Question

If the differential equation representing the family of all circles touching $x-$axis at the origin is $(x^2-y^2)\frac{dy}{dx}=g\left(x\right)y$, then $g\left(x\right)$ equals :

• A. $\frac{1}{2}x$
• B. $2x^2$
• C. $2x$
• D. $\frac{1}{2}{x}^2$

Answer 1

The correct option is C $2x$

$(x^2-y^2)y^{\prime }=g\left(x\right)y$
$x^2+(y-a{)}^2=a^2$
Differentiating the equation with respect to x
$2sx+2(y-a)y^{\prime }=0$ .
$a=\frac{x+y{y}^{\prime }}{y^{\prime }}$ .....(1)
Put 'a' in original equation, we get
$x^2+(y-\left(\frac{x+y{y}^{\prime }}{y^{\prime }}\right))2={\left(\frac{x+y{y}^{\prime }}{y^{\prime }}\right)}^2$
$y^{\prime 2}{x}^2+(y{y}^{\prime }-(x+y{y}^{\prime 2}){)}^2=(x+y{y}^{\prime }{)}^2$
By solving this equation
$(x^2-y^2)y^{\prime }=2xy$
$g\left(x\right)=2x$