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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^2I got this question in quiz.This intriguing question comes from Linear First Order Differential Equations topic in division Differential Equations of Mathematics – Class 12 |
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Answer» 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 |
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