Question

(y + xy) dx + (x − xy²) dy = 0

Answer 1

We have, 
\[\left( y + xy \right) dx + \left( x - x y^2 \right) dy = 0\]
\[ \Rightarrow y\left( 1 + x \right)dx = x\left( y^2 - 1 \right)dy\]
\[ \Rightarrow \frac{1 + x}{x}dx = \frac{y^2 - 1}{y}dy\]
Integrating both sides, we get 
\[\int\frac{1 + x}{x}dx = \int\frac{y^2 - 1}{y}dy\]
\[ \Rightarrow \int\frac{1}{x}dx + \int dx = \int y dy - \int\frac{1}{y}dy\]
\[ \Rightarrow \log \left| x \right| + x = \frac{y^2}{2} - \log \left| y \right| + C\]
\[ \Rightarrow \log \left| x \right| + x - \frac{y^2}{2} + \log \left| y \right| = C\]
\[\text{ Hence, } \log \left| x \right| + x - \frac{y^2}{2} + \log \left| y \right| =\text{ C is the required solution .} \]