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The general solution of the differential equation $\cos (x+y) d y=d x$ is
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The correct answer is:
$y=\tan \left(\frac{x+y}{2}\right)+c$
$\frac{d x}{d y}=\cos (x+y)$
Let $\quad x+y=v \quad \Rightarrow \frac{d x}{d y}+1=\frac{d v}{d y}$
So, $\quad \frac{d x}{d y}=\frac{d v}{d y}-1 \Rightarrow \frac{d v}{d y}-1=\cos v$
$$
\begin{aligned}
& \Rightarrow \int \frac{d v}{\cos v+1}=\int d y \Rightarrow \int \frac{d v}{2 \cos ^2 \frac{v}{2}}=\int d y \\
& \Rightarrow \frac{1}{2} \int \sec ^2 \frac{v}{2} d v=\int d y \Rightarrow \tan \frac{v}{2}=y+k \\
& \Rightarrow y=\tan \frac{x+y}{2}-k \Rightarrow y=\tan \frac{x+y}{2}+c
\end{aligned}
$$
Let $\quad x+y=v \quad \Rightarrow \frac{d x}{d y}+1=\frac{d v}{d y}$
So, $\quad \frac{d x}{d y}=\frac{d v}{d y}-1 \Rightarrow \frac{d v}{d y}-1=\cos v$
$$
\begin{aligned}
& \Rightarrow \int \frac{d v}{\cos v+1}=\int d y \Rightarrow \int \frac{d v}{2 \cos ^2 \frac{v}{2}}=\int d y \\
& \Rightarrow \frac{1}{2} \int \sec ^2 \frac{v}{2} d v=\int d y \Rightarrow \tan \frac{v}{2}=y+k \\
& \Rightarrow y=\tan \frac{x+y}{2}-k \Rightarrow y=\tan \frac{x+y}{2}+c
\end{aligned}
$$
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