Question

The absolute value of the constant term in the solution of the differential equation $y(2x^4+y)\frac{dy}{dx}=(1-4x{y}^2)x^2$,
if the curve passes through the center of the circle $x^2+y^2-6y=0$, is

Answer 1

$y(2x^4+y)\frac{dy}{dx}=(1-4x{y}^2)x^2\Rightarrow 2x^{4}y\frac{dy}{dx}+y^2\frac{dy}{dx}=x^2-4x^3{y}^2\Rightarrow 2x^{4}y\frac{dy}{dx}+4x^3{y}^2+y^2\frac{dy}{dx}=x^2\Rightarrow \frac{d}{dx}\left(x^4{y}^2\right)+\frac{d}{dx}\left(\frac{y^3}{3}\right)=x^2$
$\Rightarrow x^4{y}^2+\frac{y^3}{3}=\frac{x^3}{3}+C$

The given circle is $x^2+y^2-6y=0$
$\Rightarrow x^2+(y-3{)}^2=9$
So, the centre of the circle is $(0,3)$
Now, $x^4{y}^2+\frac{y^3}{3}=\frac{x^3}{3}+C$
$\Rightarrow 0+3^2=C\Rightarrow C=9$