Solve this following


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


  1. 4

  2. 12

  3. 8

  4. 16

Correct Option: , 2


$\left(x-x^{3}\right) d y=\left(y+y x^{2}-3 x^{4}\right) d x$

$\Rightarrow x d y-y d x=\left(y x^{2}-3 x^{4}\right) d x+x^{3} d y$

$\Rightarrow \frac{x d y-y d x}{x^{2}}=(y d x+x d y)-3 x^{2} d x$

$\Rightarrow \mathrm{d}\left(\frac{\mathrm{y}}{\mathrm{x}}\right)=\mathrm{d}(\mathrm{xy})-\mathrm{d}\left(\mathrm{x}^{3}\right)$


$\Rightarrow \frac{y}{x}=x y-x^{3}+c$

given $f(3)=3$

$\Rightarrow \frac{3}{3}=3 \times 3-3^{3}+\mathrm{c}$

$\Rightarrow \mathrm{c}=19$

$\therefore \frac{y}{x}=x y-x^{3}+19$

at $x=4, \frac{y}{4}=4 y-64+19$

$15 y=4 \times 45$

$\Rightarrow \mathrm{y}=12$


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