Question

Solve the following differential equation:
\[xy\frac{dy}{dx} = 1 + x + y + xy\]

 

Answer 1

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