Let y=y(x) be the solution of the differential equation (x-x^3)dy=(y+yx^2-3x^4)dx,x>2 If y(3)=3, then y(4) is equal to:

$$(x-{x}^{3})dy=(y+y{x}^{2}-3{x}^{4})dx$$(x−x3)dy=(y+yx2−3x4)dx

$$\Rightarrow xdy-ydx=(y{x}^{2}-3{x}^{4})dx+{x}^{3}dy$$⇒xdy−ydx=(yx2−3x4)dx+x3dy

$$\Rightarrow \frac{xdy-ydx}{{x}^{2}}=(ydx+xdy)-3{x}^{2}dx$$⇒x2xdy−ydx​=(ydx+xdy)−3x2dx

$$\Rightarrow d(\frac{y}{x})=d(xy)-d({x}^{3})$$⇒d(xy​)=d(xy)−d(x3)

Integrate

$$\Rightarrow \frac{y}{x}=xy-{x}^{3}+c$$⇒xy​=xy−x3+c

given $f(3)=3$f(3)=3

$$\Rightarrow \frac{3}{3}=3\times 3-{3}^{3}+c$$⇒33​=3×3−33+c

$$\Rightarrow c=19$$⇒c=19

$$\therefore \frac{y}{x}=xy-{x}^{3}+19$$∴xy​=xy−x3+19

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

$$15y=4\times 45$$15y=4×45

$$\Rightarrow y=12$$⇒y=12