What is the solution of the differential equation \[x\left( {{e^{2y}} - 1} \right) dy + \left( {{x^2} - 1} \right){e^y}dx = 0\]?
A. \[{e^y} + {e^{ - y}} = \log x - \dfrac{{{x^2}}}{2} + c\]
B. \[{e^y} - {e^{ - y}} = \log x - \dfrac{{{x^2}}}{2} + c\]
C. \[{e^y} + {e^{ - y}} = \log x + \dfrac{{{x^2}}}{2} + c\]
D. None of these

Hint: Here, the first order differential equation is given. First, simplify the given equation. Then, integrate both sides of the equation with respect to the corresponding variables. After that, solve both integrals by using the standard integral rules and U-substitution method to get the solution of the given differential equation.

Formula used:
\[\int {\dfrac{1}{x}dx} = \log x\]
\[\int {{x^n}dx} = \dfrac{{{x^{n + 1}}}}{{n + 1}}\]
\[\int {{e^x}dx} = {e^x}\]

Complete step by step solution:
The given differential equation is \[x\left( {{e^{2y}} - 1} \right) dy + \left( {{x^2} - 1} \right){e^y}dx = 0\].
Let’s find out the solution of the given differential equation.
Simplify the given equation.
\[x\left( {{e^{2y}} - 1} \right) dy = - \left( {{x^2} - 1} \right){e^y}dx\]
\[ \Rightarrow \dfrac{{\left( {{e^{2y}} - 1} \right)}}{{{e^y}}} dy = \dfrac{{\left( {1 - {x^2}} \right)}}{x}dx\]
Now integrate both sides with respect to the corresponding variables.
\[\int {\dfrac{{\left( {{e^{2y}} - 1} \right)}}{{{e^y}}} dy} = \int {\dfrac{{\left( {1 - {x^2}} \right)}}{x}dx} \]
Simplify the integrals.
\[\int {\left[ {{e^y} - \dfrac{1}{{{e^y}}}} \right] dy} = \int {\left[ {\dfrac{1}{x} - x} \right]dx} \]
Apply the integration rule.
\[\int {{e^y} dy - } \int {\dfrac{1}{{{e^y}}}dy} = \int {\dfrac{1}{x}dx} - \int {xdx} \]
\[ \Rightarrow \int {{e^y} dy - } \int {{e^{ - y}}dy} = \int {\dfrac{1}{x}dx} - \int {xdx} \] \[.....\left( 1 \right)\]
Solve the integrals by applying the formulas \[\int {\dfrac{1}{x}dx} = \log x\], \[\int {{e^x}dx} = {e^x}\], and \[\int {{x^n}dx} = \dfrac{{{x^{n + 1}}}}{{n + 1}}\].
Also apply U-substitution method to solve \[\int {{e^{ - y}}dy} \].
Substitute \[ - y = u\].
Differentiate the substitute equation.
We get,
 \[ - dy = du\]

Then, we get the equation \[\left( 1 \right)\] as follows:
\[{e^y} - \int { - {e^u}} du = \log\left( x \right) - \dfrac{{{x^2}}}{2} + c\]
\[ \Rightarrow {e^y} + {e^u} = \log\left( x \right) - \dfrac{{{x^2}}}{2} + c\]
Resubstitute the value of \[u\].
\[{e^y} + {e^{ - y}} = \log\left( x \right) - \dfrac{{{x^2}}}{2} + c\]
Hence the correct option is A.

Note: Students often apply a wrong formula to integrate \[\dfrac {1}{x}\]. They integrate it by using the power formula of integration. But the correct formula is \[\int {\dfrac{1}{x}dx} = \log x + c\].