Question
Find the general solution of the differential equation $\left( {{e^x} + {\text{ }}{e^{ - x}}} \right){\text{ }}dy{\text{ }} - {\text{ }}\left( {{e^x} - {\text{ }}{e^{ - x}}} \right){\text{ }}dx{\text{ }} = {\text{ }}0$

Answer

$({e^x} + {e^{ - x}})dy = ({e^x} - {e^{ - x}})dx$

$\int {dy} = \int {\frac{{({e^x} - {e^{ - x}})}}{{({e^x} + {e^{ - x}})}}} dx $ Since $\int {\frac{{f'(x)}}{{f(x)}}} dx = \ ln f(x) +c$

$y = \log \left| {({e^x} + {e^{ - x}})} \right| + C$

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