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प्रश्न
(x2 + 1) dy + (2y − 1) dx = 0
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उत्तर
We have,
\[\left( 1 + x^2 \right)dy + \left( 2y - 1 \right)dx = 0\]
\[ \Rightarrow \left( 1 + x^2 \right)dy = \left( 1 - 2y \right)dx\]
\[ \Rightarrow \frac{dy}{\left( 1 - 2y \right)} = \frac{1}{\left( 1 + x^2 \right)}dx\]
Integrating both sides, we get
\[\int\frac{1}{\left( 1 - 2y \right)}dy = \int\frac{1}{\left( 1 + x^2 \right)}dx\]
\[ \Rightarrow - \frac{1}{2}\log\left| 1 - 2y \right| = \tan^{- 1} x - \log \sqrt{C}\]
\[ \Rightarrow - \log\left| 1 - 2y \right| = 2 \tan^{- 1} x - 2\log \sqrt{C}\]
\[ \Rightarrow - 2 \tan^{- 1} x = - \log C + \log\left| 1 - 2y \right|\]
\[ \Rightarrow - 2 \tan^{- 1} x = \log \left| \frac{1 - 2y}{C} \right|\]
\[ \Rightarrow e^{- 2 \tan^{- 1} x} = \frac{1 - 2y}{C}\]
\[ \Rightarrow C e^{- 2 \tan^{- 1} x} = \left( 1 - 2y \right)\]
\[ \Rightarrow 1 - C e^{- 2 \tan^{- 1} x} = 2y\]
\[ \Rightarrow \frac{1}{2} - \frac{C}{2} e^{- 2 \tan^{- 1} x} = y\]
\[ \Rightarrow y = \frac{1}{2} + K e^{- 2 \tan^{- 1} x},\text{ where }K = - \frac{C}{2}\]
