$\therefore v + x\frac{{dv}}{{dx}} = \frac{{v - 1}}{{v + 1}}$ ==> $x\frac{{dv}}{{dx}} = \frac{{v - 1}}{{v + 1}} - v$
==> $x\frac{{dv}}{{dx}} = - \frac{{{v^2} + 1}}{{v + 1}}$ ==> $\int_{}^{} {\frac{{dx}}{x}} = - \int_{}^{} {\frac{{v + 1}}{{{v^2} + 1}}} {\rm{ }}dv$
==> $ - {\log _e}x = \frac{1}{2}\int_{}^{} {\frac{{2v}}{{{v^2} + 1}}} dv + \int_{}^{} {\frac{1}{{{v^2} + 1}}} dv$
==> $ - {\log _e}x = \frac{1}{2}\log ({v^2} + 1) + {\tan ^{ - 1}}v + c$
==> $ - 2{\log _e}x = \log \left( {\frac{{{x^2} + {y^2}}}{{{x^2}}}} \right) + 2{\tan ^{ - 1}}\left( {\frac{y}{x}} \right) + c$
==> ${\log _e}({x^2} + {y^2}) + 2{\tan ^{ - 1}}\frac{y}{x} + c = 0$.
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