What is the solution of (y(dy/dx) + 2x)^2 = (y^2 + 2x^2)[1 + (dy/dx)^2]?

(a) cx^±1/√2 = y/x + √(y^2 – 2x^2)/x^2

(b) cx^±√2 = y/x + √(y^2 + 2x^2)/x^2

(c) cx^±1/2√2 = y/x + √(y^2 – 2x^2)/x^2

(d) cx^±1/√2 = y/x + √(y^2 + 2x^2)/x^2

The question was asked during an internship interview.

My enquiry is from Linear First Order Differential Equations in portion Differential Equations of Mathematics – Class 12

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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