A function $f : R \rightarrow R$ satisfies x cos y (f (2x+2y)-f(2x-2y))=cos x sin y (f(2x+2y)+f(2x-2y)). If $f'(0)=\dfrac{1}{2}$, then-

$\begin{array}{l}\frac{f(2 x+2 y)-f(2 x-2 y)}{f(2 x+2 y)+f(2 x-2 y)}=\frac{\cos x \sin y}{\sin x \cos y} \\ \frac{2 f(2 x+2 y)}{-2 f(2 x-2 y)}=\frac{\cos x \sin y+\sin x \cos y}{\cos x \sin y-\sin x \cos y} \\ \frac{f(2 x+2 y)}{f(2 x-2 y)}=\frac{\sin (x+y)}{\sin (x-y)} \\ \frac{f(2 x+2 y)}{\sin (x+y)}=\frac{f(2 x-2 y)}{\sin (x-y)}\end{array}$

$\frac{f(\alpha)}{\sin \frac{\alpha}{2}}=\frac{f(\beta)}{\sin \frac{\beta}{2}} \quad\left[\begin{array}{l}\text {ধরি,} \\ 2 x+2 y=\alpha \\ 2 x-2 y=\beta\end{array}\right]$

$\begin{array}{l}f(\alpha)=k \sin \frac{\alpha}{2} \\ f(x)=k \sin \left(\frac{x}{2}\right) \\ f^{\prime}(x)=\frac{1}{2} k \cos \left(\frac{x}{2}\right) \\ f^{\prime \prime}(x)=\frac{k}{2}\left[-\sin \frac{x}{2} \cdot \frac{1}{2}\right] \\ f^{\prime \prime}(x)+\frac{k}{4} \sin \frac{x}{2}=0 \\ f^{\prime \prime}(x)+\frac{f(x)}{4}=0 \\ 4 f^{\prime \prime}(x)+f(x)=0\end{array}$