Question

First find the differential equation then find the degree of that differential equation
$\sqrt{1-x^2}+\sqrt{1+y^2}=a(x-y)$ is

• A. $1$
• B. $2$
• C. $3$
• D. $4$

Answer 1

The correct option is A $1$

We need to find degree of equation $\sqrt{1-x^2}+\sqrt{1+y^2}=a(x-y)$

Put $x=\sin \theta ;y=\sin \varphi$
$\Rightarrow \cos \theta +\cos \varphi =a(\sin \theta -\sin \varphi )$
$\Rightarrow 2\cos \frac{\theta +\varphi }{2}\cos \frac{\theta -\varphi }{2}=2a\cos \frac{\theta -\varphi }{2}\sin \frac{\theta -\varphi }{2}$
$\Rightarrow \cot \frac{\theta -\varphi }{2}=a\Rightarrow \theta -\varphi =2{\cos }^{-1}a$
$\Rightarrow {\sin }^{-1}x-{\sin }^{-1}y=2{\cot }^{-1}a$
Differentiate $\frac{1}{\sqrt{1-x^2}}-\frac{1}{\sqrt{1-y^2}}\frac{dy}{dx}=0$
So, the degree is of the differential equation is $1$.