The general solution of $\frac{\mathrm{d} y}{\mathrm{~d} x}+\sin \left(\frac{x+y}{2}\right)=\sin…

The general solution of $\frac{\mathrm{d} y}{\mathrm{~d} x}+\sin \left(\frac{x+y}{2}\right)=\sin \left(\frac{x-y}{2}\right)$ is
  1. $\log \tan \left(\frac{y}{2}\right)=\mathrm{C}-2 \sin x$
  2. $\log \tan \left(\frac{y}{4}\right)=\mathrm{C}-2 \sin \left(\frac{x}{2}\right)$
  3. $\log \tan \left(\frac{y}{2}+\frac{\pi}{4}\right)=\mathrm{C}-2 \sin x$
  4. $\quad \log \tan \left(\frac{y}{2}+\frac{\pi}{4}\right)=\mathrm{C}-2 \sin \left(\frac{x}{2}\right)$

Solution

$\begin{aligned} & \frac{\mathrm{d} y}{\mathrm{~d} x}+\sin \left(\frac{x+y}{2}\right)=\sin \left(\frac{x-y}{2}\right) \\ & \Rightarrow \frac{\mathrm{d} y}{\mathrm{~d} x}=\sin \left(\frac{x-y}{2}\right)-\sin \left(\frac{x+y}{2}\right) \\ & \Rightarrow \frac{\mathrm{d} y}{\mathrm{~d} x}=-2 \sin \left(\frac{y}{2}\right) \cdot \cos \left(\frac{x}{2}\right) \end{aligned}$
Integrating on both sides, we get $\begin{aligned} & \int \operatorname{cosec}\left(\frac{y}{2}\right) \mathrm{d} y=-\int 2 \cos \left(\frac{x}{2}\right) \mathrm{d} x+\mathrm{c}_1 \\ & \Rightarrow \frac{\log \tan \left(\frac{y}{4}\right)}{\frac{1}{2}}=-\frac{2 \sin \left(\frac{x}{2}\right)}{\frac{1}{2}}+\mathrm{c}_1 \\ & \Rightarrow \log \tan \left(\frac{y}{4}\right)=C-2 \sin \left(\frac{x}{2}\right), \text { where } C=\frac{1}{2} c_1 \end{aligned}$

Asked in: MHT CET 2024 (11 May Shift 1)

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