\(\text { If } 2 \cos x \frac{d y}{d x}=\sin 2 x-4 y \sin x \cdot y\left(\frac{\pi}{3}\right)=0 \text { find } y^{\prime…

\(2\cos x\frac{dy}{dx}=\sin 2x-4y.\sin x\\ 2\cos x\frac{dy}{dx}+4y\sin x=2\sin x\cos x\\ \frac{dy}{dx}+2y\tan x=\sin x\\ NowI.F={e}^{\int 2\tan xdx}={e}^{\ln se{c}^{2}x}=se{c}^{2}x\\ ForGeneralSolution\\ y.IF=\int Q.IFdx+c\\ y.se{c}^{2}x=\int \frac{\sin x}{{\cos }^{2}x}dx+c\\ let,\cos x=t\Rightarrow \sin xdx=-dt\\ y.se{c}^{2}x=-\int \frac{1}{{t}^{2}}dt+c\\ y.se{c}^{2}x=secx+c\\ y=\frac{1}{secx}+\frac{c}{se{c}^{2}x}\\ y=\cos x+{\cos }^{2}x.c\\ y(\frac{\pi }{3})=0(Given)\\ 0=\frac{1}{2}+\frac{c}{4}\Rightarrow c=-2\\ So,y=\cos x-2{\cos }^{2}x\\ y(\frac{\pi }{4})=\frac{1}{\sqrt{2}}-1...........................(i)\\ andy'=-\sin x+4\sin x\cos x\\ y'(\frac{\pi }{4})=-\frac{1}{\sqrt{2}}+2...................(ii)\\ Addingequation(i)&(ii),weget\\ y'(\frac{\pi }{4})+y(\frac{\pi }{4})=1\)