The degree of differential equation satisfying the relation1+x2+1+y2=λx1+y2-y1+x2 is

1+x2+1+y2=λx1+y2−y1+x2⇒1+x2(1+λy)=1+y2(λx−1)⇒1+x21+y2=λx−1λy+1x2+1y2+1=λ2x2−2λx+1λ2y2+2λy+1y2+1λ2x2−2λx+1=x2+1λ2y2+2λy+1⇒λ2x2y2−2λxy2+y2+λ2x2−2λx+1=λ2x2y2+2λx2y+x2+λ2y2+2λy+1  ⇒λ2x2−y2−2λxy2+x2y+x+y=0⇒λ2(x+y)(x−y)−2λ[xy(x+y)+(x+y)]=0⇒λ(x+y)[λ(x−y)−2xy−2]=0⇒(x+y)[λ(x−y)−2xy−2]=0⇒λ(x−y)−2xy−2=0⇒2xy+2x−y=λ⇒xy+1x−y=λ2 ⇒xdydx+y(x−y)−(xy+1)1−dydx(x−y)2=1This is the first order differential equation and clearly degree of dydx is 1Hence degree of the differential equation is 1.