Question

\[\frac{dy}{dx} = \frac{1 + y^2}{y^3}\]

Answer 1

We have, 
\[\frac{dy}{dx} = \frac{1 + y^2}{y^3}\]
\[\Rightarrow \frac{dx}{dy} = \frac{y^3}{1 + y^2}\]
\[ \Rightarrow dx = \frac{y^3}{1 + y^2}dy\]
Integrating both sides, we get
\[\int dx = \int\frac{y^3}{1 + y^2}dy\]
\[ \Rightarrow x = \int\frac{y + y^3 - y}{1 + y^2}dy\]
\[ \Rightarrow x = \int\frac{\left( 1 + y^2 \right)y - y}{1 + y^2}dy\]
\[ \Rightarrow x = \int y dy - \int\frac{y}{1 + y^2}dy\]
\[ \Rightarrow x = \frac{y^2}{2} - \int\frac{y}{1 + y^2}dy\]
\[\text{ Putting }1 + y^2 = t \text{ we get }\]
\[2y dy = dt\]
\[ \therefore x = \frac{y^2}{2} - \frac{1}{2}\int\frac{1}{t}dt\]
\[ \Rightarrow x = \frac{y^2}{2} - \frac{1}{2}\log\left| t \right| + C\]
\[ \Rightarrow x = \frac{y^2}{2} - \frac{1}{2}\log\left| 1 + y^2 \right| + C ...........\left( \because t = 1 + y^2 \right)\]
\[\text{ Hence, }x = \frac{y^2}{2} - \frac{1}{2}\log\left| 1 + y^2 \right| +\text{ C is the required solution.}\]