Question

Solve the following differential equation:-

y dx + (x − y²) dy = 0

Answer 1

We have,

\[y dx + \left( x - y^2 \right)dy = 0\]

\[ \Rightarrow y dx = - \left( x - y^2 \right)dy \]

\[ \Rightarrow \frac{dx}{dy} = - \frac{1}{y}\left( x - y^2 \right) \]

\[ \Rightarrow \frac{dx}{dy} + \frac{1}{y}x = y . . . . . . . . \left( 1 \right)\]

Clearly, it is a linear differential equation of the form

\[\frac{dx}{dy} + Px = Q\]

\[\text{where }P = \frac{1}{y}\text{ and }Q = y\]

\[ \therefore I . F . = e^{\int P\ dy} \]

\[ = e^{\int\frac{1}{y}dy} \]

\[ = e^{\log y = y}\]

Multiplying both sides of (1) by I . F . = y, we get

\[y\left( \frac{dx}{dy} + \frac{1}{y}x \right) = y \times y\]

\[ \Rightarrow y\frac{dx}{dy} + x = y^2 \]

Integrating both sides with respect to y, we get

\[xy = \int y^2 dy + C\]

\[ \Rightarrow xy = \frac{y^3}{3} + C\]

\[ \Rightarrow x = \frac{y^2}{3} + \frac{C}{y}\]

\[\text{Hence, }x = \frac{y^2}{3} + \frac{C}{y}\text{ is the required solution.}\]