Question

The equation of the curve satisfying the differential equation y (x + y³) dx = x (y³ − x) dy and passing through the point (1, 1) is
(a) y³ − 2x + 3x² y = 0
(b) y³ + 2x + 3x² y = 0
(c) y³ + 2x −3x² y = 0
(d) none of these

Answer 1

(c) y³ + 2x −3x² y = 0

We have,

$$\begin{gathered} y\left(x+y^3\right)dx=x\left(y^3-x\right)dy \\ \mathrm{Here},\left(xy+y^4\right)dx=\left(x{y}^3-x^2\right)dy \\ \Rightarrow xydx+y^{4}dx-x{y}^{3}dy+x^{2}dy=0 \\ \Rightarrow x\left(ydx+xdy\right)+y^3\left(ydx-xdy\right)=0 \\ \Rightarrow xd\left(xy\right)+x^2{y}^3\frac{\left(ydx-xdy\right)}{x^2}=0 \\ \Rightarrow xd\left(xy\right)-x^2{y}^3\frac{\left(xdy-ydx\right)}{x^2}=0 \\ \Rightarrow \frac{d\left(xy\right)}{x^2{y}^2}-\frac{y}{x}d\left(\frac{y}{x}\right)=0\left(\because \mathrm{Dividing}\mathrm{the}\mathrm{whole}\mathrm{equation}\mathrm{b}y{x}^3{y}^2\right) \\ \Rightarrow \frac{d\left(xy\right)}{x^2{y}^2}=\frac{y}{x}d\left(\frac{y}{x}\right) \end{gathered}$$
Integrating both sides we get,
$$\begin{gathered} \Rightarrow \int \frac{d\left(xy\right)}{x^2{y}^2}=\int \frac{y}{x}d\left(\frac{y}{x}\right) \\ \Rightarrow -\frac{1}{xy}=\frac{{\left({\displaystyle \frac{y}{x}}\right)}^2}{2}-c \\ \therefore -\frac{1}{xy}-\frac{1}{2}{\left(\frac{y}{x}\right)}^2-c=0 \\ \therefore \frac{1}{xy}+\frac{1}{2}\left(\frac{y^2}{x^2}\right)+c=0 \\ \therefore y^3+2x+2c{x}^{2}y=0 \end{gathered}$$

It is given that the curves passes through (1, 1).
Hence,
$$\begin{gathered} y^3+2x+2c{x}^{2}y=0 \\ {\left(1\right)}^3+2\left(1\right)+2c\left(1\right)\left(1\right)=0 \\ 1+2+2c=0 \\ 2c=-3 \\ c=-\frac{3}{2} \end{gathered}$$
∴ The required curve is $y^3+2x-2\times \frac{3}{2}{x}^{2}y=0$
$\therefore y^3+2x-3x^{2}y=0$