Question

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

Answer 1

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