Right answer is (d) cx^±1/√2 = y/x + √(y^2 + 2x^2)/x^2
Explanation: Here, y^2(dy/dx)^2 + 4x^2 + 4xy(dy/dx) = (y^2 + 2x^2)[1 + (dy/dx)^2]
=>dy/dx = y/x ± √(1/2(y/x)^2) + 1
Let, y = vx
=> v + x dv/dx = v ± √(1/2(v)^2) + 1
Integrating both sides,
±√dv/(√(1/2(v)^2) + 1) = ∫dx/x
cx^±1/√2 = y/x + √(y^2 + 2x^2)/x^2