Question

The order and degree of the differential equation $\sqrt{1-x^2}+\sqrt{1-y^2}=k(x-y)$ is

• A. degree $1$, order $1$
• B. degree $1$, order $2$
• C. degree $2$, order $1$
• D. degree $2$, order $2$

Answer 1

The correct option is A degree $1$, order $1$

Let $x=\sin \theta ,y=\sin \varphi$
So, the given equation can be rewritten as
$\sqrt{1-{\sin }^2\theta }+\sqrt{1-{\sin }^2\varphi }=k(\sin \theta -\sin \varphi )$
$\Rightarrow (\cos \theta +\cos \varphi )=k(\sin \theta -\sin \varphi )$
$\Rightarrow 2\cos \frac{\theta +\varphi }{2}\cos \frac{\theta -\varphi }{2}=2k\cos \frac{\theta +\varphi }{2}\sin \frac{\theta -\varphi }{2}$
$\Rightarrow \cot \frac{\theta -\varphi }{2}=k[\because \theta +\varphi \ne \pi ,\text{as}x\ne y]$
$\Rightarrow \theta -\varphi =2{\cot }^{-1}k$
$\Rightarrow {\sin }^{-1}x-{\sin }^{-1}y=2{\cot }^{-1}k$
Differentiating both sides w. r. t. $x$, we get
$\frac{1}{\sqrt{1-x^2}}-\frac{1}{\sqrt{1-y^2}}\frac{dy}{dx}=0$
So, the degree of the differential equation is $1$ and it is a first order differential equation.