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STD 12 - 9. differential equations — JEE STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceJEESTD 12 - 9. differential equations4 Marks
MCQ
The differential equation of all circles which passes through the origin and whose centre lies on $y$-axis, is
  • ✓
    $({x^2} - {y^2})\frac{{dy}}{{dx}} - 2xy = 0$
  • B
    $({x^2} - {y^2})\frac{{dy}}{{dx}} + 2xy = 0$
  • C
    $({x^2} - {y^2})\frac{{dy}}{{dx}} - xy = 0$
  • D
    $({x^2} - {y^2})\frac{{dy}}{{dx}} + xy = 0$
✓

Answer

Correct option: A.
$({x^2} - {y^2})\frac{{dy}}{{dx}} - 2xy = 0$
a
(a) The system of circles pass through origin and centre lies on $y$-axis is ${x^2} + {y^2} - 2ay = 0$

==> $2x + 2y\frac{{dy}}{{dx}} - 2a\frac{{dy}}{{dx}} = 0$ ==> $2a = 2y + 2x\frac{{dx}}{{dy}}$

Therefore, the required differential equation is

${x^2} + {y^2} - 2{y^2} - 2xy\frac{{dx}}{{dy}} = 0$==>$({x^2} - {y^2})\frac{{dy}}{{dx}} - 2xy = 0$.

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