1) The solution of $(y-3x^{2})dx+xdy=0$ is A) $y(x)= \sin x+\frac{1}{x^{2}}+C$ B) $y(x)= \cos x-\frac{1}{x^{2}}+C$ C) $y(x)= x^{2}+\frac{C}{x}$ D) $y(x)= \sqrt{x}+\frac{C}{x}$ Answer: Option CExplanation:We have, $(y-3x^{2})dx+xdy=0$ $\Rightarrow$ $ydx-3x^{2}dx+xdy=0$ $\Rightarrow$ $ydx+xdy=3x^{2}dx$ $\Rightarrow$ $dxy=3x^{2}dx$ On integrating both sides , we get $xy=x^{3}+C \Rightarrow y=x^{2}+\frac{C}{x}$
