Question

(1 − x²) dy + xy dx = xy² dx

Answer 1

We have,

\[\left( 1 - x^2 \right)dy + xy dx = x y^2 dx\]

\[ \Rightarrow \left( 1 - x^2 \right)dy = x y^2 dx - xy dx\]
\[ \Rightarrow \left( 1 - x^2 \right)dy = x\left( y^2 - y \right)dx\]

\[ \Rightarrow \frac{1}{\left( y^2 - y \right)}dy = \frac{x}{\left( 1 - x^2 \right)}dx\]

Integrating both sides, we get

\[\int\frac{1}{y^2 - y}dy = \int\frac{x}{1 - x^2}dx\]
\[ \Rightarrow \int\frac{1}{y^2 - y + \frac{1}{4} - \frac{1}{4}}dy = \int\frac{x}{1 - x^2}dx\]
\[ \Rightarrow \int\frac{1}{\left( y - \frac{1}{2} \right)^2 - \left( \frac{1}{2} \right)^2}dy = - \frac{1}{2}\int\frac{- 2x}{1 - x^2}dx\]
\[ \Rightarrow \frac{1}{2 \times \frac{1}{2}}\log \left| \frac{y - \frac{1}{2} - \frac{1}{2}}{y - \frac{1}{2} + \frac{1}{2}} \right| = - \frac{1}{2}\log \left| 1 - x^2 \right| + \log C\]
\[ \Rightarrow \log \left| \frac{y - 1}{y} \right| = - \frac{1}{2}\log \left| 1 - x^2 \right| + \log C\]
\[ \Rightarrow 2 \log \left| \frac{y - 1}{y} \right| = - \log \left| 1 - x^2 \right| + 2 \log C\]
\[ \Rightarrow \log \left| \frac{\left( y - 1 \right)^2}{y^2} \right| = - \log\left| 1 - x^2 \right| + 2 \log C\]
\[ \Rightarrow \log \left| \frac{\left( y - 1 \right)^2}{y^2} \right| + \log \left| 1 - x^2 \right| = \log C^2 \]
\[ \Rightarrow \log\left| \frac{\left( y - 1 \right)^2 \left| \left( 1 - x^2 \right) \right|}{y^2} \right| = \log C^2 \]
\[ \Rightarrow \frac{\left( y - 1 \right)^2 \left| \left( 1 - x^2 \right) \right|}{y^2} = C^2 \]
\[ \Rightarrow \left( y - 1 \right)^2 \left| \left( 1 - x^2 \right) \right| = y^2 C^2\]