A) \[{{e}^{y}}+{{e}^{-y}}=\log x-\frac{{{x}^{2}}}{2}+c\]
B) \[{{e}^{y}}-{{e}^{-y}}=\log x-\frac{{{x}^{2}}}{2}+c\]\[\]
C) \[{{e}^{y}}+{{e}^{-y}}=\log x+\frac{{{x}^{2}}}{2}+c\]
D) None of these
Correct Answer: A
Solution :
\[x({{e}^{2y}}-1)dy+({{x}^{2}}-1){{e}^{y}}dx=0\] Þ \[\int_{{}}^{{}}{\frac{{{e}^{2y}}-1}{{{e}^{y}}}}dy=\int_{{}}^{{}}{\frac{1-{{x}^{2}}}{x}dx}\]Þ \[{{e}^{y}}+{{e}^{-y}}=\log x-\frac{{{x}^{2}}}{2}+c\].You need to login to perform this action.
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