The solution of the differential equation x(e | vedclass

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(A) Given differential equation is $x(e^{2y} - 1)dy + (x^2 - 1)e^y dx = 0$. Rearranging the terms to separate the variables,we get: $x(e^{2y} - 1)dy =

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$e^y + e^{-y} = \log x - \frac{x^2}{2} + c$

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(A) Given differential equation is $x(e^{2y} - 1)dy + (x^2 - 1)e^y dx = 0$. Rearranging the terms to separate the variables,we get: $x(e^{2y} - 1)dy = -(x^2 - 1)e^y dx$ $\frac{e^{2y} - 1}{e^y} dy = \frac{1 - x^2}{x} dx$ $(e^y - e^{-y}) dy = (\frac{1}{x} - x) dx$ Integrating both sides: $\int (e^y - e^{-y}) dy = \int (\frac{1}{x} - x) dx$ $e^y - (-e^{-y}) = \log |x| - \frac{x^2}{2} + c$ $e^y + e^{-y} = \log |x| - \frac{x^2}{2} + c$.

The solution of the differential equation $x(e^{2y} - 1)dy + (x^2 - 1)e^y dx = 0$ is

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