Question
For each of the differential equations in find the general solution:
$(e^x + e^{–x}) ~dy – (e^x – e^{–x}) dx = 0$

Answer

Given: Differential equation $(e^x + e^{–x}) ~dy – (e^x – e^{–x}) dx = 0$
$\Rightarrow \ (\text{e}^\text{x} + \text{e}^{-\text{x}}) \ \text{dy} = (\text{e}^\text{x} – \text{e}^{–\text{x}}) \ \text{dx} \ $ $\Rightarrow \ \text{dy}=\Bigg(\frac{\text{e}^\text{x}-\text{e}^{-\text{x}}}{{\text{e}^\text{x}+\text{e}^{-\text{x}}}}\Bigg) \text{dx}$
Integrating both sides, $\ \int\text{dy}=\int\Bigg(\frac{\text{e}^\text{x}-\text{e}^{-\text{x}}}{\text{e}^\text{x}+\text{e}^{-\text{x}}}\Bigg)\text{dx}$
$\Rightarrow\ \text{y}=\text{log}\Big|\text{e}^\text{x}+\text{e}^{-\text{x}}\Big|+\text{c} \ \Bigg[\because\int\frac{f'(\text{x})}{f(\text{x})}\text{dx}=\text{log}\big|f(\text{x})\big|\Bigg]$

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