Practicing Success
The degree of the differential equation satisfying $\sqrt{1-x^2}+\sqrt{1-y^2}=a(x-y)$, is |
1 2 3 none of these |
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. |