Question

x (e^2y − 1) dy + (x² − 1) e^y dx = 0

Answer 1

We have,

\[x\left( e^{2y} - 1 \right)dy + \left( x^2 - 1 \right) e^y dx = 0\]

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

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

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

Integrating both sides, we get

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

\[ \Rightarrow e^y + e^{- y} = \log \left| x \right| - \frac{1}{2} x^2 + C\]

\[ \Rightarrow e^y + e^{- y} - \log \left| x \right| + \frac{1}{2} x^2 = C\]