$\text{The degree of differential equation satisfying the relation}$$\sqrt{1+x^2}+\sqrt{1+y^2}=$$\lambda \left(x\sqrt{1+y^2}-y\sqrt{1+x^2}\right)\text{ is }$

$\begin{array}{l}\sqrt{1+x^2}+\sqrt{1+y^2}=\lambda \left(x\sqrt{1+y^2}-y\sqrt{1+x^2}\right)\\ \Rightarrow \sqrt{1+x^2}(1+\lambda y)=\sqrt{1+y^2}(\lambda x-1)\\ \Rightarrow \frac{\sqrt{1+x^2}}{\sqrt{1+y^2}}=\frac{\lambda x-1}{\lambda y+1}\\ \frac{x^2+1}{y^2+1}=\frac{{\lambda }^2{x}^2-2\lambda x+1}{{\lambda }^2{y}^2+2\lambda y+1}\\ \left(y^2+1\right)\left({\lambda }^2{x}^2-2\lambda x+1\right)=\left(x^2+1\right)\left({\lambda }^2{y}^2+2\lambda y+1\right)\\ \Rightarrow {\lambda }^2{x}^2{y}^2-2\lambda x{y}^2+y^2+{\lambda }^2{x}^2-2\lambda x+1={\lambda }^2{x}^2{y}^2+2\lambda x^{2}y+x^2+{\lambda }^2{y}^2+2\lambda y+1\Rightarrow {\lambda }^2\left(x^2-y^2\right)-2\lambda \left(x{y}^2+x^{2}y+x+y\right)=0\Rightarrow {\lambda }^2(x+y)(x-y)-2\lambda \left[xy\right(x+y)+(x+y\left)\right]=0\Rightarrow \lambda (x+y)\left[\lambda \right(x-y)-2xy-2]=0\Rightarrow (x+y)\left[\lambda \right(x-y)-2xy-2]=0\Rightarrow \lambda (x-y)-2xy-2=0\Rightarrow \frac{2xy+2}{x-y}=\lambda \Rightarrow \frac{xy+1}{x-y}=\frac{\lambda }{2}\\ \\ \Rightarrow \frac{\left(x\frac{dy}{dx}+y\right)(x-y)-(xy+1)\left(1-\frac{dy}{dx}\right)}{{(x-y)}^2}=1\\ \text{This is the first order differential equation and clearly degree of }\frac{\text{dy}}{\text{dx}}\text{ is 1}\\ \text{Hence degree of the differential equation is 1.}\\ \end{array}$