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

विकल्प

• y³ − 2x + 3x² y = 0

• y³ + 2x + 3x² y = 0

• y³ + 2x −3x² y = 0

• none of these

Answer 1

y³ + 2x −3x² y = 0

 

We have,

\[y\left( x + y^3 \right)dx = x\left( y^3 - x \right)dy\]

\[\text{ Here, }\left( xy + y^4 \right)dx = \left( x y^3 - x^2 \right)dy\]

\[ \Rightarrow xydx + y^4 dx - x y^3 dy + x^2 dy = 0\]

\[ \Rightarrow x\left( ydx + xdy \right) + y^3 \left( ydx - xdy \right) = 0\]

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

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

\[ \Rightarrow \frac{d\left( xy \right)}{x^2 y^2} - \frac{y}{x}d\left( \frac{y}{x} \right) = 0 ...........\left( \because\text{Dividing the whole equation by }x^3 y^2 \right)\]

\[ \Rightarrow \frac{d\left( xy \right)}{x^2 y^2} = \frac{y}{x}d\left( \frac{y}{x} \right)\]
Integrating both sides we get,
\[\Rightarrow \int\frac{d\left( xy \right)}{x^2 y^2} = \int\frac{y}{x}d\left( \frac{y}{x} \right)\]
\[ \Rightarrow - \frac{1}{xy} = \frac{\left( \frac{y}{x} \right)^2}{2} - c\]
\[ \therefore - \frac{1}{xy} - \frac{1}{2} \left( \frac{y}{x} \right)^2 - c = 0\]
\[ \therefore \frac{1}{xy} + \frac{1}{2}\left( \frac{y^2}{x^2} \right) + c = 0\]
\[ \therefore y^3 + 2x + 2c x^2 y = 0\]
It is given that the curves passes through (1, 1).
Hence,
\[y^3 + 2x + 2c x^2 y = 0\]
\[ \left( 1 \right)^3 + 2\left( 1 \right) + 2c\left( 1 \right)\left( 1 \right) = 0\]
\[1 + 2 + 2c = 0\]
\[2c = - 3\]
\[c = - \frac{3}{2}\]
∴ The required curve is \[y^3 + 2x - 2 \times \frac{3}{2} x^2 y = 0\]
\[y^3 + 2x - 2 \times \frac{3}{2} x^2 y = 0\]