The degree of differential equation satisfying the relation1+x2−−−−−√+1+y2−−−−−√=λ(x1+y2−−−−−√−y1+x2−−−−−√) is

[a] 1+x2−−−−−√+1+y2−−−−−√=λ(x1+y2−−−−−√−y1+x2−−−−−√)$\Rightarrow 1+x2-----\surd \left(1+\lambda y\right)=1+y2-----\surd \left(\lambda x-1\right)$⇒1+x2−−−−−√1+y2−−−−−√=λx−1λy+1⇒x2+1y2+1=λ2x2−2λx+1λ2y2+2λy+1$\Rightarrow \left(y2+1\right)\left(\lambda 2x2-2\lambda x+1\right)$=(x2+1)(λ2y2+2λy+1)$\Rightarrow \lambda 2x2y2-2\lambda xy2+y2+\lambda 2x2-2\lambda x+1$=λ2x2y2+2λx2y+x2+λ2y2+2λy+1$\Rightarrow \lambda 2\left(x2-y2\right)-2\lambda \left(xy2+x2y+x+y\right)=0$⇒λ2(x+y)(x−y)−2λ[xy(x+y)+(x+y)]=0$\Rightarrow \lambda \left(x+y\right)\left[\lambda \left(x-y\right)-2xy-2\right]=0$⇒(x+y)[λ(x−y)−2xy−2]=0$\Rightarrow \lambda \left(x-y\right)-2xy-2=0$⇒2xy+2x−y=λ⇒xy+1x−y=λ2$\Rightarrow \left(xdydx+y\right)\left(x-y\right)-\left(xy+1\right)\left(1-dydx\right)\left(x-y\right)2=1$This is the first order differential equation and clearly degree of dydx is 1. Hence degree of the differential equation is 1.