The degree of the differential equation satisfying the relation `sqrt(1+x^2) + sqrt(1+y^2) = lambda (x sqrt(1+y^2)- ysqrt(1+x^2))` is

Answer

`(dy)/(dx) = (1+y^(2))/(1+x^(2))`

Explanation

Putting x=tanA and y=tanB in the given relation, we get cosA + cosB=`lambda`(sinA-sinB)
or `tan^(-1)x-tan^(-1)y=2tan^(-1)(1/lambda)`
Differentiating w.r.t. to x, we get
`1/(1+x^(2))-1/(1+y^(2))(dy)/(dx)=0`
or `(dy)/(dx)=(1+y^(2))/(1+x^(2))`