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प्रश्न
Solve the following differential equation:-
y dx + (x − y2) dy = 0
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उत्तर
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
\[xy = \int y^2 dy + C\]
\[ \Rightarrow xy = \frac{y^3}{3} + C\]
\[ \Rightarrow x = \frac{y^2}{3} + \frac{C}{y}\]
\[\text{Hence, }x = \frac{y^2}{3} + \frac{C}{y}\text{ is the required solution.}\]
