Question

The equation of the curve satisfying the differential equation $y(x+y^3)dx=x(y^3-x)dy$ and passing through the point $(1,1)$ is

• A. $y^3-2x+3x^{2}y=0$
• B. $y^3+2x+3x^{2}y=0$
• C. $y^3+2x-3x^{2}y=0$
• D. None of these

Answer 1

The correct option is C $y^3+2x-3x^{2}y=0$

$y(x+y^3)dx=x(y^3-x)dy$

Here, $(xy+y^4)dx=(x{y}^3-x^2)dy$

$\Rightarrow xydx+y^{4}dx-x{y}^{3}dy+x^{2}dy=0$

$\Rightarrow x(ydx+xdy)+y^3(ydx-xdy)=0$

$\Rightarrow xd\left(xy\right)+[-x^2{y}^3\frac{(ydx-xdy)}{x^2}]=0$

$\Rightarrow xd\left(xy\right)+[-x^2{y}^{3}d\left(\frac{y}{x}\right)]=0$

Dividing the whole equation by $x^3{y}^2$, we get,

$\Rightarrow \frac{d\left(xy\right)}{x^2{y}^2}-\frac{y}{x}d\left(\frac{y}{x}\right)=0$

$\Rightarrow \frac{d\left(xy\right)}{x^2{y}^2}=\frac{y}{x}d\left(\frac{y}{x}\right)$

Integrating both sides, we get

$\Rightarrow \int \frac{d\left(xy\right)}{x^2{y}^2}=\int \frac{y}{x}d\left(\frac{y}{x}\right)$

$\Rightarrow -\frac{1}{xy}=\frac{(y/x{)}^2}{2}-c$

$\Rightarrow y^3+2x+2c{x}^{2}y=0$

at $(1,1)$

$\Rightarrow 1+2+2c=0$

$\Rightarrow c=\frac{-3}{2}$

then $y^3+2x-2\cdot \frac{3}{2}{x}^{2}y=0$

$\Rightarrow y^3+2x-3x^{2}y=0$.