Answer 1

We have,

(1 + x) y dx + (1 + y) x dy = 0

\[\frac{dy}{dx} = - \frac{y\left( 1 + x \right)}{x\left( 1 + y \right)}\]

\[ \Rightarrow \left( \frac{1 + y}{y} \right)dy = - \left( \frac{1 + x}{x} \right)dx\]

\[ \Rightarrow \left( \frac{1}{y} + y \right)dy = - \left( \frac{1}{x} + 1 \right)dx\]

Integrating both sides, we get

\[\int\left( \frac{1}{y} + 1 \right)dy = - \int\left( \frac{1}{x} + 1 \right)dx\]

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

\[ \Rightarrow \log\left| y \right| + y = - \log\left| x \right| - x + C\]

\[ \Rightarrow \log\left| xy \right| + y + x = C\]

\[ \Rightarrow x + y + \log\left| xy \right| = C\]