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)

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