left(\sqrt{ e }+\frac{\pi}{6}-1\right)}$
$\Rightarrow \alpha=\frac{1}{2} e ^{\left(\frac{\pi}{6}+\sqrt{ e }-1\right)}$