Question

(1 + x²) dy = xy dx

Answer 1

$$\begin{gathered} \mathrm{We}\mathrm{have}, \\ \left(1+x^2\right)dy=xydx \\ \Rightarrow \frac{1}{y}dy=\frac{x}{1+x^2}dx \\ \mathrm{Integrating}\mathrm{both}\mathrm{sides},\mathrm{we}\mathrm{get} \\ \int \frac{1}{y}dy\mathit{=}\int \frac{x}{\mathit{1}+x^{\mathit{2}}}dx \\ \mathrm{Substituting}1+x^2=t,\mathrm{we}\mathrm{get} \\ 2xdx=dt \\ \therefore \int \frac{1}{y}dy=\frac{1}{2}\int \frac{1}{t}dt \\ \Rightarrow \log \left|y\right|=\frac{1}{2}\log \left|t\right|+\log C \\ \Rightarrow \log \left|y\right|=\frac{1}{2}\log \left|1+x^2\right|+\log C(\because t=1+x^2) \\ \Rightarrow \log \left|y\right|=\log \left[\mathrm{C}\sqrt{1+x^2}\right] \\ \Rightarrow y=C\sqrt{1+x^2} \\ \mathrm{Hence},y=C\sqrt{1+x^2}\mathrm{is}\mathrm{the}\mathrm{required}\mathrm{solution}. \end{gathered}$$