Question

\[\frac{dy}{dx} = x^2 + x - \frac{1}{x}, x \neq 0\]

Answer 1

We have, 
\[\frac{dy}{dx} = x^2 + x - \frac{1}{x}\]
\[ \Rightarrow dy = \left( x^2 + x - \frac{1}{x} \right)dx\]
Integrating both sides, we get
\[ \Rightarrow \int dy = \int\left( x^2 + x - \frac{1}{x} \right)dx\]
\[ \Rightarrow y = \frac{x^3}{3} + \frac{x^2}{2} - \log\left| x \right| + C\]
\[\text{ Clearly, } y = \frac{x^3}{3} + \frac{x^2}{2} - \log\left| x \right| +\text{ C is defined for all } x \in\text{ R except }x = 0 . \]
\[\text{ Hence, } y = \frac{x^3}{3} + \frac{x^2}{2} - \log\left| x \right| + C,\text{ where }x \in R- \left\{ 0 \right\}, \text{ is the solution to the given differential equation }.\]