Question
Solve the following equation:
$(\text{e}^\text{y}+1)\cos\text{x dx}+\text{e}^\text{y}\sin\text{x}\text{dy}=0$
$(\text{e}^\text{y}+1)\cos\text{x dx}+\text{e}^\text{y}\sin\text{x}\text{dy}=0$
✓
Answer
$(\text{e}^\text{y}+1)\cos\text{x dx}+\text{e}^\text{y}\sin\text{x}\text{dy}=0$
$(\text{e}^\text{y}+1)\cos\text{x dx}=-\text{e}^\text{y}\sin\text{x}\text{dy}$
$\int\frac{\cos\text{x}}{\sin\text{x}}\ \text{dx}=-\int\frac{\text{e}^\text{y}}{\text{e}^\text{y}+1}\ \text{dy}$
$\int\cot\text{x dx}=-\int\frac{\text{e}^\text{y}}{\text{e}^\text{y}+1}\ \text{dy}$
$\log|\sin\text{x}|=-\log|\text{e}^\text{y}+1|+\log\text|C|$
$\sin\text{x}=\frac{\text{C}}{\text{e}^\text{y}+1}$
$\sin\text{x}(\text{e}^\text{y}+1)=\text{C}$
$(\text{e}^\text{y}+1)\cos\text{x dx}=-\text{e}^\text{y}\sin\text{x}\text{dy}$
$\int\frac{\cos\text{x}}{\sin\text{x}}\ \text{dx}=-\int\frac{\text{e}^\text{y}}{\text{e}^\text{y}+1}\ \text{dy}$
$\int\cot\text{x dx}=-\int\frac{\text{e}^\text{y}}{\text{e}^\text{y}+1}\ \text{dy}$
$\log|\sin\text{x}|=-\log|\text{e}^\text{y}+1|+\log\text|C|$
$\sin\text{x}=\frac{\text{C}}{\text{e}^\text{y}+1}$
$\sin\text{x}(\text{e}^\text{y}+1)=\text{C}$
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