Question

The differential equation of all circles which pass through the origin and whose centers lie on the y-axis is

• A. $(x^2-y^2)\frac{dy}{dx}-2xy=0$
• 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 1

The correct option is A $(x^2-y^2)\frac{dy}{dx}-2xy=0$

Such family of circles is given by
$x^2+(y-a{)}^2=a^2$
or $x^2+y^2-2ay=0$
Differentiating, $2x+2y\frac{dy}{dx}=2a\frac{dy}{dx}$
or $x+y\frac{dy}{dx}=a\frac{dy}{dx}$
Substituting the value of $a$ in equation $\left(1\right),$ we get
$(x^2-y^2)\frac{dy}{dx}=2xy$ (order in one again and degree 1)