Let $y=y\left(x\right)$ be the solution of the differential equation $(x-x^3)dy=(y+y{x}^2-3x^4)dx,$ $x>2.$ If $y\left(3\right)=3$, then $y\left(4\right)$ is equal to

The correct option is A $12$
$(x-x^3)dy=y(1+x^2)dx-3x^{4}dx$
$\Rightarrow \frac{dy}{dx}+y\frac{1+x^2}{x(x^2-1)}=\frac{3x^3}{x^2-1}$
$\therefore$ I.F.$=e^{\int \frac{1+x^2}{x(x^2-1)}dx}=e^{\int (\frac{1}{x-1}+\frac{1}{x+1}-\frac{1}{x})dx}$
$=e^{\ln \left(\frac{x^2-1}{x}\right)}=\frac{x^2-1}{x}$
$\therefore$ Solution is $y\cdot \left(\frac{x^2-1}{x}\right)=\int \frac{3x^3}{x^2-1}\cdot \frac{x^2-1}{x}dx\Rightarrow y\left(\frac{x^2-1}{x}\right)=x^3+c$
$\because y\left(3\right)=3$ then $c=-19$
$\therefore y\left(x\right)=\frac{x(x^3-19)}{x^2-1}$
$\therefore y\left(4\right)=\frac{45\times 4}{15}=12$