Question

\[\frac{dy}{dx} = 1 - x + y - xy\]

Answer 1

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