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प्रश्न
x (e2y − 1) dy + (x2 − 1) ey dx = 0
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उत्तर
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
\[\int\left( e^y - e^{- y} \right)dy = \int\left( \frac{1}{x} - x \right)dx\]
\[ \Rightarrow e^y + e^{- y} = \log \left| x \right| - \frac{1}{2} x^2 + C\]
\[ \Rightarrow e^y + e^{- y} - \log \left| x \right| + \frac{1}{2} x^2 = C\]
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