Question

\[\frac{dy}{dx} + \frac{1 + y^2}{y} = 0\]

Answer 1

We have, 
\[\frac{dy}{dx} + \frac{1 + y^2}{y} = 0\]
\[\Rightarrow \frac{dy}{dx} = - \frac{\left( 1 + y^2 \right)}{y}\]
\[ \Rightarrow \frac{dx}{dy} = - \frac{y}{1 + y^2}\]
\[ \Rightarrow dx = \left( - \frac{y}{1 + y^2} \right)dy\]
Integrating both sides, we get
\[\int dx = \int\left( - \frac{y}{1 + y^2} \right)dy\]
\[ \Rightarrow x = \int\left( - \frac{y}{1 + y^2} \right)dy\]
\[\text{ Putting }1 + y^2 = t, \text{ we get }\]
\[2y dy = dt\]
\[ \therefore x = - \frac{1}{2}\int\frac{1}{t}dt\]
\[ \Rightarrow x = - \frac{1}{2}\log\left| t \right| + C\]
\[ \Rightarrow x = - \frac{1}{2}\log\left| 1 + y^2 \right| + C\]
\[ \Rightarrow x + \frac{1}{2}\log\left| 1 + y^2 \right| = C\]
\[\text{ Hence, }x + \frac{1}{2}\log\left| 1 + y^2 \right| =\text{ C is the required solution }.\]