Differential Equations — Maths STD 12 Science — Question

Given,
$\frac{d y}{d x}=\cos (x+y) \ldots$ (i)
Put $x+y=v$...(ii)
$\therefore y=v-x$
$\therefore \frac{d y}{d x}=\frac{d v}{d x}-1 \ldots$..(iii)
Substituting (ii) and (iii) in (i), we get
$\begin{aligned} & \frac{d v}{d x}-1=\cos v \\ & \therefore \frac{d v}{d x}=1+\cos v \\ & \therefore \frac{d v}{d x}=2 \cos ^2\left(\frac{v}{2}\right) \\ & \therefore \frac{1}{\cos ^2\left(\frac{v}{2}\right)} d v=2 d x \\ & \therefore \sec ^2\left(\frac{v}{2}\right) d v=2 d x\end{aligned}$
Integrating on both sides, we get
$\begin{aligned} & \int \sec ^2\left(\frac{v}{2}\right) d v=2 \int d x \\ & \therefore 2 \tan \left(\frac{v}{2}\right)=2 x+c \prime \\ & \therefore \tan \left(\frac{v}{2}\right)=x+\frac{c \prime}{2} \\ & \therefore \tan \left(\frac{x+y}{2}\right)=x+c, \text { where } c=\frac{c \prime}{2}\end{aligned}$