The curve satisfying the differential equation, (x^2-y^2 ) d x+2 x y d y=0 and passing through the point (1,1) is

Step-by-step solution

(x^2-y^2 ) d x+2 x y d y=0 d y d x = y^2-x^2 2 x y Let y=v x aligned & d y d x =v+x d v d x & v+x d v d x = v^2 x^2-x^2 2 v x^2 & v+x d v d x = v^2-1 2 v & x d v d x = -v^2-1 2 v & 2 v d v v^2+1 =- d x x aligned After integrating, we get aligned & |v^2+1 |=- |x|+ c & y^2 x^2 +1= c x aligned As curve passes through the point (1,1) , so 1+1=c c=2x^2+y^2-2 x=0 , which is a circle of radius one.