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Question
(1 + x) y dx + (1 + y) x dy = 0
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Solution
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
\[\int\left( \frac{1}{y} + 1 \right)dy = - \int\left( \frac{1}{x} + 1 \right)dx\]
\[ \Rightarrow \int\frac{1}{y}dy + \int dy = - \int\frac{1}{x}dx - \int dx\]
\[ \Rightarrow \log\left| y \right| + y = - \log\left| x \right| - x + C\]
\[ \Rightarrow \log\left| xy \right| + y + x = C\]
\[ \Rightarrow x + y + \log\left| xy \right| = C\]
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