Question

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

• A. $y^2=-2x$
• B. $y=-2x$
• C. $y^3=-2x$
• D. None of these

Answer 1

Answer: (C)

$\left(x{y}^3-x^2\right)dy-\left(xy+y^4\right)dx=0$
$\Rightarrow y^3(xdy-ydx)-x(xdy+ydx)=0$
$\Rightarrow x^2{y}^3\frac{(xdy-ydx)}{x^2}-x(xdy+ydx)=0$
$\Rightarrow x^2{y}^{3}d\left(\frac{y}{x}\right)-xd(xy)=0$
$\Rightarrow$ Dividing by $x^3{y}^2,$ we get
$\frac{y}{x}d\left(\frac{y}{x}\right)-\frac{d(xy)}{x^2{y}^2}=0$
Now, integrating $\frac{1}{2}{\left(\frac{y}{x}\right)}^2+\frac{1}{xy}=c$
It passes through the point $(4,-2)$ .
$\Rightarrow \frac{1}{8}-\frac{1}{8}=c$
$\Rightarrow c=0$
$\therefore y^3=-2x$