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]?

Category: QuestionsWhat is the equation of the curve passing through (1, 0) of (y(dy/dx) + 2x)^2 = (y^2 + 2x^2)[1 + (dy/dx)^2]?
Editor">Editor Staff asked 11 months ago

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 
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1 Answers
Editor">Editor Staff answered 11 months ago

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