Question
Solve the following differential equations:$\cos\text{x}\cos\text{y}\frac{\text{dy}}{\text{dx}}=-\sin\text{x}\sin\text{y}$
✓
Answer
We have,
$\cos\text{x}\cos\text{y}\frac{\text{dy}}{\text{dx}}=-\sin\text{x}\sin\text{y}$
$\Rightarrow\frac{\cos\text{y}}{\sin\text{y}}\text{dy}=\frac{-\sin\text{x}}{\cos\text{x}}\text{dx}$
$\Rightarrow\cot\text{y dy}=-\tan\text{x dx}$
Integrating both sides, we get
$\int\cot\text{y dy}=-\int\tan\text{x dx}$
$\Rightarrow\log|\sin\text{y}|=-\log|\sec\text{x}|+\log\text{C}$
$\Rightarrow\log |\sin\text{y}|=\log|\cos\text{x}|+\log\text{C}$
$\Rightarrow\sin\text{y}=\text{C}\cos\text{x}$
Hence, $\sin\text{y =C}\cos\text{x}$ is the reguired solution.
$\cos\text{x}\cos\text{y}\frac{\text{dy}}{\text{dx}}=-\sin\text{x}\sin\text{y}$
$\Rightarrow\frac{\cos\text{y}}{\sin\text{y}}\text{dy}=\frac{-\sin\text{x}}{\cos\text{x}}\text{dx}$
$\Rightarrow\cot\text{y dy}=-\tan\text{x dx}$
Integrating both sides, we get
$\int\cot\text{y dy}=-\int\tan\text{x dx}$
$\Rightarrow\log|\sin\text{y}|=-\log|\sec\text{x}|+\log\text{C}$
$\Rightarrow\log |\sin\text{y}|=\log|\cos\text{x}|+\log\text{C}$
$\Rightarrow\sin\text{y}=\text{C}\cos\text{x}$
Hence, $\sin\text{y =C}\cos\text{x}$ is the reguired solution.
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