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NDA 2020 Maths · Mathematics · Differential Equations · hard

The curves satisfying dy/dx=(x^2+y^2/(2xy) form the family

Single correct option

The curves satisfying dydx=x2+y22xy\dfrac{dy}{dx}=\dfrac{x^2+y^2}{2xy} form the family
  1. A. x2+y2=Cxx^2+y^2=Cx
  2. B. x2−y2=Cxx^2-y^2=Cx
  3. C. x2−y2=Cx^2-y^2=C
  4. D. x2−y2=Cx2x^2-y^2=Cx^2
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Answer: (B) x2−y2=Cxx^2-y^2=Cx

The equation is homogeneous, so put y=vxy=vx: v+xdvdx=1+v22vv+x\dfrac{dv}{dx}=\dfrac{1+v^2}{2v}, giving xdvdx=1−v22vx\dfrac{dv}{dx}=\dfrac{1-v^2}{2v}. Separating, 2v dv1−v2=dxx\dfrac{2v\,dv}{1-v^2}=\dfrac{dx}{x}, so −ln⁡∣1−v2∣=ln⁡∣x∣+c-\ln|1-v^2|=\ln|x|+c, i.e. x(1−v2)=Cx(1-v^2)=C, which becomes x2−y2=Cxx^2-y^2=Cx after multiplying by xx. The option x2−y2=Cx^2-y^2=C results from forgetting the factor xx left in x(1−v2)=Cx(1-v^2)=C.
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