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