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The curve amongst the family of curves represented by the differential equation, (x2 – y2)dx + 2xy dy = 0 which passes through (1, 1) is :

JEE · Math · previous-year question

  1. A.a circle with centre on the y-axis
  2. B.an ellipse with major axis along the y-axis
  3. C.a circle with centre on the x-axiscorrect
  4. D.a hyperbola with transverse axis along the x-axis

Answer

C. a circle with centre on the x-axis

Explanation

(x2 $$-$$ y2) dx + 2xy dy = 0 $${{dy} \over {dx}} = {{{y^2} - {x^2}} \over {2xy}}$$ Put $$y = vx \Rightarrow {{dy} \over {dx}} = v + x{{dv} \over {dx}}$$ Solving we get, $$\int {{{2v} \over {{v^2} + 1}}dv = \int { - {{dx} \over x}} } $$ ln(v2 + 1) = $$-$$ ln x + C (y2 + x2) = Cx 1 + 1 = C $$ \Rightarrow $$ C = 2 y2 + x2 = 2x

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