Question

(x² + 1) dy + (2y − 1) dx = 0

Answer 1

$$\begin{gathered} \mathrm{We}\mathrm{have}, \\ \left(1+x^2\right)dy+\left(2y-1\right)dx=0 \\ \Rightarrow \left(1+x^2\right)dy=\left(1-2y\right)dx \\ \Rightarrow \frac{dy}{\left(1-2y\right)}=\frac{1}{\left(1+x^2\right)}dx \\ \mathrm{Integrating}\mathrm{both}\mathrm{sides},\mathrm{we}\mathrm{get} \\ \int \frac{1}{\left(1-2y\right)}dy=\int \frac{1}{\left(1+x^2\right)}dx \\ \Rightarrow -\frac{1}{2}\log \left|1-2y\right|={\tan }^{-1}x-\log \sqrt{C} \\ \Rightarrow -\log \left|1-2y\right|=2{\tan }^{-1}x-2\log \sqrt{C} \\ \Rightarrow -2{\tan }^{-1}x=-\log C+\log \left|1-2y\right| \\ \Rightarrow -2{\tan }^{-1}x=\log \left|\frac{1-2y}{C}\right| \\ \Rightarrow e^{-2{\tan }^{-1}x}=\frac{1-2y}{C} \\ \Rightarrow C{e}^{-2{\tan }^{-1}x}=\left(1-2y\right) \\ \Rightarrow 1-C{e}^{-2{\tan }^{-1}x}=2y \\ \Rightarrow \frac{1}{2}-\frac{C}{2}{e}^{-2{\tan }^{-1}x}=y \\ \Rightarrow y=\frac{1}{2}+K{e}^{-2{\tan }^{-1}x},\mathrm{where}K=-\frac{C}{2} \end{gathered}$$