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

The given differential equation is,

 $\left(x-x^3\right)dy=\left(y+y{x}^2-3x^4\right)dx,x>2$

$\Rightarrow \left(xdy-ydx\right)+3x^{4}dx=y{x}^{2}dx+x^{3}dy$

$\Rightarrow \left(\frac{xdy-ydx}{x^2}\right)+3x^{2}dx=ydx+xdy$

$\Rightarrow \int d\left(y/x\right)+\int 3x^{2}dx=\int d\left(xy\right)\Rightarrow \frac{y}{x}+x^3=xy+c$

$\Rightarrow x^4=y\left(x^2-1\right)+cx$

Now, $y\left(3\right)=3$ (given) $\Rightarrow 81=24+3c\Rightarrow c=\frac{57}{3}=19$

$\therefore$ Solution is $x^4=y\left(x^2-1\right)+19x$

Putting $x=4$, we get

${\left(4\right)}^4=y\left(4^2-1\right)+19\times 4\Rightarrow y\left(4\right)=\frac{256-76}{15}=12$