Question

The curve satisfies the equation $\frac{dy}{dx}=\frac{y \left(x + y^{3}\right)}{x \left(y^{3} - x\right)}$ and passing through the point $\left(4 , \, - 2\right)$ , is

• A. $y^{2}=-2x+12$
• B. $2y=-x$
• C. $y^{3}=-2x$
• D. $y^{2}=x$

Answer 1

Answer: (C)

$\frac{dy}{dx}=\frac{y \left(x + y^{3}\right)}{x \left(y^{3} - x\right)}$
$\Rightarrow dy\left(\left(xy\right)^{3} - x^{2}\right) \, =\left(xy + y^{4}\right) \, dx$
$\Rightarrow y^{3}\left(xdy - ydx\right)=x\left(ydx + xdy\right)$
$\Rightarrow x^{2}y^{3}d\left(\frac{y}{x}\right)=xd\left(xy\right)$
$\Rightarrow \frac{y}{x}d\left(\frac{y}{x}\right)=\frac{d \left(xy\right)}{\left(\right. xy \left(\left.\right) \, \right)^{2}}$
$\Rightarrow \frac{1}{2}\left(\frac{y}{x}\right)^{2}=-\frac{1}{xy}+c$
Passes through $\left(4 , - 2\right)\Rightarrow c=0$
So $y^{3}=-2x$