Question

If the order and degree of the differential equation satisfying $\sqrt{1-x^2}+\sqrt{1-y^2}=b(x-y),$ where $b$ is a parameter, is $\alpha$ and $\beta$ respectively, then $\alpha +\beta$ is

Answer 1

$\sqrt{1-x^2}+\sqrt{1-y^2}=b(x-y)\cdots \left(1\right)$
Clearly, the order is one as there is only one independent parameter $b$.
Put $x=\sin \alpha ,y=\sin \beta$ in eqn. $\left(1\right)$
$\Rightarrow \alpha ={\sin }^{-1}x,\beta ={\sin }^{-1}y$
$\cos \alpha +\cos \beta =b(\sin \alpha -\sin \beta )$
$\Rightarrow 2\cos \left(\frac{\alpha +\beta }{2}\right)\cos \left(\frac{\alpha -\beta }{2}\right)=2b\cos \left(\frac{\alpha +\beta }{2}\right)\sin \left(\frac{\alpha -\beta }{2}\right)$
$\Rightarrow \cot \left(\frac{\alpha -\beta }{2}\right)=b$
$\Rightarrow \alpha -\beta =2{\cot }^{-1}b$
$\Rightarrow {\sin }^{-1}x-{\sin }^{-1}y=2{\cot }^{-1}b$
Differentiating w.r.t. $x,$ we get
$\frac{1}{\sqrt{1-x^2}}-\frac{1}{\sqrt{1-y^2}}\frac{dy}{dx}=0$
Degree of above differential equation is one.
$\therefore \alpha =\beta =1\Rightarrow \alpha +\beta =2$