Form the differential equation of all circles which pass through origin and whose centres lie on Y-axis.

It is given that, circles pass through origin and their centreslie on Y-axis. Let $(0,k)$ be the centre of the circle and radius is $k$.
So, the equation of circle is

${(x - 0)^2} + {(y - k)^2} = {k^2}$

$\Rightarrow$ ${x^2} + {(y - k)^2} = {k^2}$

$\Rightarrow$ ${x^2} + {y^2} - 2ky = 0$
$\Rightarrow$ $\frac{{{x^2} + {y^2}}}{{2y}} = k$

…….(i)
On differentiating Eq. (i) w.r.t. $x$,

we get $\frac{{2y\left( {2x + 2y\frac{{dy}}{{dx}}} \right) - \left( {{x^2} + {y^2}} \right)\frac{{2dy}}{{dx}}}}{{4{y^2}}} = 0$

$\Rightarrow$ $4y\left( {x + y\frac{{dy}}{{dx}}} \right) - 2\left( {{x^2} + {y^2}} \right)\frac{{dy}}{{dx}} = 0$

$\Rightarrow$ $4xy + 4{y^2}\frac{{dy}}{{dx}} - 2\left( {{x^2} + {y^2}} \right)\frac{{dy}}{{dx}} = 0$

$\Rightarrow$ $\left[ {4{y^2} - 2\left( {{x^2} + {y^2}} \right)} \right]\frac{{dy}}{{dx}} + 4xy = 0$

$\Rightarrow$ $\left( {4{y^2} - 2{x^2} - 2{y^2}} \right)\frac{{dy}}{{dx}} + 4xy = 0$

$\Rightarrow$ $\left( {2{y^2} - 2{x^2}} \right)\frac{{dy}}{{dx}} + 4xy = 0$

$\Rightarrow$ $\left( {{y^2} - {x^2}} \right)\frac{{dy}}{{dx}} + 2xy = 0$

$\Rightarrow$ $\left( {{x^2} - {y^2}} \right)\frac{{dy}}{{dx}} - 2xy = 0$