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JEE Main 20245 AprilEvening Shift
Exam
JEE Main
Year
2024
Date
5 April
Shift
Evening Shift
Subject
Mathematics
Chapter
Differential Equations

JEE Main 2024 · 5 April · Evening Shift · Mathematics · Differential Equations

The differential equation of the family of circles passing through...

The differential equation of the family of circles passing through the origin and having centre at the line y=xy=x is :

Options

  1. A(x2−y2+2xy)dx=(x2−y2−2xy)dy\left(x^2-y^2+2 x y\right) \mathrm{d} x=\left(x^2-y^2-2 x y\right) \mathrm{d} y
  2. B(x2+y2+2xy)dx=(x2+y2−2xy)dy\left(x^2+y^2+2 x y\right) \mathrm{d} x=\left(x^2+y^2-2 x y\right) \mathrm{d} y
  3. C(x2+y2−2xy)dx=(x2+y2+2xy)dy\left(x^2+y^2-2 x y\right) \mathrm{d} x=\left(x^2+y^2+2 x y\right) \mathrm{d} y
  4. D(x2−y2+2xy)dx=(x2−y2+2xy)dy\left(x^2-y^2+2 x y\right) \mathrm{d} x=\left(x^2-y^2+2 x y\right) \mathrm{d} y

Correct answer

A. (x2−y2+2xy)dx=(x2−y2−2xy)dy\left(x^2-y^2+2 x y\right) \mathrm{d} x=\left(x^2-y^2-2 x y\right) \mathrm{d} y

Step-by-step solution

C≡x2+y2+gx+gy=0....(1)2x+2yy′+g+gy′=0g=−(2x+2yy′1+y′)\begin{aligned} & C \equiv x^2+y^2+g x+g y=0....(1)\\ & 2 x+2 y y^{\prime}+g+g y^{\prime}=0 \\ & g=-\left(\frac{2 x+2 y y^{\prime}}{1+y^{\prime}}\right) \end{aligned} Put in (1) x2+y2−(2x+2yy′1+y′)(x+y)=0(x2−y2−2xy)y′=x2−y2+2xy\begin{aligned} & x^2+y^2-\left(\frac{2 x+2 y y^{\prime}}{1+y^{\prime}}\right)(x+y)=0 \\ & \left(x^2-y^2-2 x y\right) y^{\prime}=x^2-y^2+2 x y \end{aligned}
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