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(B) Given differential equation: $(x-x^{3}) dy = (y+yx^{2}-3x^{4}) dx$Rearranging the terms:$(x-x^{3}) dy - y(1+x^{2}) dx = -3x^{4} dx$Divide by $x(1-

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$12$

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(B) Given differential equation: $(x-x^{3}) dy = (y+yx^{2}-3x^{4}) dx$Rearranging the terms:$(x-x^{3}) dy - y(1+x^{2}) dx = -3x^{4} dx$Divide by $x(1-x^{2})$ or rearrange to identify the structure:$x dy - x^{3} dy = y dx + yx^{2} dx - 3x^{4} dx$$x dy - y dx = x^{3} dy + yx^{2} dx - 3x^{4} dx$Divide by $x^{2}$:$\frac{x dy - y dx}{x^{2}} = \frac{x^{3} dy + yx^{2} dx}{x^{2}} - 3x^{2} dx$$d(\frac{y}{x}) = d(xy) - d(x^{3})$Integrating both sides:$\frac{y}{x} = xy - x^{3} + C$Given $y(3)=3$,substitute $x=3, y=3$:$\frac{3}{3} = 3(3) - 3^{3} + C$$1 = 9 - 27 + C$$1 = -18 + C \Rightarrow C = 19$So,the equation is $\frac{y}{x} = xy - x^{3} + 19$For $x=4$:$\frac{y}{4} = 4y - 64 + 19$$\frac{y}{4} = 4y - 45$$y = 16y - 180$$15y = 180$$y = 12$

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 :

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