Mathematics | TS EAMCET 2017 | Discussion Page

The solution of $(y-3x^{2})dx+xdy=0$ is

$y(x)= \sin x+\frac{1}{x^{2}}+C$

$y(x)= \cos x-\frac{1}{x^{2}}+C$

$y(x)= x^{2}+\frac{C}{x}$

$y(x)= \sqrt{x}+\frac{C}{x}$

Option C

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

$xy=x^{3}+C \Rightarrow y=x^{2}+\frac{C}{x}$

Discussion Forum