MCQ
The general solution of differential equation $\frac{d x}{d y}=\cos (x+y)$ is
- A$\tan \left(\frac{x+y}{2}\right)=y+c$
- B$\tan \left(\frac{x+y}{2}\right)=x+c$
- C$\cot \left(\frac{x+y}{2}\right)=y+c$
- D$\cot \left(\frac{x+y}{2}\right)=x+c$
(a) : We have $d x / d y=\cos (x+y)$...(i)
Let $x+y=v$
On differentiating both sides w.r.t. $y$, we get
$
\frac{d x}{d y}+1=\frac{d v}{d y} \Rightarrow \frac{d x}{d y}=\frac{d v}{d y}-1
$
Put in (i), we get
$
\therefore \quad \frac{d v}{d y}-1=\cos v \Rightarrow \frac{d v}{d y}=1+\cos v \Rightarrow \frac{d v}{d y}=2 \cos ^2 \frac{v}{2}
$
On integrating both sides, we get
$
\begin{aligned}
& \int \frac{d v}{\cos ^2 \frac{v}{2}}=2 \int d y \Rightarrow \int \sec ^2 \frac{v}{2} d v=2 y+K \\
& \Rightarrow \quad 2 \tan \left(\frac{v}{2}\right)=2 y+K \Rightarrow 2 \tan \left(\frac{x+y}{2}\right)=2 y+K \\
& \Rightarrow \quad \tan \left(\frac{x+y}{2}\right)=y+c \quad \quad(\text { where } c=K / 2)
\end{aligned}
$
(where $c=K / 2$ )
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