What is the equation of the curve passing through (1, 0) of (y(dy/dx) + 2x)^2 = (y^2 + 2x^2)[1 + (dy/dx)^2]?

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

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

(c) √2x^√2 = y/x + √(y^2 + 2x^2)/x^2

(d) √2x = y/x + √(y^2 + 2x^2)/x^2

I got this question in quiz.

This intriguing question comes from Linear First Order Differential Equations topic in division Differential Equations of Mathematics – Class 12

NCERT Solutions for Subject Clas 12 Math Select the correct answer from above options

Interview Questions and Answers, Database Interview Questions and Answers for Freshers and Experience

Right choice is (a) √2x^±1/√2 = y/x + √(y^2 + 2x^2)/x^2

To explain I would say: 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 (put v/√2 = tan t)

putting x = 1, y = 0, we get c = √2

So, the curve is given by,

√2x^±1/√2 = y/x + √(y^2 + 2x^2)/x^2