Question

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

• A. 2
• B. 3
• C. 4
• D. None of these

Answer 1

Answer: (D)

None of these

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
$sec\theta +sec\varphi =A(tan\theta sec\varphi -tan\varphi sec\theta )$
$\Rightarrow \frac{cos\varphi +cos\theta }{cos\theta cos\varphi }=A\left(\frac{sin\theta -sin\varphi }{cos\theta cos\varphi }\right)\Rightarrow 2cos\frac{\theta +\varphi }{2}cos\frac{\theta -\varphi }{2}=2Asin\frac{\theta -\varphi }{2}cos\frac{\theta -\varphi }{2}\Rightarrow cot\frac{\theta -\varphi }{2}=A\Rightarrow \theta -\varphi =2co{t}^{-1}A\Rightarrow ta{n}^{-1}x-ta{n}^{-1}y=2co{t}^{-1}A$
Differentiating this, we get $\frac{1}{1+x^2}-\left(\frac{1}{1+y^2}\right)\frac{dy}{dx}=0$, which is a differential equation of degree 1.