The degree of the differential equation satisfying $\sqrt{1-x^2}+\sqrt{1-y^2}=a(x-y)$, is
Answer & explanation
Correct answer: option 1
We have,
$a(x-y)=\sqrt{1-y^2}+\sqrt{1-x^2}$
Putting $x=\sin A, y=\sin B$, we get
$\cos A+\cos B=a(\sin A-\sin B)$
$\Rightarrow \cot \frac{A-B}{2}=a$
$\Rightarrow A-B=2 \cot ^{-1} a$
$\Rightarrow \sin ^{-1} x-\sin ^{-1} y=2 \cot ^{-1} a$
Differentiating w.r.to $x$, we get
$\frac{1}{\sqrt{1-x^2}}-\frac{1}{\sqrt{1-y^2}} \frac{d y}{d x}=0 \Rightarrow \frac{d y}{d x}=\sqrt{\frac{1-y^2}{1-x^2}}$
Clearly, it is differential equation of first order and first degree.