Question

(x² + 1) dy + (2y − 1) dx = 0

Answer 1

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

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

\[ \Rightarrow - \frac{1}{2}\log\left| 1 - 2y \right| = \tan^{- 1} x - \log \sqrt{C}\]

\[ \Rightarrow - \log\left| 1 - 2y \right| = 2 \tan^{- 1} x - 2\log \sqrt{C}\]

\[ \Rightarrow - 2 \tan^{- 1} x = - \log C + \log\left| 1 - 2y \right|\]

\[ \Rightarrow - 2 \tan^{- 1} x = \log \left| \frac{1 - 2y}{C} \right|\]

\[ \Rightarrow e^{- 2 \tan^{- 1} x} = \frac{1 - 2y}{C}\]

\[ \Rightarrow C e^{- 2 \tan^{- 1} x} = \left( 1 - 2y \right)\]

\[ \Rightarrow 1 - C e^{- 2 \tan^{- 1} x} = 2y\]

\[ \Rightarrow \frac{1}{2} - \frac{C}{2} e^{- 2 \tan^{- 1} x} = y\]

\[ \Rightarrow y = \frac{1}{2} + K e^{- 2 \tan^{- 1} x},\text{ where }K = - \frac{C}{2}\]