Question

The degree of the differential equation satisfying
$\sqrt{1+x^2}+\sqrt{1+y^2}=A(x\sqrt{1+y^2}+y\sqrt{1+x^2})$ is

• A. $2$
• B. $3$
• C. $4$
• D. none of these

Answer 1

The correct option is D none of these

$\sqrt{1+x^2}+\sqrt{1+y^2}=A(x\sqrt{1+y^2}+y\sqrt{1+x^2})$
Put $x=\tan \theta$ and $y=\tan \varphi$
Then $\sqrt{1+x^2}=\sec \theta ,\sqrt{1+y^2}=\sec \varphi$
and the equation becomes
$\sec \varphi +\sec \theta =A\left(\tan \theta \sec \varphi -\tan \varphi \sec \theta \right)\Rightarrow \cos \theta +\cos \varphi =A\left(\sin \theta -\sin \varphi \right)$
$\Rightarrow 2\cos \left(\frac{\varphi +\theta }{2}\right)\cos \left(\frac{\theta -\varphi }{2}\right)=2A\sin \left(\frac{\theta -\varphi }{2}\right)\cos \left(\frac{\varphi +\theta }{2}\right)$
$\Rightarrow \cot \left(\frac{\theta -\varphi }{2}\right)=A\Rightarrow \theta -\varphi =2{\cot }^{-1}A\Rightarrow {\tan }^{-1}x-{\tan }^{-1}y=2{\cot }^{-1}A$
Differentiating this, we get
$\frac{1}{1+x^2}-\frac{1}{1+y^2}\frac{dy}{dx}=0$
which is a differential equation of degree $1$.